Remove Only One Duplicate from An Array
2
I'm trying to only remove one of the 2s from an array, but my code removes all of them. My code is as follows:
var arr = [2,7,9,5,2]
arr.filter(item => ((item !== 2)));
and:
var arr = [2,7,9,2,2,5,2]
arr.filter(item => ((item !== 2)));
Both remove all the 2s. I thought about removing duplicates, where it works if there's only one duplicate - e.g.:
Array.from(new Set([2,7,9,5,2]));
function uniq(a) {
return Array.from(new Set(a))
}
But fails if there's multiple duplicates as it just removes them all, including any other duplicated numbers:
Array.from(new Set([2,7,9,9,2,2,5,2]));
function uniq(a) {
return Array.from(new Set(a))
}
Does anyone know how to only remove one of the 2s? Thanks for any help here.
javascript
asked Nov 19 '18 at 10:32
user8758206user8758206
456110
add a comment |
2
I'm trying to only remove one of the 2s from an array, but my code removes all of them. My code is as follows:
var arr = [2,7,9,5,2]
arr.filter(item => ((item !== 2)));
and:
var arr = [2,7,9,2,2,5,2]
arr.filter(item => ((item !== 2)));
Both remove all the 2s. I thought about removing duplicates, where it works if there's only one duplicate - e.g.:
Array.from(new Set([2,7,9,5,2]));
function uniq(a) {
return Array.from(new Set(a))
}
But fails if there's multiple duplicates as it just removes them all, including any other duplicated numbers:
Array.from(new Set([2,7,9,9,2,2,5,2]));
function uniq(a) {
return Array.from(new Set(a))
}
Does anyone know how to only remove one of the 2s? Thanks for any help here.
javascript
asked Nov 19 '18 at 10:32
user8758206user8758206
456110
1
create a simple loop, when found an index with value of 2:.splice(idx, 1)this index andbreakthe loop
– Calvin Nunes
Nov 19 '18 at 10:34
1
Which one do you want to remove ? The first2, or last, or ...
– Mihai Alexandru-Ionut
Nov 19 '18 at 10:34
1
Possible duplicate of How do I remove a particular element from an array in JavaScript?
– Haroldo_OK
Nov 19 '18 at 10:38
add a comment |
2
2
2
I'm trying to only remove one of the 2s from an array, but my code removes all of them. My code is as follows:
var arr = [2,7,9,5,2]
arr.filter(item => ((item !== 2)));
and:
var arr = [2,7,9,2,2,5,2]
arr.filter(item => ((item !== 2)));
Both remove all the 2s. I thought about removing duplicates, where it works if there's only one duplicate - e.g.:
Array.from(new Set([2,7,9,5,2]));
function uniq(a) {
return Array.from(new Set(a))
}
But fails if there's multiple duplicates as it just removes them all, including any other duplicated numbers:
Array.from(new Set([2,7,9,9,2,2,5,2]));
function uniq(a) {
return Array.from(new Set(a))
}
Does anyone know how to only remove one of the 2s? Thanks for any help here.
javascript
asked Nov 19 '18 at 10:32
user8758206user8758206
456110
I'm trying to only remove one of the 2s from an array, but my code removes all of them. My code is as follows:
var arr = [2,7,9,5,2]
arr.filter(item => ((item !== 2)));
and:
var arr = [2,7,9,2,2,5,2]
arr.filter(item => ((item !== 2)));
Both remove all the 2s. I thought about removing duplicates, where it works if there's only one duplicate - e.g.:
Array.from(new Set([2,7,9,5,2]));
function uniq(a) {
return Array.from(new Set(a))
}
But fails if there's multiple duplicates as it just removes them all, including any other duplicated numbers:
Array.from(new Set([2,7,9,9,2,2,5,2]));
function uniq(a) {
return Array.from(new Set(a))
}
Does anyone know how to only remove one of the 2s? Thanks for any help here.
javascript
javascript
asked Nov 19 '18 at 10:32
user8758206user8758206
456110
asked Nov 19 '18 at 10:32
user8758206user8758206
456110
asked Nov 19 '18 at 10:32
user8758206user8758206
456110
asked Nov 19 '18 at 10:32
user8758206user8758206
456110
asked Nov 19 '18 at 10:32
user8758206user8758206
456110
456110
1
create a simple loop, when found an index with value of 2:.splice(idx, 1)this index andbreakthe loop
– Calvin Nunes
Nov 19 '18 at 10:34
1
Which one do you want to remove ? The first2, or last, or ...
