how to assign object in object
how to assign the object in object and filter the value which pass and fail;
the input is:
[
{
name: 'John',
score: 90,
time: 'evening'
},
{
name: 'Doni',
score: 68,
time: 'morning'
},
{
name: 'Jiu',
score: 50,
time: 'evening'
},
{
name: 'Shin',
score: 92,
time: 'morning'
},
];
and i want the output like this :
{
"evening": {
"pass": [
{
"name": "John",
"score": 90
}
],
"fail": [
{
"name": "jiu",
"score": 50
}
]
},
"morning": {
"pass": [
{
"name": "Shin",
"score": 92
}
],
"fail": [
{
"name": "Doni",
"score": 68
}
]
}
}
do we need to use Object.assign for this ? and how many loop we use for this ??
i do love to know how to add another string in the object beside that ouput,
thanks
javascript arrays object iterator
add a comment |
how to assign the object in object and filter the value which pass and fail;
the input is:
[
{
name: 'John',
score: 90,
time: 'evening'
},
{
name: 'Doni',
score: 68,
time: 'morning'
},
{
name: 'Jiu',
score: 50,
time: 'evening'
},
{
name: 'Shin',
score: 92,
time: 'morning'
},
];
and i want the output like this :
{
"evening": {
"pass": [
{
"name": "John",
"score": 90
}
],
"fail": [
{
"name": "jiu",
"score": 50
}
]
},
"morning": {
"pass": [
{
"name": "Shin",
"score": 92
}
],
"fail": [
{
"name": "Doni",
"score": 68
}
]
}
}
do we need to use Object.assign for this ? and how many loop we use for this ??
i do love to know how to add another string in the object beside that ouput,
thanks
javascript arrays object iterator
do we need to use Object.assign for this? Not necessarily. and how many loop we use for this? You do not want to know how many loops you need, let the code handle that for you.
– Abana Clara
Nov 20 '18 at 0:25
yepp, do u know how ?
– Zr Classic
Nov 20 '18 at 0:26
add a comment |
how to assign the object in object and filter the value which pass and fail;
the input is:
[
{
name: 'John',
score: 90,
time: 'evening'
},
{
name: 'Doni',
score: 68,
time: 'morning'
},
{
name: 'Jiu',
score: 50,
time: 'evening'
},
{
name: 'Shin',
score: 92,
time: 'morning'
},
];
and i want the output like this :
{
"evening": {
"pass": [
{
"name": "John",
"score": 90
}
],
"fail": [
{
"name": "jiu",
"score": 50
}
]
},
"morning": {
"pass": [
{
"name": "Shin",
"score": 92
}
],
"fail": [
{
"name": "Doni",
"score": 68
}
]
}
}
do we need to use Object.assign for this ? and how many loop we use for this ??
i do love to know how to add another string in the object beside that ouput,
thanks
javascript arrays object iterator
how to assign the object in object and filter the value which pass and fail;
the input is:
[
{
name: 'John',
score: 90,
time: 'evening'
},
{
name: 'Doni',
score: 68,
time: 'morning'
},
{
name: 'Jiu',
score: 50,
time: 'evening'
},
{
name: 'Shin',
score: 92,
time: 'morning'
},
];
and i want the output like this :
{
"evening": {
"pass": [
{
"name": "John",
"score": 90
}
],
"fail": [
{
"name": "jiu",
"score": 50
}
]
},
"morning": {
"pass": [
{
"name": "Shin",
"score": 92
}
],
"fail": [
{
"name": "Doni",
"score": 68
}
]
}
}
do we need to use Object.assign for this ? and how many loop we use for this ??
i do love to know how to add another string in the object beside that ouput,
thanks
javascript arrays object iterator
javascript arrays object iterator
asked Nov 20 '18 at 0:20
Zr ClassicZr Classic
756
756
do we need to use Object.assign for this? Not necessarily. and how many loop we use for this? You do not want to know how many loops you need, let the code handle that for you.
– Abana Clara
Nov 20 '18 at 0:25
yepp, do u know how ?
– Zr Classic
Nov 20 '18 at 0:26
add a comment |
do we need to use Object.assign for this? Not necessarily. and how many loop we use for this? You do not want to know how many loops you need, let the code handle that for you.
– Abana Clara
Nov 20 '18 at 0:25
yepp, do u know how ?
– Zr Classic
Nov 20 '18 at 0:26
do we need to use Object.assign for this? Not necessarily. and how many loop we use for this? You do not want to know how many loops you need, let the code handle that for you.
– Abana Clara
Nov 20 '18 at 0:25
do we need to use Object.assign for this? Not necessarily. and how many loop we use for this? You do not want to know how many loops you need, let the code handle that for you.
