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10

2

Suppose there are two objects.

const a = [
 id: '1-1-1', name: 'a111' ,
 id: '1-1-2', name: 'a112' ,
 id: '1-2-1', name: 'a121' ,
 id: '1-2-2', name: 'a122' ,
 id: '2-1-1', name: 'a211' ,
 id: '2-1-2', name: 'a212' 
]
const b = ['1-1', '1-2', '2-1']

and the result

 

 '1-1':[
 id: '1-1-1', name: 'a111' ,
 id: '1-1-2', name: 'a112' ,
 ],
 '1-2':[
 id: '1-2-1', name: 'a121' ,
 id: '1-2-2', name: 'a122' ,
 ],
 '2-1':[
 id: '2-1-1', name: 'a211' ,
 id: '2-1-2', name: 'a212' ,
 ]
 

Basically, I want to group the data.

I use includes to check if the item from b to match the id from a. Then construct the new array.

This is my attempt(fiddle):

return b.map(item => a.map(jtem => 
 if(jtem.id.includes(item))
 return 
 [item]: jtem
 
 
))

For somehow, it doesn't work.

and, is there a clever way to avoid the nested for loop or map function?

javascript ecmascript-6 ecmascript-7

share|improve this question

edited Mar 25 at 7:04

SPG

asked Mar 25 at 7:01

SPGSPG

2,452103359

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Shouldn't the result be an object?
  
  – Jack Bashford
  Mar 25 at 7:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @JackBashford Hey man, sry, u r right, I just updated it.
  
  – SPG
  Mar 25 at 7:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Do you really want includes? I'd recommend startsWith
  
  – Bergi
  Mar 25 at 8:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Bergi thx, I think startWith is better
  
  – SPG
  Mar 26 at 0:30
  

  
  
  
  

add a comment | 

10

2

Suppose there are two objects.

const a = [
 id: '1-1-1', name: 'a111' ,
 id: '1-1-2', name: 'a112' ,
 id: '1-2-1', name: 'a121' ,
 id: '1-2-2', name: 'a122' ,
 id: '2-1-1', name: 'a211' ,
 id: '2-1-2', name: 'a212' 
]
const b = ['1-1', '1-2', '2-1']

and the result

 

 '1-1':[
 id: '1-1-1', name: 'a111' ,
 id: '1-1-2', name: 'a112' ,
 ],
 '1-2':[
 id: '1-2-1', name: 'a121' ,
 id: '1-2-2', name: 'a122' ,
 ],
 '2-1':[
 id: '2-1-1', name: 'a211' ,
 id: '2-1-2', name: 'a212' ,
 ]
 

Basically, I want to group the data.

I use includes to check if the item from b to match the id from a. Then construct the new array.

This is my attempt(fiddle):

return b.map(item => a.map(jtem => 
 if(jtem.id.includes(item))
 return 
 [item]: jtem
 
 
))

For somehow, it doesn't work.

and, is there a clever way to avoid the nested for loop or map function?

javascript ecmascript-6 ecmascript-7

share|improve this question

edited Mar 25 at 7:04

SPG

asked Mar 25 at 7:01

SPGSPG

2,452103359

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Shouldn't the result be an object?
  
  – Jack Bashford
  Mar 25 at 7:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @JackBashford Hey man, sry, u r right, I just updated it.
  
  – SPG
  Mar 25 at 7:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Do you really want includes? I'd recommend startsWith
  
  – Bergi
  Mar 25 at 8:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Bergi thx, I think startWith is better
  
  – SPG
  Mar 26 at 0:30
  

  
  
  
  

add a comment | 

Suppose there are two objects.

const a = [
 id: '1-1-1', name: 'a111' ,
 id: '1-1-2', name: 'a112' ,
 id: '1-2-1', name: 'a121' ,
 id: '1-2-2', name: 'a122' ,
 id: '2-1-1', name: 'a211' ,
 id: '2-1-2', name: 'a212' 
]
const b = ['1-1', '1-2', '2-1']

and the result

 

 '1-1':[
 id: '1-1-1', name: 'a111' ,
 id: '1-1-2', name: 'a112' ,
 ],
 '1-2':[
 id: '1-2-1', name: 'a121' ,
 id: '1-2-2', name: 'a122' ,
 ],
 '2-1':[
 id: '2-1-1', name: 'a211' ,
 id: '2-1-2', name: 'a212' ,
 ]
 

Basically, I want to group the data.