– Mihai Alexandru-Ionut
Nov 19 '18 at 10:34
1
Possible duplicate of How do I remove a particular element from an array in JavaScript?
– Haroldo_OK
Nov 19 '18 at 10:38
add a comment |
1
create a simple loop, when found an index with value of 2:.splice(idx, 1)this index andbreakthe loop
– Calvin Nunes
Nov 19 '18 at 10:34
1
Which one do you want to remove ? The first2, or last, or ...
– Mihai Alexandru-Ionut
Nov 19 '18 at 10:34
1
Possible duplicate of How do I remove a particular element from an array in JavaScript?
– Haroldo_OK
Nov 19 '18 at 10:38
1
1
create a simple loop, when found an index with value of 2:
.splice(idx, 1) this index and break the loop– Calvin Nunes
Nov 19 '18 at 10:34
create a simple loop, when found an index with value of 2:
.splice(idx, 1) this index and break the loop– Calvin Nunes
Nov 19 '18 at 10:34
1
1
Which one do you want to remove ? The first
2, or last, or ...– Mihai Alexandru-Ionut
Nov 19 '18 at 10:34
Which one do you want to remove ? The first
2, or last, or ...– Mihai Alexandru-Ionut
Nov 19 '18 at 10:34
1
1
Possible duplicate of How do I remove a particular element from an array in JavaScript?
– Haroldo_OK
Nov 19 '18 at 10:38
Possible duplicate of How do I remove a particular element from an array in JavaScript?
– Haroldo_OK
Nov 19 '18 at 10:38
add a comment |
6 Answers
6
active
oldest
votes
3
You could use indexOf method in combination with splice.
var arr = [2,7,9,5,2]
var idx = arr.indexOf(2)
if (idx >= 0) {
arr.splice(idx, 1);
}
console.log(arr);edited Nov 19 '18 at 10:49
Haroldo_OK
3,32521847
answered Nov 19 '18 at 10:39
Mihai Alexandru-IonutMihai Alexandru-Ionut
30.3k63971
Close, but it won't work correctly if the number isn't present on the array; checking ifindexOfis non-negative would make it more robust.
– Haroldo_OK
Nov 19 '18 at 10:47
1
brilliant, thank you
– user8758206
Nov 19 '18 at 12:05
add a comment |
2
You could take a closure with a counter and remove only the first 2.
var array = [2, 7, 9, 2, 3, 2, 5, 2],
result = array.filter((i => v => v !== 2 || --i)(1));
console.log(result);For any other 2, you could adjust the start value for decrementing.
var array = [2, 7, 9, 2, 3, 2, 5, 2],
result = array.filter((i => v => v !== 2 || --i)(2));
console.log(result);answered Nov 19 '18 at 10:42
Nina ScholzNina Scholz
182k1495164
interesting answer. What does the v and (1) and (2) represent? I'm new to JS and haven't seen this type of syntax before - can you link to anything I can read to learn this?
– user8758206
Nov 19 '18 at 16:44
1
one is to remove the first occurence of2and two for the second, and so on. it works, because the number gets decremented and if zero, the value is filterd out. the technique's name is closure with an IIFE (immediately-invoked function expression).
– Nina Scholz
Nov 19 '18 at 17:00
you're obviously a very advanced coder - thank you
– user8758206
Nov 19 '18 at 20:41
add a comment |
1
you can follow the following method
var arr= [2,3,4,2,4,5];
var unique = ;
$.each(arr, function(i, el){
if($.inArray(el, unique) === -1) unique.push(el);
})
answered Nov 19 '18 at 10:42
Tijo JohnTijo John
372516
add a comment |
1
There are various ways to do that; one relatively simple way would be to use indexOf; see this other post: https://stackoverflow.com/a/5767357/679240
var array = [2, 7, 9, 5, 2];
console.log(array)
var index = array.indexOf(2);
if (index > -1) {
array.splice(index, 1);
}
// array = [7, 9, 5, 2]
console.log(array);answered Nov 19 '18 at 10:38
Haroldo_OKHaroldo_OK
3,32521847
add a comment |
1
You can do:
const arr = [2, 7, 9, 2, 2, 5, 2];
const result = arr
.reduce((a, c) => {
a.temp[c] = ++a.temp[c] || 1;
if (a.temp[c] !== 2) {
a.array.push(c);
}
return a;
}, {temp: {}, array: })
.array;
console.log(result);answered Nov 19 '18 at 10:38
Yosvel QuinteroYosvel Quintero
11.1k42429
add a comment |
1
Most simple way to filter all duplicates from array:
arr.filter((item, position) => arr.indexOf(item) === position)
This method skip element if another element with the same value already exist.