– Abana Clara
Nov 20 '18 at 0:25
yepp, do u know how ?
– Zr Classic
Nov 20 '18 at 0:26
yepp, do u know how ?
– Zr Classic
Nov 20 '18 at 0:26
add a comment |
3 Answers
3
active
oldest
votes
You can do something like this:
const data = [{ name: 'John', score: 90, time: 'evening' }, { name: 'Doni', score: 68, time: 'morning' }, { name: 'Jiu', score: 50, time: 'evening' }, { name: 'Shin', score: 92, time: 'morning' }, ];
const grp = (d, p) => d.reduce((r,c) => (r[c[p]] = [...r[c[p]] || , c], r), {})
const grpV = (d, rng) => d.reduce((r,{name, score}) => {
let key = score > rng ? 'pass' : 'fail'
r[key] = [...r[key] || , {name, score}]
return r
}, {})
const r = Object.entries(grp(data, 'time')).map(([k,v]) => ({[k]: grpV(v, 75)}))
console.log(r)
The idea is the group 2 times one on the time
and 2nd on the score
.
grp
: function to group by a property (in this case 'time') which returns an object with 2 properties: evening
and morning
each of which is an array containing the classes.
grpV
: function to group by value (in this case 75
) which returns an object with 2 properties: pass
and fail
each of which is an array containing the classes.
On the end once we have those tools
we are saying ... give me the entries
of the grouped by time
object and for each of the groups ... group by score.
Here how something like this could look like if we ware using lodash
:
const data = [{ name: 'John', score: 90, time: 'evening' }, { name: 'Doni', score: 68, time: 'morning' }, { name: 'Jiu', score: 50, time: 'evening' }, { name: 'Shin', score: 92, time: 'morning' }, ];
const partition = (x, p) => _(x)
.partition(y => y.score > p)
.map((x,i) => ({ [i==0 ? 'pass': 'fail']: _.omit(x[0], 'time')}))
.value()
const r = _(data)
.groupBy('time')
.mapValues(x => partition(x, 75))
.value()
console.log(r)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.min.js"></script>
Adding it as an example since it does help with readability of what the ES6 example is doing to some extend.
hmmm it seems difficult for newbie to understand those syntax
– Zr Classic
Nov 20 '18 at 10:17
add a comment |
There's a lot of ways to do this. The simplest is probably to make a base object that represent your empty results. Then loop over the students and fill the arrays:
let students = [{name: 'John',score: 90,time: 'evening'},{name: 'Doni',score: 68,time: 'morning'},{name: 'Jiu',score: 50,time: 'evening'},{name: 'Shin',score: 92,time: 'morning'},];
// Empty case
let base = {
"evening": {"pass": , "fail": },
"morning": {"pass": , "fail": }
}
const PASSING = 70
students.forEach(({name, score, time}) => {
let key = score >= PASSING ? 'pass' : 'fail'
base[time][key].push({name, score})
})
console.log(base)
This makes is easy to have empty arrays, which is probably what you want if there are no students in a particular category.
EDIT based on comment:
To support arbitrary times, you can just create the times on the object as you find them. reduce()
is good for this, but you could also use a regular loop. For example with an added afternoon
time:
let students = [{name: 'Mark',score: 95,time: 'afternoon'}, {name: 'John',score: 90,time: 'evening'},{name: 'Doni',score: 68,time: 'morning'},{name: 'Jiu',score: 50,time: 'evening'},{name: 'Shin',score: 92,time: 'morning'},];
const PASSING = 70
let result = students.reduce((obj, {name, score, time}) => {
if (!obj[time]) obj[time] = {'pass': , 'fail': }
let key = score >= PASSING ? 'pass' : 'fail'
obj[time][key].push({name, score})
return obj
}, {})
console.log(result)
how about if we other times, midnight, noon? should we create it on base variable? can u use that by function
– Zr Classic
Nov 20 '18 at 1:27
@ZrClassic If you are going to have arbitrary time categories, you will be better off building the object as you go. That will be a more general solution, but also more complicated.
– Mark Meyer
Nov 20 '18 at 1:29
add a comment |
I'm sure there are more elegant ways to do this. But this one is probably one of the simplest beginner-friendly ways you can go about this.
I loop through the input array, check the existence of the .time
values as keys on the output
object and create the pass
and fail
keys. Then evaluate the .score
against the passingScore
and push the necessary data to it.