I use includes to check if the item from b to match the id from a. Then construct the new array.

This is my attempt(fiddle):

return b.map(item => a.map(jtem => 
 if(jtem.id.includes(item))
 return 
 [item]: jtem
 
 
))

For somehow, it doesn't work.

and, is there a clever way to avoid the nested for loop or map function?

javascript ecmascript-6 ecmascript-7

share|improve this question

edited Mar 25 at 7:04

SPG

asked Mar 25 at 7:01

SPGSPG

2,452103359

const a = [
 id: '1-1-1', name: 'a111' ,
 id: '1-1-2', name: 'a112' ,
 id: '1-2-1', name: 'a121' ,
 id: '1-2-2', name: 'a122' ,
 id: '2-1-1', name: 'a211' ,
 id: '2-1-2', name: 'a212' 
]
const b = ['1-1', '1-2', '2-1']

 

 '1-1':[
 id: '1-1-1', name: 'a111' ,
 id: '1-1-2', name: 'a112' ,
 ],
 '1-2':[
 id: '1-2-1', name: 'a121' ,
 id: '1-2-2', name: 'a122' ,
 ],
 '2-1':[
 id: '2-1-1', name: 'a211' ,
 id: '2-1-2', name: 'a212' ,
 ]
 

return b.map(item => a.map(jtem => 
 if(jtem.id.includes(item))
 return 
 [item]: jtem
 
 
))

share|improve this question

edited Mar 25 at 7:04

SPG

asked Mar 25 at 7:01

SPGSPG

2,452103359

share|improve this question

edited Mar 25 at 7:04

SPG

asked Mar 25 at 7:01

SPGSPG

2,452103359

asked Mar 25 at 7:01

SPGSPG

2,452103359

asked Mar 25 at 7:01

SPGSPG

2,452103359

1 Answer
1

active

oldest

votes

13

You can do that in following steps:

• Apply reduce() on the array b




• During each iteration use filter() on the the array a



• Get all the items from a which starts with item of b using String.prototype.startsWith()
  


• At last set it as property of the ac and return ac
  

const a = [
 id: '1-1-1', name: 'a111' ,
 id: '1-1-2', name: 'a112' ,
 id: '1-2-1', name: 'a121' ,
 id: '1-2-2', name: 'a122' ,
 id: '2-1-1', name: 'a211' ,
 id: '2-1-2', name: 'a212' 
]
const b = ['1-1', '1-2', '2-1']

let res = b.reduce((ac,b) => 
 
 ac[b] = a.filter(x => x.id.startsWith(b));
 return ac;

,)
console.log(res)

As suggested by @Falco is the comments that It would be better to scan over the a once as its large. So here is that version.Actually its better regarding performance

const a = [
 id: '1-1-1', name: 'a111' ,
 id: '1-1-2', name: 'a112' ,
 id: '1-2-1', name: 'a121' ,
 id: '1-2-2', name: 'a122' ,
 id: '2-1-1', name: 'a211' ,
 id: '2-1-2', name: 'a212' 
]
const b = ['1-1', '1-2', '2-1']


let res = a.reduce((ac,x) => 
 let temp = b.find(y => x.id.startsWith(y))
 if(!ac[temp]) ac[temp] = [];
 ac[temp].push(x);
 return ac;
,)

console.log(res)

Note: startsWith is not supported by I.E. So you can create polyfill using indexOf

if(!String.prototype.startWith)
 String.prototype.startsWith = function(str)
 return this.indexOf(str) === 0
 

share|improve this answer

edited Mar 25 at 15:19

answered Mar 25 at 7:04

Maheer AliMaheer Ali

10.6k825

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  While it is specifically said in the question that the op wants to use es6, and that IE don't support es6 features, I just want to mention that startsWith() don't work in IE (while reduce, filter, and setting a property of an object is totally fine if IE > 9) and if someone wants to do the same thing that startsWith do, they can implment their own with some substring :)
  
  – Neyt
  Mar 25 at 10:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  For big a and small b I would probably go with a.reduce(...) because of locality and only scan over the big array once.
  