If you need to filter only first duplicate, you can use additional bool key:
arr.filter((item, position) => {
if (!already && arr.indexOf(item) !== position) {
already = true
return false
} else return true
})
But this method have overheaded. Smartest way is use for loop:
for (let i = 0; i < arr.length; i++) {
if (arr.indexOf(arr[i]) !== i) {
arr.splice(i,1);
break;
}
}
answered Nov 19 '18 at 11:21
XeelleyXeelley
314
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
You could use indexOf method in combination with splice.
var arr = [2,7,9,5,2]
var idx = arr.indexOf(2)
if (idx >= 0) {
arr.splice(idx, 1);
}
console.log(arr);edited Nov 19 '18 at 10:49
Haroldo_OK
3,32521847
answered Nov 19 '18 at 10:39
Mihai Alexandru-IonutMihai Alexandru-Ionut
30.3k63971
Close, but it won't work correctly if the number isn't present on the array; checking ifindexOfis non-negative would make it more robust.
– Haroldo_OK
Nov 19 '18 at 10:47
1
brilliant, thank you
– user8758206
Nov 19 '18 at 12:05
add a comment |
3
You could use indexOf method in combination with splice.
var arr = [2,7,9,5,2]
var idx = arr.indexOf(2)
if (idx >= 0) {
arr.splice(idx, 1);
}
console.log(arr);edited Nov 19 '18 at 10:49
Haroldo_OK
3,32521847
answered Nov 19 '18 at 10:39
Mihai Alexandru-IonutMihai Alexandru-Ionut
30.3k63971
Close, but it won't work correctly if the number isn't present on the array; checking ifindexOfis non-negative would make it more robust.
– Haroldo_OK
Nov 19 '18 at 10:47
1
brilliant, thank you
– user8758206
Nov 19 '18 at 12:05
add a comment |
3
3
3
You could use indexOf method in combination with splice.
var arr = [2,7,9,5,2]
var idx = arr.indexOf(2)
if (idx >= 0) {
arr.splice(idx, 1);
}
console.log(arr);edited Nov 19 '18 at 10:49
Haroldo_OK
3,32521847
answered Nov 19 '18 at 10:39
Mihai Alexandru-IonutMihai Alexandru-Ionut
30.3k63971
You could use indexOf method in combination with splice.
var arr = [2,7,9,5,2]
var idx = arr.indexOf(2)
if (idx >= 0) {
arr.splice(idx, 1);
}
console.log(arr);var arr = [2,7,9,5,2]
var idx = arr.indexOf(2)
if (idx >= 0) {
arr.splice(idx, 1);
}
console.log(arr);var arr = [2,7,9,5,2]
var idx = arr.indexOf(2)
if (idx >= 0) {
arr.splice(idx, 1);
}
console.log(arr);edited Nov 19 '18 at 10:49
Haroldo_OK
3,32521847
answered Nov 19 '18 at 10:39
Mihai Alexandru-IonutMihai Alexandru-Ionut
30.3k63971
edited Nov 19 '18 at 10:49
Haroldo_OK
3,32521847
edited Nov 19 '18 at 10:49
Haroldo_OK
3,32521847
edited Nov 19 '18 at 10:49
Haroldo_OK
3,32521847
3,32521847
answered Nov 19 '18 at 10:39
Mihai Alexandru-IonutMihai Alexandru-Ionut
30.3k63971
answered Nov 19 '18 at 10:39
Mihai Alexandru-IonutMihai Alexandru-Ionut
30.3k63971
answered Nov 19 '18 at 10:39
Mihai Alexandru-IonutMihai Alexandru-Ionut
30.3k63971
30.3k63971
Close, but it won't work correctly if the number isn't present on the array; checking ifindexOfis non-negative would make it more robust.