Should be pretty easy to understand once you see and try the code below:
const data = [
{name: 'John',score: 90, time: 'evening'},
{name: 'Doni',score: 68, time: 'morning'},
{name: 'Jiu',score: 50, time: 'evening'},
{name: 'Shin',score: 92, time: 'morning'},
{name: 'Fubar',score: 75, time: 'noon'},
];
function formatData(data){
const passingScore = 75;
const output = {};
data.forEach(function(item){
if(!output[item.time]) output[item.time] = {pass: , fail: };
const stud = { name: item.name, score: item.score };
if(item.score >= passingScore) output[item.time]['pass'].push(stud)
else output[item.time]['fail'].push(stud)
});
return output;
}
console.log(formatData(data));
this is the perfect one for beginner
– Zr Classic
Nov 20 '18 at 4:29
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can do something like this:
const data = [{ name: 'John', score: 90, time: 'evening' }, { name: 'Doni', score: 68, time: 'morning' }, { name: 'Jiu', score: 50, time: 'evening' }, { name: 'Shin', score: 92, time: 'morning' }, ];
const grp = (d, p) => d.reduce((r,c) => (r[c[p]] = [...r[c[p]] || , c], r), {})
const grpV = (d, rng) => d.reduce((r,{name, score}) => {
let key = score > rng ? 'pass' : 'fail'
r[key] = [...r[key] || , {name, score}]
return r
}, {})
const r = Object.entries(grp(data, 'time')).map(([k,v]) => ({[k]: grpV(v, 75)}))
console.log(r)
The idea is the group 2 times one on the time
and 2nd on the score
.
grp
: function to group by a property (in this case 'time') which returns an object with 2 properties: evening
and morning
each of which is an array containing the classes.
grpV
: function to group by value (in this case 75
) which returns an object with 2 properties: pass
and fail
each of which is an array containing the classes.
On the end once we have those tools
we are saying ... give me the entries
of the grouped by time
object and for each of the groups ... group by score.
Here how something like this could look like if we ware using lodash
:
const data = [{ name: 'John', score: 90, time: 'evening' }, { name: 'Doni', score: 68, time: 'morning' }, { name: 'Jiu', score: 50, time: 'evening' }, { name: 'Shin', score: 92, time: 'morning' }, ];
const partition = (x, p) => _(x)
.partition(y => y.score > p)
.map((x,i) => ({ [i==0 ? 'pass': 'fail']: _.omit(x[0], 'time')}))
.value()
const r = _(data)
.groupBy('time')
.mapValues(x => partition(x, 75))
.value()
console.log(r)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.min.js"></script>
Adding it as an example since it does help with readability of what the ES6 example is doing to some extend.
hmmm it seems difficult for newbie to understand those syntax
– Zr Classic
Nov 20 '18 at 10:17
add a comment |
You can do something like this:
const data = [{ name: 'John', score: 90, time: 'evening' }, { name: 'Doni', score: 68, time: 'morning' }, { name: 'Jiu', score: 50, time: 'evening' }, { name: 'Shin', score: 92, time: 'morning' }, ];
const grp = (d, p) => d.reduce((r,c) => (r[c[p]] = [...r[c[p]] || , c], r), {})
const grpV = (d, rng) => d.reduce((r,{name, score}) => {
let key = score > rng ? 'pass' : 'fail'
r[key] = [...r[key] || , {name, score}]
return r
}, {})
const r = Object.entries(grp(data, 'time')).map(([k,v]) => ({[k]: grpV(v, 75)}))
console.log(r)
The idea is the group 2 times one on the time
and 2nd on the score
.
grp
: function to group by a property (in this case 'time') which returns an object with 2 properties: evening
and morning
each of which is an array containing the classes.
grpV
: function to group by value (in this case 75
) which returns an object with 2 properties: pass
and fail
each of which is an array containing the classes.
On the end once we have those tools
we are saying ... give me the entries
of the grouped by time
object and for each of the groups ... group by score.
Here how something like this could look like if we ware using lodash
:
const data = [{ name: 'John', score: 90, time: 'evening' }, { name: 'Doni', score: 68, time: 'morning' }, { name: 'Jiu', score: 50, time: 'evening' }, { name: 'Shin', score: 92, time: 'morning' }, ];
const partition = (x, p) => _(x)
.partition(y => y.score > p)
.map((x,i) => ({ [i==0 ? 'pass': 'fail']: _.omit(x[0], 'time')}))
.value()
const r = _(data)
.groupBy('time')
.mapValues(x => partition(x, 75))
.value()
console.log(r)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.min.js"></script>
Adding it as an example since it does help with readability of what the ES6 example is doing to some extend.
