Removing duplicates from array is only returning one object
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1
down vote
favorite
I am trying to remove duplicate entry's from this json but it is only returning one object I am not understanding where I am going wrong.
The code is as follows.
// exemplary array of objects (id 'NewLive' occurs twice)
var arr = [
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}},
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}}
],
obj = {}, new_arr = ;
// in the end the last unique object will be considered
arr.forEach(function(v){
obj[v['id']] = v;
console.log(JSON.stringify(new_arr));
});
new_arr = Object.keys(obj).map(function(id) { return obj[id]; });
console.log(JSON.stringify(new_arr));
I am attaching codepen also with this.
https://codepen.io/anon/pen/oQXJWK
javascript arrays node.js json
|
show 5 more comments
up vote
1
down vote
favorite
I am trying to remove duplicate entry's from this json but it is only returning one object I am not understanding where I am going wrong.
The code is as follows.
// exemplary array of objects (id 'NewLive' occurs twice)
var arr = [
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}},
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}}
],
obj = {}, new_arr = ;
// in the end the last unique object will be considered
arr.forEach(function(v){
obj[v['id']] = v;
console.log(JSON.stringify(new_arr));
});
new_arr = Object.keys(obj).map(function(id) { return obj[id]; });
console.log(JSON.stringify(new_arr));
I am attaching codepen also with this.
https://codepen.io/anon/pen/oQXJWK
javascript arrays node.js json
Are you just trying to return the same object without duplicates?
– Jacques
Nov 9 at 5:26
@Jacques Yes I am
– Arun VM
Nov 9 at 5:27
Possible duplicate of Get all unique values in a JavaScript array (remove duplicates)
– Frank Fajardo
Nov 9 at 5:28
obj[v['id']] = v;
should beobj[v['jobcodeid']] = v;
. Since there is noid
, you get undefined for all objects and hence you are basically replacing all values by next. Hence you get only 1
– Rajesh
Nov 9 at 5:30
for (const job of arr) { if (!obj[job.jobcodeid.S]) { obj[job.jobcodeid.S] = true; new_arr.push(job); } }
– Jacques
Nov 9 at 5:30
|
show 5 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I am trying to remove duplicate entry's from this json but it is only returning one object I am not understanding where I am going wrong.
The code is as follows.
// exemplary array of objects (id 'NewLive' occurs twice)
var arr = [
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}},
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}}
],
obj = {}, new_arr = ;
// in the end the last unique object will be considered
arr.forEach(function(v){
obj[v['id']] = v;
console.log(JSON.stringify(new_arr));
});
new_arr = Object.keys(obj).map(function(id) { return obj[id]; });
console.log(JSON.stringify(new_arr));
I am attaching codepen also with this.
https://codepen.io/anon/pen/oQXJWK
javascript arrays node.js json
I am trying to remove duplicate entry's from this json but it is only returning one object I am not understanding where I am going wrong.
The code is as follows.
// exemplary array of objects (id 'NewLive' occurs twice)
var arr = [
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}},
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}}
],
obj = {}, new_arr = ;
// in the end the last unique object will be considered
arr.forEach(function(v){
obj[v['id']] = v;
console.log(JSON.stringify(new_arr));
});
new_arr = Object.keys(obj).map(function(id) { return obj[id]; });
console.log(JSON.stringify(new_arr));
I am attaching codepen also with this.
https://codepen.io/anon/pen/oQXJWK
javascript arrays node.js json
javascript arrays node.js json
edited Nov 9 at 8:29
lucascaro
3,30611430
3,30611430
asked Nov 9 at 5:22
Arun VM
689
689
Are you just trying to return the same object without duplicates?
– Jacques
Nov 9 at 5:26
@Jacques Yes I am
– Arun VM
Nov 9 at 5:27
Possible duplicate of Get all unique values in a JavaScript array (remove duplicates)
– Frank Fajardo
Nov 9 at 5:28
obj[v['id']] = v;
should beobj[v['jobcodeid']] = v;
. Since there is noid
, you get undefined for all objects and hence you are basically replacing all values by next. Hence you get only 1
– Rajesh
Nov 9 at 5:30
for (const job of arr) { if (!obj[job.jobcodeid.S]) { obj[job.jobcodeid.S] = true; new_arr.push(job); } }
– Jacques
Nov 9 at 5:30
|
show 5 more comments
Are you just trying to return the same object without duplicates?
– Jacques
Nov 9 at 5:26
@Jacques Yes I am
– Arun VM
Nov 9 at 5:27
Possible duplicate of Get all unique values in a JavaScript array (remove duplicates)
– Frank Fajardo
Nov 9 at 5:28
obj[v['id']] = v;
should beobj[v['jobcodeid']] = v;
. Since there is noid
, you get undefined for all objects and hence you are basically replacing all values by next. Hence you get only 1
– Rajesh
Nov 9 at 5:30
for (const job of arr) { if (!obj[job.jobcodeid.S]) { obj[job.jobcodeid.S] = true; new_arr.push(job); } }
– Jacques
Nov 9 at 5:30
Are you just trying to return the same object without duplicates?
– Jacques
Nov 9 at 5:26
Are you just trying to return the same object without duplicates?
– Jacques
Nov 9 at 5:26
@Jacques Yes I am
– Arun VM
Nov 9 at 5:27
@Jacques Yes I am
– Arun VM
Nov 9 at 5:27
Possible duplicate of Get all unique values in a JavaScript array (remove duplicates)
– Frank Fajardo
Nov 9 at 5:28
Possible duplicate of Get all unique values in a JavaScript array (remove duplicates)
– Frank Fajardo
Nov 9 at 5:28
obj[v['id']] = v;
should be obj[v['jobcodeid']] = v;
. Since there is no id
, you get undefined for all objects and hence you are basically replacing all values by next. Hence you get only 1– Rajesh
Nov 9 at 5:30
obj[v['id']] = v;
should be obj[v['jobcodeid']] = v;
. Since there is no id
, you get undefined for all objects and hence you are basically replacing all values by next. Hence you get only 1– Rajesh
Nov 9 at 5:30
for (const job of arr) { if (!obj[job.jobcodeid.S]) { obj[job.jobcodeid.S] = true; new_arr.push(job); } }
– Jacques
Nov 9 at 5:30
for (const job of arr) { if (!obj[job.jobcodeid.S]) { obj[job.jobcodeid.S] = true; new_arr.push(job); } }
– Jacques
Nov 9 at 5:30
|
show 5 more comments
5 Answers
5
active
oldest
votes
up vote
1
down vote
accepted
Fist of all, you have to use obj[v['jobcodeid']] = v;
instead of obj[v['id']] = v;
.
but as v[jobcodeid]
is an object, js will convert it to a string i.e. [object Object]
and you will only one element in final array.