– Haroldo_OK
Nov 19 '18 at 10:47
1
brilliant, thank you
– user8758206
Nov 19 '18 at 12:05
add a comment |
Close, but it won't work correctly if the number isn't present on the array; checking ifindexOfis non-negative would make it more robust.
– Haroldo_OK
Nov 19 '18 at 10:47
1
brilliant, thank you
– user8758206
Nov 19 '18 at 12:05
Close, but it won't work correctly if the number isn't present on the array; checking if
indexOf is non-negative would make it more robust.– Haroldo_OK
Nov 19 '18 at 10:47
Close, but it won't work correctly if the number isn't present on the array; checking if
indexOf is non-negative would make it more robust.– Haroldo_OK
Nov 19 '18 at 10:47
1
1
brilliant, thank you
– user8758206
Nov 19 '18 at 12:05
brilliant, thank you
– user8758206
Nov 19 '18 at 12:05
add a comment |
2
You could take a closure with a counter and remove only the first 2.
var array = [2, 7, 9, 2, 3, 2, 5, 2],
result = array.filter((i => v => v !== 2 || --i)(1));
console.log(result);For any other 2, you could adjust the start value for decrementing.
var array = [2, 7, 9, 2, 3, 2, 5, 2],
result = array.filter((i => v => v !== 2 || --i)(2));
console.log(result);answered Nov 19 '18 at 10:42
Nina ScholzNina Scholz
182k1495164
interesting answer. What does the v and (1) and (2) represent? I'm new to JS and haven't seen this type of syntax before - can you link to anything I can read to learn this?
– user8758206
Nov 19 '18 at 16:44
1
one is to remove the first occurence of2and two for the second, and so on. it works, because the number gets decremented and if zero, the value is filterd out. the technique's name is closure with an IIFE (immediately-invoked function expression).
– Nina Scholz
Nov 19 '18 at 17:00
you're obviously a very advanced coder - thank you
– user8758206
Nov 19 '18 at 20:41
add a comment |
2
You could take a closure with a counter and remove only the first 2.
var array = [2, 7, 9, 2, 3, 2, 5, 2],
result = array.filter((i => v => v !== 2 || --i)(1));
console.log(result);For any other 2, you could adjust the start value for decrementing.
var array = [2, 7, 9, 2, 3, 2, 5, 2],
result = array.filter((i => v => v !== 2 || --i)(2));
console.log(result);answered Nov 19 '18 at 10:42
Nina ScholzNina Scholz
182k1495164
interesting answer. What does the v and (1) and (2) represent? I'm new to JS and haven't seen this type of syntax before - can you link to anything I can read to learn this?
– user8758206
Nov 19 '18 at 16:44
1
one is to remove the first occurence of2and two for the second, and so on. it works, because the number gets decremented and if zero, the value is filterd out. the technique's name is closure with an IIFE (immediately-invoked function expression).
– Nina Scholz
Nov 19 '18 at 17:00
you're obviously a very advanced coder - thank you
– user8758206
Nov 19 '18 at 20:41
add a comment |
2
2
2
You could take a closure with a counter and remove only the first 2.
var array = [2, 7, 9, 2, 3, 2, 5, 2],
result = array.filter((i => v => v !== 2 || --i)(1));
console.log(result);For any other 2, you could adjust the start value for decrementing.
var array = [2, 7, 9, 2, 3, 2, 5, 2],
result = array.filter((i => v => v !== 2 || --i)(2));
console.log(result);answered Nov 19 '18 at 10:42
Nina ScholzNina Scholz
182k1495164
You could take a closure with a counter and remove only the first 2.
var array = [2, 7, 9, 2, 3, 2, 5, 2],
result = array.filter((i => v => v !== 2 || --i)(1));
console.log(result);For any other 2, you could adjust the start value for decrementing.