hmmm it seems difficult for newbie to understand those syntax
– Zr Classic
Nov 20 '18 at 10:17
add a comment |
You can do something like this:
const data = [{ name: 'John', score: 90, time: 'evening' }, { name: 'Doni', score: 68, time: 'morning' }, { name: 'Jiu', score: 50, time: 'evening' }, { name: 'Shin', score: 92, time: 'morning' }, ];
const grp = (d, p) => d.reduce((r,c) => (r[c[p]] = [...r[c[p]] || , c], r), {})
const grpV = (d, rng) => d.reduce((r,{name, score}) => {
let key = score > rng ? 'pass' : 'fail'
r[key] = [...r[key] || , {name, score}]
return r
}, {})
const r = Object.entries(grp(data, 'time')).map(([k,v]) => ({[k]: grpV(v, 75)}))
console.log(r)
The idea is the group 2 times one on the time
and 2nd on the score
.
grp
: function to group by a property (in this case 'time') which returns an object with 2 properties: evening
and morning
each of which is an array containing the classes.
grpV
: function to group by value (in this case 75
) which returns an object with 2 properties: pass
and fail
each of which is an array containing the classes.
On the end once we have those tools
we are saying ... give me the entries
of the grouped by time
object and for each of the groups ... group by score.
Here how something like this could look like if we ware using lodash
:
const data = [{ name: 'John', score: 90, time: 'evening' }, { name: 'Doni', score: 68, time: 'morning' }, { name: 'Jiu', score: 50, time: 'evening' }, { name: 'Shin', score: 92, time: 'morning' }, ];
const partition = (x, p) => _(x)
.partition(y => y.score > p)
.map((x,i) => ({ [i==0 ? 'pass': 'fail']: _.omit(x[0], 'time')}))
.value()
const r = _(data)
.groupBy('time')
.mapValues(x => partition(x, 75))
.value()
console.log(r)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.min.js"></script>
Adding it as an example since it does help with readability of what the ES6 example is doing to some extend.
You can do something like this:
const data = [{ name: 'John', score: 90, time: 'evening' }, { name: 'Doni', score: 68, time: 'morning' }, { name: 'Jiu', score: 50, time: 'evening' }, { name: 'Shin', score: 92, time: 'morning' }, ];
const grp = (d, p) => d.reduce((r,c) => (r[c[p]] = [...r[c[p]] || , c], r), {})
const grpV = (d, rng) => d.reduce((r,{name, score}) => {
let key = score > rng ? 'pass' : 'fail'
r[key] = [...r[key] || , {name, score}]
return r
}, {})
const r = Object.entries(grp(data, 'time')).map(([k,v]) => ({[k]: grpV(v, 75)}))
console.log(r)
The idea is the group 2 times one on the time
and 2nd on the score
.
grp
: function to group by a property (in this case 'time') which returns an object with 2 properties: evening
and morning
each of which is an array containing the classes.
grpV
: function to group by value (in this case 75
) which returns an object with 2 properties: pass
and fail
each of which is an array containing the classes.
On the end once we have those tools
we are saying ... give me the entries
of the grouped by time
object and for each of the groups ... group by score.
Here how something like this could look like if we ware using lodash
:
const data = [{ name: 'John', score: 90, time: 'evening' }, { name: 'Doni', score: 68, time: 'morning' }, { name: 'Jiu', score: 50, time: 'evening' }, { name: 'Shin', score: 92, time: 'morning' }, ];
const partition = (x, p) => _(x)
.partition(y => y.score > p)
.map((x,i) => ({ [i==0 ? 'pass': 'fail']: _.omit(x[0], 'time')}))
.value()
const r = _(data)
.groupBy('time')
.mapValues(x => partition(x, 75))
.value()
console.log(r)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.min.js"></script>
Adding it as an example since it does help with readability of what the ES6 example is doing to some extend.