// exemplary array of objects (id 'NewLive' occurs twice)
var arr=[{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}], obj = {}, new_arr = ;
// in the end the last unique object will be considered
arr.forEach(function(v){
obj[v['jobcodeid']] = v;
});
new_arr = Object.keys(obj).map(function(id) { return obj[id]; });
console.log(JSON.stringify(new_arr));
You should use v.jobcodeid.S
as keys of object.
// exemplary array of objects (id 'NewLive' occurs twice)
var arr=[{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}], obj = {}, new_arr = ;
// in the end the last unique object will be considered
arr.forEach(function(v){
obj[v.jobcodeid.S] = v;
});
new_arr = Object.keys(obj).map(function(id) { return obj[id]; });
console.log(JSON.stringify(new_arr));
I missed that in my comment. +1
– Rajesh
Nov 9 at 5:33
Have you tested the code? Can u update the same on codepen?
– Arun VM
Nov 9 at 5:34
Yeah i have updated the same pen any way
– Arun VM
Nov 9 at 5:36
@ArunVM added code snippets.
– nrgwsth
Nov 9 at 5:39
I have accepted the answer Peace :), Thank you
– Arun VM
Nov 9 at 5:39
|
show 1 more comment
up vote
4
down vote
The reason why your code is returning a single element is because you are using v['id']
but there is no id
property on the objects, therefore throughout your loop you are setting obj[undefined]
over and over.
In your jsfiddle code though this looks correct and the code seems to be working as intended.
In case someone got to this question to find out how to remove duplicates from an array in javascript, here are a few options:
The classic way: good old for loop
This is essentially the solution you used, iterate over the array, check if the key has been added to the result array, and if it's not there, add the element to the result.
Example:
const result = ;
const knownIDs = new Set();
for (const item of input) {
if (!knownIDs.has(item.jobcodeid.S)) {
result.push(item);
knownIDs.add(item.jobcodeid.S);
}
}
To Map and back
To filter duplicates you can convert the elements to a Map
of key -> value, and then convert back into an array. This works because keys are unique in a Map
, and duplicates will be automatically eliminated. The main advantage of this method is that due to the simplicity of the code it will have fewer bugs.
console.log(
Array.from(
new Map(
input.map(i => [i.jobcodeid.S, i])
).values()
)
)
filter and Set
Another option is to use a Set
to record known ids and filter
to remove items with known ids. The advantage of this method is that it might be easier to read since the intention is explicit. Also this is more performant than converting to Map
and back.
const knownKeys = new Set();
console.log(
input.filter(i => {
if (!knownKeys.has(i.jobcodeid.S)) {
knownKeys.add(i.jobcodeid.S);
return true;
}
})
);
See them in action:
const input = [{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}];
// Classic for loop
const result = ;
const knownIDs = new Set();
for (const item of input) {
if (!knownIDs.has(item.jobcodeid.S)) {
result.push(item);
knownIDs.add(item.jobcodeid.S);
}
}
console.log(result.map(r => r.jobcodeid.S));
// To Map and back
console.log(
Array.from(
new Map(
input.map(i => [i.jobcodeid.S, i])
).values()
)
)
// filter and set
const knownKeys = new Set();
console.log(
input.filter(i => {
if (!knownKeys.has(i.jobcodeid.S)) {
knownKeys.add(i.jobcodeid.S);
return true;
}
})
);
For the record, I ran benchmarks on the accepted solution, mine and the performance improvements from Jacques' answer
accepted solution x 1,892,585 ops/sec ±3.48% (89 runs sampled)
Map and back x 495,116 ops/sec ±2.27% (90 runs sampled)
Set and filter x 1,600,833 ops/sec ±1.98% (90 runs sampled)
Jacques x 2,110,510 ops/sec ±0.98% (92 runs sampled)
Fastest is Jacques
As you can see, Jacques' solution is indeed twice as fast so if you are aiming to filter huge arrays or if performance is key, you should definitely choose that!
add a comment |
up vote
1
down vote
Posting an answer to show another way to do it with greater efficiency.
var arr = [
{"jobcodeid":{"S":"Etc_new"}
},
{"jobcodeid":{"S":"NewLive"}
},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}},
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}}
],
obj = {}, new_arr = ;
// in the end the last unique object will be considered
for (const job of arr) {
if (!obj[job.jobcodeid.S]) {
obj[job.jobcodeid.S] = true;
new_arr.push(job);
}
}
console.log(JSON.stringify(new_arr));
This answer is always runs N iterations. When you loop through the keys after setting the unique values, it can run up to 2N iterations. (Changed from talking about Big O/Complexity to be more clear)
O(2n)
is exactly the same asO(n)
-- en.wikipedia.org/wiki/Big_O_notation
– lucascaro
Nov 9 at 5:57
You are correct, I shouldn't use big o to explain this as it is technically used to explain how complexity grows as quantity increases. My point is that the one solution can take twice as long to complete as the other.
– Jacques
Nov 9 at 6:01
1
Even though looping twice doesn't necessarily mean slower, you are running fewer operations total, so you are right, this code would perform faster. I ran some quick benchmarks that confirm that this solution works roughly 2x faster than the next best :)
– lucascaro
Nov 9 at 6:18
add a comment |
up vote
1
down vote
All you need is to use Set
!