var array = [2, 7, 9, 2, 3, 2, 5, 2],
result = array.filter((i => v => v !== 2 || --i)(2));
console.log(result);var array = [2, 7, 9, 2, 3, 2, 5, 2],
result = array.filter((i => v => v !== 2 || --i)(1));
console.log(result);var array = [2, 7, 9, 2, 3, 2, 5, 2],
result = array.filter((i => v => v !== 2 || --i)(1));
console.log(result);var array = [2, 7, 9, 2, 3, 2, 5, 2],
result = array.filter((i => v => v !== 2 || --i)(2));
console.log(result);var array = [2, 7, 9, 2, 3, 2, 5, 2],
result = array.filter((i => v => v !== 2 || --i)(2));
console.log(result);answered Nov 19 '18 at 10:42
Nina ScholzNina Scholz
182k1495164
edited Nov 19 '18 at 10:48
answered Nov 19 '18 at 10:42
Nina ScholzNina Scholz
182k1495164
answered Nov 19 '18 at 10:42
Nina ScholzNina Scholz
182k1495164
answered Nov 19 '18 at 10:42
Nina ScholzNina Scholz
182k1495164
182k1495164
interesting answer. What does the v and (1) and (2) represent? I'm new to JS and haven't seen this type of syntax before - can you link to anything I can read to learn this?
– user8758206
Nov 19 '18 at 16:44
1
one is to remove the first occurence of2and two for the second, and so on. it works, because the number gets decremented and if zero, the value is filterd out. the technique's name is closure with an IIFE (immediately-invoked function expression).
– Nina Scholz
Nov 19 '18 at 17:00
you're obviously a very advanced coder - thank you
– user8758206
Nov 19 '18 at 20:41
add a comment |
interesting answer. What does the v and (1) and (2) represent? I'm new to JS and haven't seen this type of syntax before - can you link to anything I can read to learn this?
– user8758206
Nov 19 '18 at 16:44
1
one is to remove the first occurence of2and two for the second, and so on. it works, because the number gets decremented and if zero, the value is filterd out. the technique's name is closure with an IIFE (immediately-invoked function expression).
– Nina Scholz
Nov 19 '18 at 17:00
you're obviously a very advanced coder - thank you
– user8758206
Nov 19 '18 at 20:41
interesting answer. What does the v and (1) and (2) represent? I'm new to JS and haven't seen this type of syntax before - can you link to anything I can read to learn this?
– user8758206
Nov 19 '18 at 16:44
interesting answer. What does the v and (1) and (2) represent? I'm new to JS and haven't seen this type of syntax before - can you link to anything I can read to learn this?
– user8758206
Nov 19 '18 at 16:44
1
1
one is to remove the first occurence of
2 and two for the second, and so on. it works, because the number gets decremented and if zero, the value is filterd out. the technique's name is closure with an IIFE (immediately-invoked function expression).– Nina Scholz
Nov 19 '18 at 17:00
one is to remove the first occurence of
2 and two for the second, and so on. it works, because the number gets decremented and if zero, the value is filterd out. the technique's name is closure with an IIFE (immediately-invoked function expression).– Nina Scholz
Nov 19 '18 at 17:00
you're obviously a very advanced coder - thank you
– user8758206
Nov 19 '18 at 20:41
you're obviously a very advanced coder - thank you
– user8758206
Nov 19 '18 at 20:41
add a comment |
1
you can follow the following method
var arr= [2,3,4,2,4,5];
var unique = ;
$.each(arr, function(i, el){
if($.inArray(el, unique) === -1) unique.push(el);
})
answered Nov 19 '18 at 10:42
Tijo JohnTijo John
372516
add a comment |
1
you can follow the following method
var arr= [2,3,4,2,4,5];
var unique = ;
$.each(arr, function(i, el){
if($.inArray(el, unique) === -1) unique.push(el);
})
answered Nov 19 '18 at 10:42
Tijo JohnTijo John
372516
add a comment |
1
1
1
you can follow the following method
var arr= [2,3,4,2,4,5];
var unique = ;
$.each(arr, function(i, el){
if($.inArray(el, unique) === -1) unique.push(el);
})
answered Nov 19 '18 at 10:42
Tijo JohnTijo John
372516
you can follow the following method
var arr= [2,3,4,2,4,5];
var unique = ;
$.each(arr, function(i, el){
if($.inArray(el, unique) === -1) unique.push(el);
})
answered Nov 19 '18 at 10:42
Tijo JohnTijo John
372516
answered Nov 19 '18 at 10:42
Tijo JohnTijo John
372516
answered Nov 19 '18 at 10:42
Tijo JohnTijo John
372516
answered Nov 19 '18 at 10:42
Tijo JohnTijo John
372516
372516