const data = [{ name: 'John', score: 90, time: 'evening' }, { name: 'Doni', score: 68, time: 'morning' }, { name: 'Jiu', score: 50, time: 'evening' }, { name: 'Shin', score: 92, time: 'morning' }, ];
const grp = (d, p) => d.reduce((r,c) => (r[c[p]] = [...r[c[p]] || , c], r), {})
const grpV = (d, rng) => d.reduce((r,{name, score}) => {
let key = score > rng ? 'pass' : 'fail'
r[key] = [...r[key] || , {name, score}]
return r
}, {})
const r = Object.entries(grp(data, 'time')).map(([k,v]) => ({[k]: grpV(v, 75)}))
console.log(r)
const data = [{ name: 'John', score: 90, time: 'evening' }, { name: 'Doni', score: 68, time: 'morning' }, { name: 'Jiu', score: 50, time: 'evening' }, { name: 'Shin', score: 92, time: 'morning' }, ];
const grp = (d, p) => d.reduce((r,c) => (r[c[p]] = [...r[c[p]] || , c], r), {})
const grpV = (d, rng) => d.reduce((r,{name, score}) => {
let key = score > rng ? 'pass' : 'fail'
r[key] = [...r[key] || , {name, score}]
return r
}, {})
const r = Object.entries(grp(data, 'time')).map(([k,v]) => ({[k]: grpV(v, 75)}))
console.log(r)
const data = [{ name: 'John', score: 90, time: 'evening' }, { name: 'Doni', score: 68, time: 'morning' }, { name: 'Jiu', score: 50, time: 'evening' }, { name: 'Shin', score: 92, time: 'morning' }, ];
const partition = (x, p) => _(x)
.partition(y => y.score > p)
.map((x,i) => ({ [i==0 ? 'pass': 'fail']: _.omit(x[0], 'time')}))
.value()
const r = _(data)
.groupBy('time')
.mapValues(x => partition(x, 75))
.value()
console.log(r)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.min.js"></script>
const data = [{ name: 'John', score: 90, time: 'evening' }, { name: 'Doni', score: 68, time: 'morning' }, { name: 'Jiu', score: 50, time: 'evening' }, { name: 'Shin', score: 92, time: 'morning' }, ];
const partition = (x, p) => _(x)
.partition(y => y.score > p)
.map((x,i) => ({ [i==0 ? 'pass': 'fail']: _.omit(x[0], 'time')}))
.value()
const r = _(data)
.groupBy('time')
.mapValues(x => partition(x, 75))
.value()
console.log(r)
<script src="https://cdnjs.cloudflare.com/ajax/libs/lodash.js/4.17.10/lodash.min.js"></script>
edited Nov 20 '18 at 1:43
answered Nov 20 '18 at 0:43
AkrionAkrion
9,45511224
9,45511224
hmmm it seems difficult for newbie to understand those syntax
– Zr Classic
Nov 20 '18 at 10:17
add a comment |
hmmm it seems difficult for newbie to understand those syntax
– Zr Classic
Nov 20 '18 at 10:17
hmmm it seems difficult for newbie to understand those syntax
– Zr Classic
Nov 20 '18 at 10:17
hmmm it seems difficult for newbie to understand those syntax
– Zr Classic
Nov 20 '18 at 10:17
add a comment |
There's a lot of ways to do this. The simplest is probably to make a base object that represent your empty results. Then loop over the students and fill the arrays:
let students = [{name: 'John',score: 90,time: 'evening'},{name: 'Doni',score: 68,time: 'morning'},{name: 'Jiu',score: 50,time: 'evening'},{name: 'Shin',score: 92,time: 'morning'},];
// Empty case
let base = {
"evening": {"pass": , "fail": },
"morning": {"pass": , "fail": }
}
const PASSING = 70
students.forEach(({name, score, time}) => {
let key = score >= PASSING ? 'pass' : 'fail'
base[time][key].push({name, score})
})
console.log(base)
This makes is easy to have empty arrays, which is probably what you want if there are no students in a particular category.
EDIT based on comment:
To support arbitrary times, you can just create the times on the object as you find them. reduce()
is good for this, but you could also use a regular loop. For example with an added afternoon
time:
let students = [{name: 'Mark',score: 95,time: 'afternoon'}, {name: 'John',score: 90,time: 'evening'},{name: 'Doni',score: 68,time: 'morning'},{name: 'Jiu',score: 50,time: 'evening'},{name: 'Shin',score: 92,time: 'morning'},];
const PASSING = 70
let result = students.reduce((obj, {name, score, time}) => {
if (!obj[time]) obj[time] = {'pass': , 'fail': }
let key = score >= PASSING ? 'pass' : 'fail'
obj[time][key].push({name, score})
return obj
}, {})
console.log(result)
how about if we other times, midnight, noon? should we create it on base variable? can u use that by function
– Zr Classic
Nov 20 '18 at 1:27
@ZrClassic If you are going to have arbitrary time categories, you will be better off building the object as you go. That will be a more general solution, but also more complicated.
– Mark Meyer
Nov 20 '18 at 1:29
add a comment |
There's a lot of ways to do this. The simplest is probably to make a base object that represent your empty results. Then loop over the students and fill the arrays:
let students = [{name: 'John',score: 90,time: 'evening'},{name: 'Doni',score: 68,time: 'morning'},{name: 'Jiu',score: 50,time: 'evening'},{name: 'Shin',score: 92,time: 'morning'},];
// Empty case
let base = {
"evening": {"pass": , "fail": },
"morning": {"pass": , "fail": }
}
const PASSING = 70
students.forEach(({name, score, time}) => {
let key = score >= PASSING ? 'pass' : 'fail'
base[time][key].push({name, score})
})
console.log(base)
This makes is easy to have empty arrays, which is probably what you want if there are no students in a particular category.