const arr = [
{ jobcodeid: { S: "Etc_new" } },
{ jobcodeid: { S: "NewLive" } },
{ jobcodeid: { S: "NewLiveVid" } },
{ jobcodeid: { S: "New_Live" } },
{ jobcodeid: { S: "New_Live_Vid" } },
{ jobcodeid: { S: "Newest" } },
{ jobcodeid: { S: "NewestLive" } },
{ jobcodeid: { S: "NewestLiveVid" } },
{ jobcodeid: { S: "Very_New_Vid" } },
{ jobcodeid: { S: "Etc_new" } },
{ jobcodeid: { S: "NewLive" } },
{ jobcodeid: { S: "NewLiveVid" } },
{ jobcodeid: { S: "New_Live" } },
{ jobcodeid: { S: "New_Live_Vid" } },
{ jobcodeid: { S: "Newest" } },
{ jobcodeid: { S: "NewestLive" } },
{ jobcodeid: { S: "NewestLiveVid" } },
{ jobcodeid: { S: "Very_New_Vid" } }
];
const uniqueItems = [...new Set(arr.map(i => i.jobcodeid.S))]
add a comment |
up vote
0
down vote
Try this also.. another way to solve this problem
var arr = [
{"jobcodeid":{"S":"Etc_new"}
},
{"jobcodeid":{"S":"NewLive"}
},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}},
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}}
],
obj = {}, new_arr = ;
arr.forEach(function(v){
obj[v['id']] = v;
for(var i=0;i< new_arr.length;i++){
if(new_arr[i].jobcodeid.S == v.jobcodeid.S) {
return;
}
}
new_arr.push(v);
});
console.log(new_arr);
While this would work, please consider code complexity here. You have a loop within a loop, which is very inefficient.
– Jacques
Nov 9 at 6:05
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Fist of all, you have to use obj[v['jobcodeid']] = v;
instead of obj[v['id']] = v;
.
but as v[jobcodeid]
is an object, js will convert it to a string i.e. [object Object]
and you will only one element in final array.
// exemplary array of objects (id 'NewLive' occurs twice)
var arr=[{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}], obj = {}, new_arr = ;
// in the end the last unique object will be considered
arr.forEach(function(v){
obj[v['jobcodeid']] = v;
});
new_arr = Object.keys(obj).map(function(id) { return obj[id]; });
console.log(JSON.stringify(new_arr));
You should use v.jobcodeid.S
as keys of object.
// exemplary array of objects (id 'NewLive' occurs twice)
var arr=[{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}], obj = {}, new_arr = ;
// in the end the last unique object will be considered
arr.forEach(function(v){
obj[v.jobcodeid.S] = v;
});
new_arr = Object.keys(obj).map(function(id) { return obj[id]; });
console.log(JSON.stringify(new_arr));
I missed that in my comment. +1
– Rajesh
Nov 9 at 5:33
Have you tested the code? Can u update the same on codepen?
– Arun VM
Nov 9 at 5:34
Yeah i have updated the same pen any way
– Arun VM
Nov 9 at 5:36
@ArunVM added code snippets.
– nrgwsth
Nov 9 at 5:39
I have accepted the answer Peace :), Thank you
– Arun VM
Nov 9 at 5:39
|
show 1 more comment
up vote
1
down vote
accepted
Fist of all, you have to use obj[v['jobcodeid']] = v;
instead of obj[v['id']] = v;
.
but as v[jobcodeid]
is an object, js will convert it to a string i.e. [object Object]
and you will only one element in final array.
// exemplary array of objects (id 'NewLive' occurs twice)
var arr=[{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}], obj = {}, new_arr = ;
// in the end the last unique object will be considered
arr.forEach(function(v){
obj[v['jobcodeid']] = v;
});
new_arr = Object.keys(obj).map(function(id) { return obj[id]; });
console.log(JSON.stringify(new_arr));
You should use v.jobcodeid.S
as keys of object.
// exemplary array of objects (id 'NewLive' occurs twice)
var arr=[{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}], obj = {}, new_arr = ;
// in the end the last unique object will be considered
arr.forEach(function(v){
obj[v.jobcodeid.S] = v;
});
new_arr = Object.keys(obj).map(function(id) { return obj[id]; });
console.log(JSON.stringify(new_arr));
I missed that in my comment. +1
– Rajesh
Nov 9 at 5:33
Have you tested the code? Can u update the same on codepen?
– Arun VM
Nov 9 at 5:34
Yeah i have updated the same pen any way
– Arun VM
Nov 9 at 5:36
@ArunVM added code snippets.
– nrgwsth
Nov 9 at 5:39
I have accepted the answer Peace :), Thank you
– Arun VM
Nov 9 at 5:39
|
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Fist of all, you have to use obj[v['jobcodeid']] = v;
instead of obj[v['id']] = v;
.
but as v[jobcodeid]
is an object, js will convert it to a string i.e. [object Object]
and you will only one element in final array.
// exemplary array of objects (id 'NewLive' occurs twice)
var arr=[{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}], obj = {}, new_arr = ;
// in the end the last unique object will be considered
arr.forEach(function(v){
obj[v['jobcodeid']] = v;
});
new_arr = Object.keys(obj).map(function(id) { return obj[id]; });
console.log(JSON.stringify(new_arr));
You should use v.jobcodeid.S
as keys of object.
// exemplary array of objects (id 'NewLive' occurs twice)
var arr=[{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}], obj = {}, new_arr = ;
// in the end the last unique object will be considered
arr.forEach(function(v){
obj[v.jobcodeid.S] = v;
});
new_arr = Object.keys(obj).map(function(id) { return obj[id]; });
console.log(JSON.stringify(new_arr));
Fist of all, you have to use obj[v['jobcodeid']] = v;
instead of obj[v['id']] = v;
.
but as v[jobcodeid]
is an object, js will convert it to a string i.e. [object Object]
and you will only one element in final array.
// exemplary array of objects (id 'NewLive' occurs twice)
var arr=[{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}], obj = {}, new_arr = ;
// in the end the last unique object will be considered
arr.forEach(function(v){
obj[v['jobcodeid']] = v;
});
new_arr = Object.keys(obj).map(function(id) { return obj[id]; });
console.log(JSON.stringify(new_arr));
You should use v.jobcodeid.S
as keys of object.