add a comment |
add a comment |
1
There are various ways to do that; one relatively simple way would be to use indexOf; see this other post: https://stackoverflow.com/a/5767357/679240
var array = [2, 7, 9, 5, 2];
console.log(array)
var index = array.indexOf(2);
if (index > -1) {
array.splice(index, 1);
}
// array = [7, 9, 5, 2]
console.log(array);answered Nov 19 '18 at 10:38
Haroldo_OKHaroldo_OK
3,32521847
add a comment |
1
There are various ways to do that; one relatively simple way would be to use indexOf; see this other post: https://stackoverflow.com/a/5767357/679240
var array = [2, 7, 9, 5, 2];
console.log(array)
var index = array.indexOf(2);
if (index > -1) {
array.splice(index, 1);
}
// array = [7, 9, 5, 2]
console.log(array);answered Nov 19 '18 at 10:38
Haroldo_OKHaroldo_OK
3,32521847
add a comment |
1
1
1
There are various ways to do that; one relatively simple way would be to use indexOf; see this other post: https://stackoverflow.com/a/5767357/679240
var array = [2, 7, 9, 5, 2];
console.log(array)
var index = array.indexOf(2);
if (index > -1) {
array.splice(index, 1);
}
// array = [7, 9, 5, 2]
console.log(array);answered Nov 19 '18 at 10:38
Haroldo_OKHaroldo_OK
3,32521847
There are various ways to do that; one relatively simple way would be to use indexOf; see this other post: https://stackoverflow.com/a/5767357/679240
var array = [2, 7, 9, 5, 2];
console.log(array)
var index = array.indexOf(2);
if (index > -1) {
array.splice(index, 1);
}
// array = [7, 9, 5, 2]
console.log(array);var array = [2, 7, 9, 5, 2];
console.log(array)
var index = array.indexOf(2);
if (index > -1) {
array.splice(index, 1);
}
// array = [7, 9, 5, 2]
console.log(array);var array = [2, 7, 9, 5, 2];
console.log(array)
var index = array.indexOf(2);
if (index > -1) {
array.splice(index, 1);
}
// array = [7, 9, 5, 2]
console.log(array);answered Nov 19 '18 at 10:38
Haroldo_OKHaroldo_OK
3,32521847
edited Nov 19 '18 at 10:43
answered Nov 19 '18 at 10:38
Haroldo_OKHaroldo_OK
3,32521847
answered Nov 19 '18 at 10:38
Haroldo_OKHaroldo_OK
3,32521847
answered Nov 19 '18 at 10:38
Haroldo_OKHaroldo_OK
3,32521847
3,32521847
add a comment |
add a comment |
1
You can do:
const arr = [2, 7, 9, 2, 2, 5, 2];
const result = arr
.reduce((a, c) => {
a.temp[c] = ++a.temp[c] || 1;
if (a.temp[c] !== 2) {
a.array.push(c);
}
return a;
}, {temp: {}, array: })
.array;
console.log(result);answered Nov 19 '18 at 10:38
Yosvel QuinteroYosvel Quintero
11.1k42429
add a comment |
1
You can do:
const arr = [2, 7, 9, 2, 2, 5, 2];
const result = arr
.reduce((a, c) => {
a.temp[c] = ++a.temp[c] || 1;
if (a.temp[c] !== 2) {
a.array.push(c);
}
return a;
}, {temp: {}, array: })
.array;
console.log(result);answered Nov 19 '18 at 10:38
Yosvel QuinteroYosvel Quintero
11.1k42429
add a comment |
1
1
1
You can do:
const arr = [2, 7, 9, 2, 2, 5, 2];
const result = arr
.reduce((a, c) => {
a.temp[c] = ++a.temp[c] || 1;
if (a.temp[c] !== 2) {
a.array.push(c);
}
return a;
}, {temp: {}, array: })
.array;
console.log(result);answered Nov 19 '18 at 10:38
Yosvel QuinteroYosvel Quintero
11.1k42429
You can do:
const arr = [2, 7, 9, 2, 2, 5, 2];
const result = arr
.reduce((a, c) => {
a.temp[c] = ++a.temp[c] || 1;
if (a.temp[c] !== 2) {
a.array.push(c);
}
return a;
}, {temp: {}, array: })
.array;
console.log(result);const arr = [2, 7, 9, 2, 2, 5, 2];
const result = arr
.reduce((a, c) => {
a.temp[c] = ++a.temp[c] || 1;
if (a.temp[c] !== 2) {
a.array.push(c);
}
return a;
}, {temp: {}, array: })
.array;
console.log(result);const arr = [2, 7, 9, 2, 2, 5, 2];
const result = arr
.reduce((a, c) => {
a.temp[c] = ++a.temp[c] || 1;
if (a.temp[c] !== 2) {
a.array.push(c);
}
return a;
}, {temp: {}, array: })
.array;
console.log(result);answered Nov 19 '18 at 10:38
Yosvel QuinteroYosvel Quintero
11.1k42429
edited Nov 19 '18 at 10:46
answered Nov 19 '18 at 10:38
Yosvel QuinteroYosvel Quintero
11.1k42429
answered Nov 19 '18 at 10:38
Yosvel QuinteroYosvel Quintero
11.1k42429
answered Nov 19 '18 at 10:38
Yosvel QuinteroYosvel Quintero
11.1k42429
11.1k42429
add a comment |
add a comment |
1
Most simple way to filter all duplicates from array:
arr.filter((item, position) => arr.indexOf(item) === position)
This method skip element if another element with the same value already exist.