EDIT based on comment:
To support arbitrary times, you can just create the times on the object as you find them. reduce()
is good for this, but you could also use a regular loop. For example with an added afternoon
time:
let students = [{name: 'Mark',score: 95,time: 'afternoon'}, {name: 'John',score: 90,time: 'evening'},{name: 'Doni',score: 68,time: 'morning'},{name: 'Jiu',score: 50,time: 'evening'},{name: 'Shin',score: 92,time: 'morning'},];
const PASSING = 70
let result = students.reduce((obj, {name, score, time}) => {
if (!obj[time]) obj[time] = {'pass': , 'fail': }
let key = score >= PASSING ? 'pass' : 'fail'
obj[time][key].push({name, score})
return obj
}, {})
console.log(result)
how about if we other times, midnight, noon? should we create it on base variable? can u use that by function
– Zr Classic
Nov 20 '18 at 1:27
@ZrClassic If you are going to have arbitrary time categories, you will be better off building the object as you go. That will be a more general solution, but also more complicated.
– Mark Meyer
Nov 20 '18 at 1:29
add a comment |
There's a lot of ways to do this. The simplest is probably to make a base object that represent your empty results. Then loop over the students and fill the arrays:
let students = [{name: 'John',score: 90,time: 'evening'},{name: 'Doni',score: 68,time: 'morning'},{name: 'Jiu',score: 50,time: 'evening'},{name: 'Shin',score: 92,time: 'morning'},];
// Empty case
let base = {
"evening": {"pass": , "fail": },
"morning": {"pass": , "fail": }
}
const PASSING = 70
students.forEach(({name, score, time}) => {
let key = score >= PASSING ? 'pass' : 'fail'
base[time][key].push({name, score})
})
console.log(base)
This makes is easy to have empty arrays, which is probably what you want if there are no students in a particular category.
EDIT based on comment:
To support arbitrary times, you can just create the times on the object as you find them. reduce()
is good for this, but you could also use a regular loop. For example with an added afternoon
time:
let students = [{name: 'Mark',score: 95,time: 'afternoon'}, {name: 'John',score: 90,time: 'evening'},{name: 'Doni',score: 68,time: 'morning'},{name: 'Jiu',score: 50,time: 'evening'},{name: 'Shin',score: 92,time: 'morning'},];
const PASSING = 70
let result = students.reduce((obj, {name, score, time}) => {
if (!obj[time]) obj[time] = {'pass': , 'fail': }
let key = score >= PASSING ? 'pass' : 'fail'
obj[time][key].push({name, score})
return obj
}, {})
console.log(result)
There's a lot of ways to do this. The simplest is probably to make a base object that represent your empty results. Then loop over the students and fill the arrays:
let students = [{name: 'John',score: 90,time: 'evening'},{name: 'Doni',score: 68,time: 'morning'},{name: 'Jiu',score: 50,time: 'evening'},{name: 'Shin',score: 92,time: 'morning'},];
// Empty case
let base = {
"evening": {"pass": , "fail": },
"morning": {"pass": , "fail": }
}
const PASSING = 70
students.forEach(({name, score, time}) => {
let key = score >= PASSING ? 'pass' : 'fail'
base[time][key].push({name, score})
})
console.log(base)
This makes is easy to have empty arrays, which is probably what you want if there are no students in a particular category.
EDIT based on comment:
To support arbitrary times, you can just create the times on the object as you find them. reduce()
is good for this, but you could also use a regular loop. For example with an added afternoon
time:
let students = [{name: 'Mark',score: 95,time: 'afternoon'}, {name: 'John',score: 90,time: 'evening'},{name: 'Doni',score: 68,time: 'morning'},{name: 'Jiu',score: 50,time: 'evening'},{name: 'Shin',score: 92,time: 'morning'},];
const PASSING = 70
let result = students.reduce((obj, {name, score, time}) => {
if (!obj[time]) obj[time] = {'pass': , 'fail': }
let key = score >= PASSING ? 'pass' : 'fail'
obj[time][key].push({name, score})
return obj
}, {})
console.log(result)
let students = [{name: 'John',score: 90,time: 'evening'},{name: 'Doni',score: 68,time: 'morning'},{name: 'Jiu',score: 50,time: 'evening'},{name: 'Shin',score: 92,time: 'morning'},];
// Empty case
let base = {
"evening": {"pass": , "fail": },
"morning": {"pass": , "fail": }
}
const PASSING = 70
students.forEach(({name, score, time}) => {
let key = score >= PASSING ? 'pass' : 'fail'
base[time][key].push({name, score})
})
console.log(base)