  – Falco
  Mar 25 at 11:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Falco Thanks for suggestion I updated.
  
  – Maheer Ali
  Mar 25 at 11:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Neyt Thanks for suggestion I updated
  
  – Maheer Ali
  Mar 25 at 11:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MaheerAli Thank you - here is a benchmark comparing the two :-) jsbench.me/dfjtoadysr/1
  
  – Falco
  Mar 25 at 11:47
  

  
  
  
  

add a comment | 

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13

You can do that in following steps:

• Apply reduce() on the array b




• During each iteration use filter() on the the array a



• Get all the items from a which starts with item of b using String.prototype.startsWith()
  


• At last set it as property of the ac and return ac
  

const a = [
 id: '1-1-1', name: 'a111' ,
 id: '1-1-2', name: 'a112' ,
 id: '1-2-1', name: 'a121' ,
 id: '1-2-2', name: 'a122' ,
 id: '2-1-1', name: 'a211' ,
 id: '2-1-2', name: 'a212' 
]
const b = ['1-1', '1-2', '2-1']

let res = b.reduce((ac,b) => 
 
 ac[b] = a.filter(x => x.id.startsWith(b));
 return ac;

,)
console.log(res)

As suggested by @Falco is the comments that It would be better to scan over the a once as its large. So here is that version.Actually its better regarding performance

const a = [
 id: '1-1-1', name: 'a111' ,
 id: '1-1-2', name: 'a112' ,
 id: '1-2-1', name: 'a121' ,
 id: '1-2-2', name: 'a122' ,
 id: '2-1-1', name: 'a211' ,
 id: '2-1-2', name: 'a212' 
]
const b = ['1-1', '1-2', '2-1']


let res = a.reduce((ac,x) => 
 let temp = b.find(y => x.id.startsWith(y))
 if(!ac[temp]) ac[temp] = [];
 ac[temp].push(x);
 return ac;
,)

console.log(res)

Note: startsWith is not supported by I.E. So you can create polyfill using indexOf

if(!String.prototype.startWith)
 String.prototype.startsWith = function(str)
 return this.indexOf(str) === 0
 

share|improve this answer

edited Mar 25 at 15:19

answered Mar 25 at 7:04

Maheer AliMaheer Ali

10.6k825

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  While it is specifically said in the question that the op wants to use es6, and that IE don't support es6 features, I just want to mention that startsWith() don't work in IE (while reduce, filter, and setting a property of an object is totally fine if IE > 9) and if someone wants to do the same thing that startsWith do, they can implment their own with some substring :)
  
  – Neyt
  Mar 25 at 10:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  For big a and small b I would probably go with a.reduce(...) because of locality and only scan over the big array once.
  
  – Falco
  Mar 25 at 11:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Falco Thanks for suggestion I updated.
  
  – Maheer Ali
  Mar 25 at 11:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Neyt Thanks for suggestion I updated
  
  – Maheer Ali
  Mar 25 at 11:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MaheerAli Thank you - here is a benchmark comparing the two :-) jsbench.me/dfjtoadysr/1
  
  – Falco
  Mar 25 at 11:47
  

  
  
  
  

add a comment | 

13

You can do that in following steps:

• Apply reduce() on the array b




• During each iteration use filter() on the the array a



• Get all the items from a which starts with item of b using String.prototype.startsWith()
  