// exemplary array of objects (id 'NewLive' occurs twice)
var arr=[{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}], obj = {}, new_arr = ;
// in the end the last unique object will be considered
arr.forEach(function(v){
obj[v.jobcodeid.S] = v;
});
new_arr = Object.keys(obj).map(function(id) { return obj[id]; });
console.log(JSON.stringify(new_arr));
// exemplary array of objects (id 'NewLive' occurs twice)
var arr=[{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}], obj = {}, new_arr = ;
// in the end the last unique object will be considered
arr.forEach(function(v){
obj[v['jobcodeid']] = v;
});
new_arr = Object.keys(obj).map(function(id) { return obj[id]; });
console.log(JSON.stringify(new_arr));
// exemplary array of objects (id 'NewLive' occurs twice)
var arr=[{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}], obj = {}, new_arr = ;
// in the end the last unique object will be considered
arr.forEach(function(v){
obj[v['jobcodeid']] = v;
});
new_arr = Object.keys(obj).map(function(id) { return obj[id]; });
console.log(JSON.stringify(new_arr));
// exemplary array of objects (id 'NewLive' occurs twice)
var arr=[{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}], obj = {}, new_arr = ;
// in the end the last unique object will be considered
arr.forEach(function(v){
obj[v.jobcodeid.S] = v;
});
new_arr = Object.keys(obj).map(function(id) { return obj[id]; });
console.log(JSON.stringify(new_arr));
// exemplary array of objects (id 'NewLive' occurs twice)
var arr=[{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}], obj = {}, new_arr = ;
// in the end the last unique object will be considered
arr.forEach(function(v){
obj[v.jobcodeid.S] = v;
});
new_arr = Object.keys(obj).map(function(id) { return obj[id]; });
console.log(JSON.stringify(new_arr));
edited Nov 9 at 5:36
answered Nov 9 at 5:32
nrgwsth
3,7281822
3,7281822
I missed that in my comment. +1
– Rajesh
Nov 9 at 5:33
Have you tested the code? Can u update the same on codepen?
– Arun VM
Nov 9 at 5:34
Yeah i have updated the same pen any way
– Arun VM
Nov 9 at 5:36
@ArunVM added code snippets.
– nrgwsth
Nov 9 at 5:39
I have accepted the answer Peace :), Thank you
– Arun VM
Nov 9 at 5:39
|
show 1 more comment
I missed that in my comment. +1
– Rajesh
Nov 9 at 5:33
Have you tested the code? Can u update the same on codepen?
– Arun VM
Nov 9 at 5:34
Yeah i have updated the same pen any way
– Arun VM
Nov 9 at 5:36
@ArunVM added code snippets.
– nrgwsth
Nov 9 at 5:39
I have accepted the answer Peace :), Thank you
– Arun VM
Nov 9 at 5:39
I missed that in my comment. +1
– Rajesh
Nov 9 at 5:33
I missed that in my comment. +1
– Rajesh
Nov 9 at 5:33
Have you tested the code? Can u update the same on codepen?
– Arun VM
Nov 9 at 5:34
Have you tested the code? Can u update the same on codepen?
– Arun VM
Nov 9 at 5:34
Yeah i have updated the same pen any way
– Arun VM
Nov 9 at 5:36
Yeah i have updated the same pen any way
– Arun VM
Nov 9 at 5:36
@ArunVM added code snippets.
– nrgwsth
Nov 9 at 5:39
@ArunVM added code snippets.
– nrgwsth
Nov 9 at 5:39
I have accepted the answer Peace :), Thank you
– Arun VM
Nov 9 at 5:39
I have accepted the answer Peace :), Thank you
– Arun VM
Nov 9 at 5:39
|
show 1 more comment
up vote
4
down vote
The reason why your code is returning a single element is because you are using v['id']
but there is no id
property on the objects, therefore throughout your loop you are setting obj[undefined]
over and over.
In your jsfiddle code though this looks correct and the code seems to be working as intended.
In case someone got to this question to find out how to remove duplicates from an array in javascript, here are a few options:
The classic way: good old for loop
This is essentially the solution you used, iterate over the array, check if the key has been added to the result array, and if it's not there, add the element to the result.
Example:
const result = ;
const knownIDs = new Set();
for (const item of input) {
if (!knownIDs.has(item.jobcodeid.S)) {
result.push(item);
knownIDs.add(item.jobcodeid.S);
}
}
To Map and back
To filter duplicates you can convert the elements to a Map
of key -> value, and then convert back into an array. This works because keys are unique in a Map
, and duplicates will be automatically eliminated. The main advantage of this method is that due to the simplicity of the code it will have fewer bugs.
console.log(
Array.from(
new Map(
input.map(i => [i.jobcodeid.S, i])
).values()
)
)
filter and Set
Another option is to use a Set
to record known ids and filter
to remove items with known ids. The advantage of this method is that it might be easier to read since the intention is explicit. Also this is more performant than converting to Map
and back.
const knownKeys = new Set();
console.log(
input.filter(i => {
if (!knownKeys.has(i.jobcodeid.S)) {
knownKeys.add(i.jobcodeid.S);
return true;
}
})
);
See them in action:
const input = [{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}];
// Classic for loop
const result = ;
const knownIDs = new Set();
for (const item of input) {
if (!knownIDs.has(item.jobcodeid.S)) {
result.push(item);
knownIDs.add(item.jobcodeid.S);
}
}
console.log(result.map(r => r.jobcodeid.S));
// To Map and back
console.log(
Array.from(
new Map(
input.map(i => [i.jobcodeid.S, i])
).values()
)
)
// filter and set
const knownKeys = new Set();
console.log(
input.filter(i => {
if (!knownKeys.has(i.jobcodeid.S)) {
knownKeys.add(i.jobcodeid.S);
return true;
}
})
);
For the record, I ran benchmarks on the accepted solution, mine and the performance improvements from Jacques' answer
accepted solution x 1,892,585 ops/sec ±3.48% (89 runs sampled)
Map and back x 495,116 ops/sec ±2.27% (90 runs sampled)
Set and filter x 1,600,833 ops/sec ±1.98% (90 runs sampled)
Jacques x 2,110,510 ops/sec ±0.98% (92 runs sampled)
Fastest is Jacques
As you can see, Jacques' solution is indeed twice as fast so if you are aiming to filter huge arrays or if performance is key, you should definitely choose that!
add a comment |
up vote
4
down vote
The reason why your code is returning a single element is because you are using v['id']
but there is no id
property on the objects, therefore throughout your loop you are setting obj[undefined]
over and over.
In your jsfiddle code though this looks correct and the code seems to be working as intended.
In case someone got to this question to find out how to remove duplicates from an array in javascript, here are a few options:
The classic way: good old for loop
This is essentially the solution you used, iterate over the array, check if the key has been added to the result array, and if it's not there, add the element to the result.