If you need to filter only first duplicate, you can use additional bool key:
arr.filter((item, position) => {
if (!already && arr.indexOf(item) !== position) {
already = true
return false
} else return true
})
But this method have overheaded. Smartest way is use for loop:
for (let i = 0; i < arr.length; i++) {
if (arr.indexOf(arr[i]) !== i) {
arr.splice(i,1);
break;
}
}
answered Nov 19 '18 at 11:21
XeelleyXeelley
314
add a comment |
1
Most simple way to filter all duplicates from array:
arr.filter((item, position) => arr.indexOf(item) === position)
This method skip element if another element with the same value already exist.
If you need to filter only first duplicate, you can use additional bool key:
arr.filter((item, position) => {
if (!already && arr.indexOf(item) !== position) {
already = true
return false
} else return true
})
But this method have overheaded. Smartest way is use for loop:
for (let i = 0; i < arr.length; i++) {
if (arr.indexOf(arr[i]) !== i) {
arr.splice(i,1);
break;
}
}
answered Nov 19 '18 at 11:21
XeelleyXeelley
314
add a comment |
1
1
1
Most simple way to filter all duplicates from array:
arr.filter((item, position) => arr.indexOf(item) === position)
This method skip element if another element with the same value already exist.
If you need to filter only first duplicate, you can use additional bool key:
arr.filter((item, position) => {
if (!already && arr.indexOf(item) !== position) {
already = true
return false
} else return true
})
But this method have overheaded. Smartest way is use for loop:
for (let i = 0; i < arr.length; i++) {
if (arr.indexOf(arr[i]) !== i) {
arr.splice(i,1);
break;
}
}
answered Nov 19 '18 at 11:21
XeelleyXeelley
314
Most simple way to filter all duplicates from array:
arr.filter((item, position) => arr.indexOf(item) === position)
This method skip element if another element with the same value already exist.
If you need to filter only first duplicate, you can use additional bool key:
arr.filter((item, position) => {
if (!already && arr.indexOf(item) !== position) {
already = true
return false
} else return true
})
But this method have overheaded. Smartest way is use for loop:
for (let i = 0; i < arr.length; i++) {
if (arr.indexOf(arr[i]) !== i) {
arr.splice(i,1);
break;
}
}
answered Nov 19 '18 at 11:21
XeelleyXeelley
314
answered Nov 19 '18 at 11:21
XeelleyXeelley
314
answered Nov 19 '18 at 11:21
XeelleyXeelley
314
answered Nov 19 '18 at 11:21
XeelleyXeelley
314
314
add a comment |
add a comment |
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1
create a simple loop, when found an index with value of 2:
.splice(idx, 1)this index andbreakthe loop– Calvin Nunes
Nov 19 '18 at 10:34
1
Which one do you want to remove ? The first
2, or last, or ...– Mihai Alexandru-Ionut
Nov 19 '18 at 10:34
1
Possible duplicate of How do I remove a particular element from an array in JavaScript?
– Haroldo_OK
Nov 19 '18 at 10:38