let students = [{name: 'John',score: 90,time: 'evening'},{name: 'Doni',score: 68,time: 'morning'},{name: 'Jiu',score: 50,time: 'evening'},{name: 'Shin',score: 92,time: 'morning'},];
// Empty case
let base = {
"evening": {"pass": , "fail": },
"morning": {"pass": , "fail": }
}
const PASSING = 70
students.forEach(({name, score, time}) => {
let key = score >= PASSING ? 'pass' : 'fail'
base[time][key].push({name, score})
})
console.log(base)
let students = [{name: 'Mark',score: 95,time: 'afternoon'}, {name: 'John',score: 90,time: 'evening'},{name: 'Doni',score: 68,time: 'morning'},{name: 'Jiu',score: 50,time: 'evening'},{name: 'Shin',score: 92,time: 'morning'},];
const PASSING = 70
let result = students.reduce((obj, {name, score, time}) => {
if (!obj[time]) obj[time] = {'pass': , 'fail': }
let key = score >= PASSING ? 'pass' : 'fail'
obj[time][key].push({name, score})
return obj
}, {})
console.log(result)
let students = [{name: 'Mark',score: 95,time: 'afternoon'}, {name: 'John',score: 90,time: 'evening'},{name: 'Doni',score: 68,time: 'morning'},{name: 'Jiu',score: 50,time: 'evening'},{name: 'Shin',score: 92,time: 'morning'},];
const PASSING = 70
let result = students.reduce((obj, {name, score, time}) => {
if (!obj[time]) obj[time] = {'pass': , 'fail': }
let key = score >= PASSING ? 'pass' : 'fail'
obj[time][key].push({name, score})
return obj
}, {})
console.log(result)
edited Nov 20 '18 at 1:35
answered Nov 20 '18 at 0:40
Mark MeyerMark Meyer
38.6k33159
38.6k33159
how about if we other times, midnight, noon? should we create it on base variable? can u use that by function
– Zr Classic
Nov 20 '18 at 1:27
@ZrClassic If you are going to have arbitrary time categories, you will be better off building the object as you go. That will be a more general solution, but also more complicated.
– Mark Meyer
Nov 20 '18 at 1:29
add a comment |
how about if we other times, midnight, noon? should we create it on base variable? can u use that by function
– Zr Classic
Nov 20 '18 at 1:27
@ZrClassic If you are going to have arbitrary time categories, you will be better off building the object as you go. That will be a more general solution, but also more complicated.
– Mark Meyer
Nov 20 '18 at 1:29
how about if we other times, midnight, noon? should we create it on base variable? can u use that by function
– Zr Classic
Nov 20 '18 at 1:27
how about if we other times, midnight, noon? should we create it on base variable? can u use that by function
– Zr Classic
Nov 20 '18 at 1:27
@ZrClassic If you are going to have arbitrary time categories, you will be better off building the object as you go. That will be a more general solution, but also more complicated.
– Mark Meyer
Nov 20 '18 at 1:29
@ZrClassic If you are going to have arbitrary time categories, you will be better off building the object as you go. That will be a more general solution, but also more complicated.
– Mark Meyer
Nov 20 '18 at 1:29
add a comment |
I'm sure there are more elegant ways to do this. But this one is probably one of the simplest beginner-friendly ways you can go about this.
I loop through the input array, check the existence of the .time
values as keys on the output
object and create the pass
and fail
keys. Then evaluate the .score
against the passingScore
and push the necessary data to it.
Should be pretty easy to understand once you see and try the code below:
const data = [
{name: 'John',score: 90, time: 'evening'},
{name: 'Doni',score: 68, time: 'morning'},
{name: 'Jiu',score: 50, time: 'evening'},
{name: 'Shin',score: 92, time: 'morning'},
{name: 'Fubar',score: 75, time: 'noon'},
];
function formatData(data){
const passingScore = 75;
const output = {};
data.forEach(function(item){
if(!output[item.time]) output[item.time] = {pass: , fail: };
const stud = { name: item.name, score: item.score };
if(item.score >= passingScore) output[item.time]['pass'].push(stud)
else output[item.time]['fail'].push(stud)
});
return output;
}
console.log(formatData(data));
this is the perfect one for beginner
– Zr Classic
Nov 20 '18 at 4:29
add a comment |
I'm sure there are more elegant ways to do this. But this one is probably one of the simplest beginner-friendly ways you can go about this.
I loop through the input array, check the existence of the .time
values as keys on the output
object and create the pass
and fail
keys. Then evaluate the .score
against the passingScore
and push the necessary data to it.