Example:
const result = ;
const knownIDs = new Set();
for (const item of input) {
if (!knownIDs.has(item.jobcodeid.S)) {
result.push(item);
knownIDs.add(item.jobcodeid.S);
}
}
To Map and back
To filter duplicates you can convert the elements to a Map
of key -> value, and then convert back into an array. This works because keys are unique in a Map
, and duplicates will be automatically eliminated. The main advantage of this method is that due to the simplicity of the code it will have fewer bugs.
console.log(
Array.from(
new Map(
input.map(i => [i.jobcodeid.S, i])
).values()
)
)
filter and Set
Another option is to use a Set
to record known ids and filter
to remove items with known ids. The advantage of this method is that it might be easier to read since the intention is explicit. Also this is more performant than converting to Map
and back.
const knownKeys = new Set();
console.log(
input.filter(i => {
if (!knownKeys.has(i.jobcodeid.S)) {
knownKeys.add(i.jobcodeid.S);
return true;
}
})
);
See them in action:
const input = [{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}];
// Classic for loop
const result = ;
const knownIDs = new Set();
for (const item of input) {
if (!knownIDs.has(item.jobcodeid.S)) {
result.push(item);
knownIDs.add(item.jobcodeid.S);
}
}
console.log(result.map(r => r.jobcodeid.S));
// To Map and back
console.log(
Array.from(
new Map(
input.map(i => [i.jobcodeid.S, i])
).values()
)
)
// filter and set
const knownKeys = new Set();
console.log(
input.filter(i => {
if (!knownKeys.has(i.jobcodeid.S)) {
knownKeys.add(i.jobcodeid.S);
return true;
}
})
);
For the record, I ran benchmarks on the accepted solution, mine and the performance improvements from Jacques' answer
accepted solution x 1,892,585 ops/sec ±3.48% (89 runs sampled)
Map and back x 495,116 ops/sec ±2.27% (90 runs sampled)
Set and filter x 1,600,833 ops/sec ±1.98% (90 runs sampled)
Jacques x 2,110,510 ops/sec ±0.98% (92 runs sampled)
Fastest is Jacques
As you can see, Jacques' solution is indeed twice as fast so if you are aiming to filter huge arrays or if performance is key, you should definitely choose that!
add a comment |
up vote
4
down vote
up vote
4
down vote
The reason why your code is returning a single element is because you are using v['id']
but there is no id
property on the objects, therefore throughout your loop you are setting obj[undefined]
over and over.
In your jsfiddle code though this looks correct and the code seems to be working as intended.
In case someone got to this question to find out how to remove duplicates from an array in javascript, here are a few options:
The classic way: good old for loop
This is essentially the solution you used, iterate over the array, check if the key has been added to the result array, and if it's not there, add the element to the result.
Example:
const result = ;
const knownIDs = new Set();
for (const item of input) {
if (!knownIDs.has(item.jobcodeid.S)) {
result.push(item);
knownIDs.add(item.jobcodeid.S);
}
}
To Map and back
To filter duplicates you can convert the elements to a Map
of key -> value, and then convert back into an array. This works because keys are unique in a Map
, and duplicates will be automatically eliminated. The main advantage of this method is that due to the simplicity of the code it will have fewer bugs.
console.log(
Array.from(
new Map(
input.map(i => [i.jobcodeid.S, i])
).values()
)
)
filter and Set
Another option is to use a Set
to record known ids and filter
to remove items with known ids. The advantage of this method is that it might be easier to read since the intention is explicit. Also this is more performant than converting to Map
and back.
const knownKeys = new Set();
console.log(
input.filter(i => {
if (!knownKeys.has(i.jobcodeid.S)) {
knownKeys.add(i.jobcodeid.S);
return true;
}
})
);
See them in action:
const input = [{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}];
// Classic for loop
const result = ;
const knownIDs = new Set();
for (const item of input) {
if (!knownIDs.has(item.jobcodeid.S)) {
result.push(item);
knownIDs.add(item.jobcodeid.S);
}
}
console.log(result.map(r => r.jobcodeid.S));
// To Map and back
console.log(
Array.from(
new Map(
input.map(i => [i.jobcodeid.S, i])
).values()
)
)
// filter and set
const knownKeys = new Set();
console.log(
input.filter(i => {
if (!knownKeys.has(i.jobcodeid.S)) {
knownKeys.add(i.jobcodeid.S);
return true;
}
})
);
For the record, I ran benchmarks on the accepted solution, mine and the performance improvements from Jacques' answer
accepted solution x 1,892,585 ops/sec ±3.48% (89 runs sampled)
Map and back x 495,116 ops/sec ±2.27% (90 runs sampled)
Set and filter x 1,600,833 ops/sec ±1.98% (90 runs sampled)
Jacques x 2,110,510 ops/sec ±0.98% (92 runs sampled)
Fastest is Jacques
As you can see, Jacques' solution is indeed twice as fast so if you are aiming to filter huge arrays or if performance is key, you should definitely choose that!
The reason why your code is returning a single element is because you are using v['id']
but there is no id
property on the objects, therefore throughout your loop you are setting obj[undefined]
over and over.
In your jsfiddle code though this looks correct and the code seems to be working as intended.
In case someone got to this question to find out how to remove duplicates from an array in javascript, here are a few options:
The classic way: good old for loop
This is essentially the solution you used, iterate over the array, check if the key has been added to the result array, and if it's not there, add the element to the result.
Example:
const result = ;
const knownIDs = new Set();
for (const item of input) {
if (!knownIDs.has(item.jobcodeid.S)) {
result.push(item);
knownIDs.add(item.jobcodeid.S);
}
}
To Map and back
To filter duplicates you can convert the elements to a Map
of key -> value, and then convert back into an array. This works because keys are unique in a Map
, and duplicates will be automatically eliminated. The main advantage of this method is that due to the simplicity of the code it will have fewer bugs.
console.log(
Array.from(
new Map(
input.map(i => [i.jobcodeid.S, i])
).values()
)
)
filter and Set
Another option is to use a Set
to record known ids and filter
to remove items with known ids. The advantage of this method is that it might be easier to read since the intention is explicit. Also this is more performant than converting to Map
and back.