Should be pretty easy to understand once you see and try the code below:
const data = [
{name: 'John',score: 90, time: 'evening'},
{name: 'Doni',score: 68, time: 'morning'},
{name: 'Jiu',score: 50, time: 'evening'},
{name: 'Shin',score: 92, time: 'morning'},
{name: 'Fubar',score: 75, time: 'noon'},
];
function formatData(data){
const passingScore = 75;
const output = {};
data.forEach(function(item){
if(!output[item.time]) output[item.time] = {pass: , fail: };
const stud = { name: item.name, score: item.score };
if(item.score >= passingScore) output[item.time]['pass'].push(stud)
else output[item.time]['fail'].push(stud)
});
return output;
}
console.log(formatData(data));
this is the perfect one for beginner
– Zr Classic
Nov 20 '18 at 4:29
add a comment |
I'm sure there are more elegant ways to do this. But this one is probably one of the simplest beginner-friendly ways you can go about this.
I loop through the input array, check the existence of the .time
values as keys on the output
object and create the pass
and fail
keys. Then evaluate the .score
against the passingScore
and push the necessary data to it.
Should be pretty easy to understand once you see and try the code below:
const data = [
{name: 'John',score: 90, time: 'evening'},
{name: 'Doni',score: 68, time: 'morning'},
{name: 'Jiu',score: 50, time: 'evening'},
{name: 'Shin',score: 92, time: 'morning'},
{name: 'Fubar',score: 75, time: 'noon'},
];
function formatData(data){
const passingScore = 75;
const output = {};
data.forEach(function(item){
if(!output[item.time]) output[item.time] = {pass: , fail: };
const stud = { name: item.name, score: item.score };
if(item.score >= passingScore) output[item.time]['pass'].push(stud)
else output[item.time]['fail'].push(stud)
});
return output;
}
console.log(formatData(data));
I'm sure there are more elegant ways to do this. But this one is probably one of the simplest beginner-friendly ways you can go about this.
I loop through the input array, check the existence of the .time
values as keys on the output
object and create the pass
and fail
keys. Then evaluate the .score
against the passingScore
and push the necessary data to it.
Should be pretty easy to understand once you see and try the code below:
const data = [
{name: 'John',score: 90, time: 'evening'},
{name: 'Doni',score: 68, time: 'morning'},
{name: 'Jiu',score: 50, time: 'evening'},
{name: 'Shin',score: 92, time: 'morning'},
{name: 'Fubar',score: 75, time: 'noon'},
];
function formatData(data){
const passingScore = 75;
const output = {};
data.forEach(function(item){
if(!output[item.time]) output[item.time] = {pass: , fail: };
const stud = { name: item.name, score: item.score };
if(item.score >= passingScore) output[item.time]['pass'].push(stud)
else output[item.time]['fail'].push(stud)
});
return output;
}
console.log(formatData(data));
const data = [
{name: 'John',score: 90, time: 'evening'},
{name: 'Doni',score: 68, time: 'morning'},
{name: 'Jiu',score: 50, time: 'evening'},
{name: 'Shin',score: 92, time: 'morning'},
{name: 'Fubar',score: 75, time: 'noon'},
];
function formatData(data){
const passingScore = 75;
const output = {};
data.forEach(function(item){
if(!output[item.time]) output[item.time] = {pass: , fail: };
const stud = { name: item.name, score: item.score };
if(item.score >= passingScore) output[item.time]['pass'].push(stud)
else output[item.time]['fail'].push(stud)
});
return output;
}
console.log(formatData(data));
const data = [
{name: 'John',score: 90, time: 'evening'},
{name: 'Doni',score: 68, time: 'morning'},
{name: 'Jiu',score: 50, time: 'evening'},
{name: 'Shin',score: 92, time: 'morning'},
{name: 'Fubar',score: 75, time: 'noon'},
];
function formatData(data){
const passingScore = 75;
const output = {};
data.forEach(function(item){
if(!output[item.time]) output[item.time] = {pass: , fail: };
const stud = { name: item.name, score: item.score };
if(item.score >= passingScore) output[item.time]['pass'].push(stud)
else output[item.time]['fail'].push(stud)
});
return output;
}
console.log(formatData(data));
edited Nov 20 '18 at 4:53
answered Nov 20 '18 at 0:33
Abana ClaraAbana Clara
1,646919
1,646919
this is the perfect one for beginner
– Zr Classic
Nov 20 '18 at 4:29
add a comment |
this is the perfect one for beginner
– Zr Classic
Nov 20 '18 at 4:29
this is the perfect one for beginner
– Zr Classic
Nov 20 '18 at 4:29
this is the perfect one for beginner
– Zr Classic
Nov 20 '18 at 4:29
add a comment |
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do we need to use Object.assign for this? Not necessarily. and how many loop we use for this? You do not want to know how many loops you need, let the code handle that for you.
– Abana Clara
Nov 20 '18 at 0:25
yepp, do u know how ?
– Zr Classic
Nov 20 '18 at 0:26