const knownKeys = new Set();
console.log(
input.filter(i => {
if (!knownKeys.has(i.jobcodeid.S)) {
knownKeys.add(i.jobcodeid.S);
return true;
}
})
);
See them in action:
const input = [{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}];
// Classic for loop
const result = ;
const knownIDs = new Set();
for (const item of input) {
if (!knownIDs.has(item.jobcodeid.S)) {
result.push(item);
knownIDs.add(item.jobcodeid.S);
}
}
console.log(result.map(r => r.jobcodeid.S));
// To Map and back
console.log(
Array.from(
new Map(
input.map(i => [i.jobcodeid.S, i])
).values()
)
)
// filter and set
const knownKeys = new Set();
console.log(
input.filter(i => {
if (!knownKeys.has(i.jobcodeid.S)) {
knownKeys.add(i.jobcodeid.S);
return true;
}
})
);
For the record, I ran benchmarks on the accepted solution, mine and the performance improvements from Jacques' answer
accepted solution x 1,892,585 ops/sec ±3.48% (89 runs sampled)
Map and back x 495,116 ops/sec ±2.27% (90 runs sampled)
Set and filter x 1,600,833 ops/sec ±1.98% (90 runs sampled)
Jacques x 2,110,510 ops/sec ±0.98% (92 runs sampled)
Fastest is Jacques
As you can see, Jacques' solution is indeed twice as fast so if you are aiming to filter huge arrays or if performance is key, you should definitely choose that!
const input = [{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}];
// Classic for loop
const result = ;
const knownIDs = new Set();
for (const item of input) {
if (!knownIDs.has(item.jobcodeid.S)) {
result.push(item);
knownIDs.add(item.jobcodeid.S);
}
}
console.log(result.map(r => r.jobcodeid.S));
// To Map and back
console.log(
Array.from(
new Map(
input.map(i => [i.jobcodeid.S, i])
).values()
)
)
// filter and set
const knownKeys = new Set();
console.log(
input.filter(i => {
if (!knownKeys.has(i.jobcodeid.S)) {
knownKeys.add(i.jobcodeid.S);
return true;
}
})
);
const input = [{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}},{"jobcodeid":{"S":"Etc_new"}},{"jobcodeid":{"S":"NewLive"}},{"jobcodeid":{"S":"NewLiveVid"}},{"jobcodeid":{"S":"New_Live"}},{"jobcodeid":{"S":"New_Live_Vid"}},{"jobcodeid":{"S":"Newest"}},{"jobcodeid":{"S":"NewestLive"}},{"jobcodeid":{"S":"NewestLiveVid"}},{"jobcodeid":{"S":"Very_New_Vid"}}];
// Classic for loop
const result = ;
const knownIDs = new Set();
for (const item of input) {
if (!knownIDs.has(item.jobcodeid.S)) {
result.push(item);
knownIDs.add(item.jobcodeid.S);
}
}
console.log(result.map(r => r.jobcodeid.S));
// To Map and back
console.log(
Array.from(
new Map(
input.map(i => [i.jobcodeid.S, i])
).values()
)
)
// filter and set
const knownKeys = new Set();
console.log(
input.filter(i => {
if (!knownKeys.has(i.jobcodeid.S)) {
knownKeys.add(i.jobcodeid.S);
return true;
}
})
);
edited Nov 11 at 6:09
answered Nov 9 at 5:34
lucascaro
3,30611430
3,30611430
add a comment |
add a comment |
up vote
1
down vote
Posting an answer to show another way to do it with greater efficiency.
var arr = [
{"jobcodeid":{"S":"Etc_new"}
},
{"jobcodeid":{"S":"NewLive"}
},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}},
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}}
],
obj = {}, new_arr = ;
// in the end the last unique object will be considered
for (const job of arr) {
if (!obj[job.jobcodeid.S]) {
obj[job.jobcodeid.S] = true;
new_arr.push(job);
}
}
console.log(JSON.stringify(new_arr));
This answer is always runs N iterations. When you loop through the keys after setting the unique values, it can run up to 2N iterations. (Changed from talking about Big O/Complexity to be more clear)
O(2n)
is exactly the same asO(n)
-- en.wikipedia.org/wiki/Big_O_notation
– lucascaro
Nov 9 at 5:57
You are correct, I shouldn't use big o to explain this as it is technically used to explain how complexity grows as quantity increases. My point is that the one solution can take twice as long to complete as the other.
– Jacques
Nov 9 at 6:01
1
Even though looping twice doesn't necessarily mean slower, you are running fewer operations total, so you are right, this code would perform faster. I ran some quick benchmarks that confirm that this solution works roughly 2x faster than the next best :)
– lucascaro
Nov 9 at 6:18
add a comment |
up vote
1
down vote
Posting an answer to show another way to do it with greater efficiency.
var arr = [
{"jobcodeid":{"S":"Etc_new"}
},
{"jobcodeid":{"S":"NewLive"}
},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}},
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}}
],
obj = {}, new_arr = ;
// in the end the last unique object will be considered
for (const job of arr) {
if (!obj[job.jobcodeid.S]) {
obj[job.jobcodeid.S] = true;
new_arr.push(job);
}
}
console.log(JSON.stringify(new_arr));
This answer is always runs N iterations. When you loop through the keys after setting the unique values, it can run up to 2N iterations. (Changed from talking about Big O/Complexity to be more clear)
O(2n)
is exactly the same asO(n)
-- en.wikipedia.org/wiki/Big_O_notation
– lucascaro
Nov 9 at 5:57
You are correct, I shouldn't use big o to explain this as it is technically used to explain how complexity grows as quantity increases. My point is that the one solution can take twice as long to complete as the other.
– Jacques
Nov 9 at 6:01
1
Even though looping twice doesn't necessarily mean slower, you are running fewer operations total, so you are right, this code would perform faster. I ran some quick benchmarks that confirm that this solution works roughly 2x faster than the next best :)
– lucascaro
Nov 9 at 6:18
add a comment |
up vote
1
down vote
up vote
1
down vote
Posting an answer to show another way to do it with greater efficiency.
var arr = [
{"jobcodeid":{"S":"Etc_new"}
},
{"jobcodeid":{"S":"NewLive"}
},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}},
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}}
],
obj = {}, new_arr = ;
// in the end the last unique object will be considered
for (const job of arr) {
if (!obj[job.jobcodeid.S]) {
obj[job.jobcodeid.S] = true;
new_arr.push(job);
}
}
console.log(JSON.stringify(new_arr));
This answer is always runs N iterations. When you loop through the keys after setting the unique values, it can run up to 2N iterations. (Changed from talking about Big O/Complexity to be more clear)
Posting an answer to show another way to do it with greater efficiency.
var arr = [
{"jobcodeid":{"S":"Etc_new"}
},
{"jobcodeid":{"S":"NewLive"}
},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}},
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}}
],
obj = {}, new_arr = ;
// in the end the last unique object will be considered
for (const job of arr) {
if (!obj[job.jobcodeid.S]) {
obj[job.jobcodeid.S] = true;
new_arr.push(job);
}
}
console.log(JSON.stringify(new_arr));
This answer is always runs N iterations. When you loop through the keys after setting the unique values, it can run up to 2N iterations. (Changed from talking about Big O/Complexity to be more clear)
var arr = [
{"jobcodeid":{"S":"Etc_new"}
},
{"jobcodeid":{"S":"NewLive"}
},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}},
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}}
],
obj = {}, new_arr = ;
// in the end the last unique object will be considered
for (const job of arr) {
if (!obj[job.jobcodeid.S]) {
obj[job.jobcodeid.S] = true;
new_arr.push(job);
}
}
console.log(JSON.stringify(new_arr));
var arr = [
{"jobcodeid":{"S":"Etc_new"}
},
{"jobcodeid":{"S":"NewLive"}
},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}},
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}}
],
obj = {}, new_arr = ;
// in the end the last unique object will be considered
for (const job of arr) {
if (!obj[job.jobcodeid.S]) {
obj[job.jobcodeid.S] = true;
new_arr.push(job);
}
}
console.log(JSON.stringify(new_arr));
edited Nov 9 at 6:04
answered Nov 9 at 5:55
Jacques
2,25721135
2,25721135
O(2n)
is exactly the same asO(n)
-- en.wikipedia.org/wiki/Big_O_notation
– lucascaro
Nov 9 at 5:57
You are correct, I shouldn't use big o to explain this as it is technically used to explain how complexity grows as quantity increases. My point is that the one solution can take twice as long to complete as the other.
– Jacques
Nov 9 at 6:01
1
Even though looping twice doesn't necessarily mean slower, you are running fewer operations total, so you are right, this code would perform faster. I ran some quick benchmarks that confirm that this solution works roughly 2x faster than the next best :)
– lucascaro
Nov 9 at 6:18
add a comment |
O(2n)
is exactly the same asO(n)
-- en.wikipedia.org/wiki/Big_O_notation
– lucascaro
Nov 9 at 5:57
You are correct, I shouldn't use big o to explain this as it is technically used to explain how complexity grows as quantity increases. My point is that the one solution can take twice as long to complete as the other.
– Jacques
Nov 9 at 6:01
1
Even though looping twice doesn't necessarily mean slower, you are running fewer operations total, so you are right, this code would perform faster. I ran some quick benchmarks that confirm that this solution works roughly 2x faster than the next best :)
– lucascaro
Nov 9 at 6:18
O(2n)
is exactly the same as O(n)
-- en.wikipedia.org/wiki/Big_O_notation– lucascaro
Nov 9 at 5:57
O(2n)
is exactly the same as O(n)
-- en.wikipedia.org/wiki/Big_O_notation– lucascaro
Nov 9 at 5:57
You are correct, I shouldn't use big o to explain this as it is technically used to explain how complexity grows as quantity increases. My point is that the one solution can take twice as long to complete as the other.
– Jacques
Nov 9 at 6:01
You are correct, I shouldn't use big o to explain this as it is technically used to explain how complexity grows as quantity increases. My point is that the one solution can take twice as long to complete as the other.
– Jacques
Nov 9 at 6:01
1
1
Even though looping twice doesn't necessarily mean slower, you are running fewer operations total, so you are right, this code would perform faster. I ran some quick benchmarks that confirm that this solution works roughly 2x faster than the next best :)
– lucascaro
Nov 9 at 6:18
Even though looping twice doesn't necessarily mean slower, you are running fewer operations total, so you are right, this code would perform faster. I ran some quick benchmarks that confirm that this solution works roughly 2x faster than the next best :)
– lucascaro
Nov 9 at 6:18
add a comment |
up vote
1
down vote
All you need is to use Set
!
const arr = [
{ jobcodeid: { S: "Etc_new" } },
{ jobcodeid: { S: "NewLive" } },
{ jobcodeid: { S: "NewLiveVid" } },
{ jobcodeid: { S: "New_Live" } },
{ jobcodeid: { S: "New_Live_Vid" } },
{ jobcodeid: { S: "Newest" } },
{ jobcodeid: { S: "NewestLive" } },
{ jobcodeid: { S: "NewestLiveVid" } },
{ jobcodeid: { S: "Very_New_Vid" } },
{ jobcodeid: { S: "Etc_new" } },
{ jobcodeid: { S: "NewLive" } },
{ jobcodeid: { S: "NewLiveVid" } },
{ jobcodeid: { S: "New_Live" } },
{ jobcodeid: { S: "New_Live_Vid" } },
{ jobcodeid: { S: "Newest" } },
{ jobcodeid: { S: "NewestLive" } },
{ jobcodeid: { S: "NewestLiveVid" } },
{ jobcodeid: { S: "Very_New_Vid" } }
];
const uniqueItems = [...new Set(arr.map(i => i.jobcodeid.S))]
add a comment |
up vote
1
down vote
All you need is to use Set
!
const arr = [
{ jobcodeid: { S: "Etc_new" } },
{ jobcodeid: { S: "NewLive" } },
{ jobcodeid: { S: "NewLiveVid" } },
{ jobcodeid: { S: "New_Live" } },
{ jobcodeid: { S: "New_Live_Vid" } },
{ jobcodeid: { S: "Newest" } },
{ jobcodeid: { S: "NewestLive" } },
{ jobcodeid: { S: "NewestLiveVid" } },
{ jobcodeid: { S: "Very_New_Vid" } },
{ jobcodeid: { S: "Etc_new" } },
{ jobcodeid: { S: "NewLive" } },
{ jobcodeid: { S: "NewLiveVid" } },
{ jobcodeid: { S: "New_Live" } },
{ jobcodeid: { S: "New_Live_Vid" } },
{ jobcodeid: { S: "Newest" } },
{ jobcodeid: { S: "NewestLive" } },
{ jobcodeid: { S: "NewestLiveVid" } },
{ jobcodeid: { S: "Very_New_Vid" } }
];
const uniqueItems = [...new Set(arr.map(i => i.jobcodeid.S))]
add a comment |
up vote
1
down vote
up vote
1
down vote
All you need is to use Set
!
const arr = [
{ jobcodeid: { S: "Etc_new" } },
{ jobcodeid: { S: "NewLive" } },
{ jobcodeid: { S: "NewLiveVid" } },
{ jobcodeid: { S: "New_Live" } },
{ jobcodeid: { S: "New_Live_Vid" } },
{ jobcodeid: { S: "Newest" } },
{ jobcodeid: { S: "NewestLive" } },
{ jobcodeid: { S: "NewestLiveVid" } },
{ jobcodeid: { S: "Very_New_Vid" } },
{ jobcodeid: { S: "Etc_new" } },
{ jobcodeid: { S: "NewLive" } },
{ jobcodeid: { S: "NewLiveVid" } },
{ jobcodeid: { S: "New_Live" } },
{ jobcodeid: { S: "New_Live_Vid" } },
{ jobcodeid: { S: "Newest" } },
{ jobcodeid: { S: "NewestLive" } },
{ jobcodeid: { S: "NewestLiveVid" } },
{ jobcodeid: { S: "Very_New_Vid" } }
];
const uniqueItems = [...new Set(arr.map(i => i.jobcodeid.S))]
All you need is to use Set
!
const arr = [
{ jobcodeid: { S: "Etc_new" } },
{ jobcodeid: { S: "NewLive" } },
{ jobcodeid: { S: "NewLiveVid" } },
{ jobcodeid: { S: "New_Live" } },
{ jobcodeid: { S: "New_Live_Vid" } },
{ jobcodeid: { S: "Newest" } },
{ jobcodeid: { S: "NewestLive" } },
{ jobcodeid: { S: "NewestLiveVid" } },
{ jobcodeid: { S: "Very_New_Vid" } },
{ jobcodeid: { S: "Etc_new" } },
{ jobcodeid: { S: "NewLive" } },
{ jobcodeid: { S: "NewLiveVid" } },
{ jobcodeid: { S: "New_Live" } },
{ jobcodeid: { S: "New_Live_Vid" } },
{ jobcodeid: { S: "Newest" } },
{ jobcodeid: { S: "NewestLive" } },
{ jobcodeid: { S: "NewestLiveVid" } },
{ jobcodeid: { S: "Very_New_Vid" } }
];
const uniqueItems = [...new Set(arr.map(i => i.jobcodeid.S))]
answered Nov 9 at 9:20
Phap Duong Dieu
692
692
add a comment |
add a comment |
up vote
0
down vote
Try this also.. another way to solve this problem
var arr = [
{"jobcodeid":{"S":"Etc_new"}
},
{"jobcodeid":{"S":"NewLive"}
},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}},
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}}
],
obj = {}, new_arr = ;
arr.forEach(function(v){
obj[v['id']] = v;
for(var i=0;i< new_arr.length;i++){
if(new_arr[i].jobcodeid.S == v.jobcodeid.S) {
return;
}
}
new_arr.push(v);
});
console.log(new_arr);
While this would work, please consider code complexity here. You have a loop within a loop, which is very inefficient.
– Jacques
Nov 9 at 6:05
add a comment |
up vote
0
down vote
Try this also.. another way to solve this problem
var arr = [
{"jobcodeid":{"S":"Etc_new"}
},
{"jobcodeid":{"S":"NewLive"}
},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}},
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}}
],
obj = {}, new_arr = ;
arr.forEach(function(v){
obj[v['id']] = v;
for(var i=0;i< new_arr.length;i++){
if(new_arr[i].jobcodeid.S == v.jobcodeid.S) {
return;
}
}
new_arr.push(v);
});
console.log(new_arr);
While this would work, please consider code complexity here. You have a loop within a loop, which is very inefficient.
– Jacques
Nov 9 at 6:05
add a comment |
up vote
0
down vote
up vote
0
down vote
Try this also.. another way to solve this problem
var arr = [
{"jobcodeid":{"S":"Etc_new"}
},
{"jobcodeid":{"S":"NewLive"}
},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}},
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}}
],
obj = {}, new_arr = ;
arr.forEach(function(v){
obj[v['id']] = v;
for(var i=0;i< new_arr.length;i++){
if(new_arr[i].jobcodeid.S == v.jobcodeid.S) {
return;
}
}
new_arr.push(v);
});
console.log(new_arr);
Try this also.. another way to solve this problem
var arr = [
{"jobcodeid":{"S":"Etc_new"}
},
{"jobcodeid":{"S":"NewLive"}
},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}},
{"jobcodeid":{"S":"Etc_new"}},
{"jobcodeid":{"S":"NewLive"}},
{"jobcodeid":{"S":"NewLiveVid"}},
{"jobcodeid":{"S":"New_Live"}},
{"jobcodeid":{"S":"New_Live_Vid"}},
{"jobcodeid":{"S":"Newest"}},
{"jobcodeid":{"S":"NewestLive"}},
{"jobcodeid":{"S":"NewestLiveVid"}},
{"jobcodeid":{"S":"Very_New_Vid"}}
],
obj = {}, new_arr = ;
arr.forEach(function(v){
obj[v['id']] = v;
for(var i=0;i< new_arr.length;i++){
if(new_arr[i].jobcodeid.S == v.jobcodeid.S) {
return;
}
}
new_arr.push(v);
});
console.log(new_arr);
answered Nov 9 at 5:56
GaneshMani
11
11
While this would work, please consider code complexity here. You have a loop within a loop, which is very inefficient.
– Jacques
Nov 9 at 6:05
add a comment |
While this would work, please consider code complexity here. You have a loop within a loop, which is very inefficient.
– Jacques
Nov 9 at 6:05
While this would work, please consider code complexity here. You have a loop within a loop, which is very inefficient.
– Jacques
Nov 9 at 6:05
While this would work, please consider code complexity here. You have a loop within a loop, which is very inefficient.
– Jacques
Nov 9 at 6:05
add a comment |
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Are you just trying to return the same object without duplicates?
– Jacques
Nov 9 at 5:26
@Jacques Yes I am
– Arun VM
Nov 9 at 5:27
Possible duplicate of Get all unique values in a JavaScript array (remove duplicates)
– Frank Fajardo
Nov 9 at 5:28
obj[v['id']] = v;
should beobj[v['jobcodeid']] = v;
. Since there is noid
, you get undefined for all objects and hence you are basically replacing all values by next. Hence you get only 1– Rajesh
Nov 9 at 5:30
for (const job of arr) { if (!obj[job.jobcodeid.S]) { obj[job.jobcodeid.S] = true; new_arr.push(job); } }
– Jacques
Nov 9 at 5:30