• At last set it as property of the ac and return ac
  

const a = [
 id: '1-1-1', name: 'a111' ,
 id: '1-1-2', name: 'a112' ,
 id: '1-2-1', name: 'a121' ,
 id: '1-2-2', name: 'a122' ,
 id: '2-1-1', name: 'a211' ,
 id: '2-1-2', name: 'a212' 
]
const b = ['1-1', '1-2', '2-1']

let res = b.reduce((ac,b) => 
 
 ac[b] = a.filter(x => x.id.startsWith(b));
 return ac;

,)
console.log(res)

As suggested by @Falco is the comments that It would be better to scan over the a once as its large. So here is that version.Actually its better regarding performance

const a = [
 id: '1-1-1', name: 'a111' ,
 id: '1-1-2', name: 'a112' ,
 id: '1-2-1', name: 'a121' ,
 id: '1-2-2', name: 'a122' ,
 id: '2-1-1', name: 'a211' ,
 id: '2-1-2', name: 'a212' 
]
const b = ['1-1', '1-2', '2-1']


let res = a.reduce((ac,x) => 
 let temp = b.find(y => x.id.startsWith(y))
 if(!ac[temp]) ac[temp] = [];
 ac[temp].push(x);
 return ac;
,)

console.log(res)

Note: startsWith is not supported by I.E. So you can create polyfill using indexOf

if(!String.prototype.startWith)
 String.prototype.startsWith = function(str)
 return this.indexOf(str) === 0
 

share|improve this answer

edited Mar 25 at 15:19

answered Mar 25 at 7:04

Maheer AliMaheer Ali

10.6k825

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  While it is specifically said in the question that the op wants to use es6, and that IE don't support es6 features, I just want to mention that startsWith() don't work in IE (while reduce, filter, and setting a property of an object is totally fine if IE > 9) and if someone wants to do the same thing that startsWith do, they can implment their own with some substring :)
  
  – Neyt
  Mar 25 at 10:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  For big a and small b I would probably go with a.reduce(...) because of locality and only scan over the big array once.
  
  – Falco
  Mar 25 at 11:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @Falco Thanks for suggestion I updated.
  
  – Maheer Ali
  Mar 25 at 11:30
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @Neyt Thanks for suggestion I updated
  
  – Maheer Ali
  Mar 25 at 11:33
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @MaheerAli Thank you - here is a benchmark comparing the two :-) jsbench.me/dfjtoadysr/1
  
  – Falco
  Mar 25 at 11:47
  

  
  
  
  

add a comment | 

You can do that in following steps:

• Apply reduce() on the array b




• During each iteration use filter() on the the array a



• Get all the items from a which starts with item of b using String.prototype.startsWith()
  


• At last set it as property of the ac and return ac
  

const a = [
 id: '1-1-1', name: 'a111' ,
 id: '1-1-2', name: 'a112' ,
 id: '1-2-1', name: 'a121' ,
 id: '1-2-2', name: 'a122' ,
 id: '2-1-1', name: 'a211' ,
 id: '2-1-2', name: 'a212' 
]
const b = ['1-1', '1-2', '2-1']

let res = b.reduce((ac,b) => 
 
 ac[b] = a.filter(x => x.id.startsWith(b));
 return ac;

,)
console.log(res)

As suggested by @Falco is the comments that It would be better to scan over the a once as its large. So here is that version.Actually its better regarding performance

const a = [
 id: '1-1-1', name: 'a111' ,
 id: '1-1-2', name: 'a112' ,
 id: '1-2-1', name: 'a121' ,
 id: '1-2-2', name: 'a122' ,
 id: '2-1-1', name: 'a211' ,
 id: '2-1-2', name: 'a212' 
]
const b = ['1-1', '1-2', '2-1']


let res = a.reduce((ac,x) => 
 let temp = b.find(y => x.id.startsWith(y))
 if(!ac[temp]) ac[temp] = [];
 ac[temp].push(x);
 return ac;
,)

console.log(res)

Note: startsWith is not supported by I.E. So you can create polyfill using indexOf

if(!String.prototype.startWith)
 String.prototype.startsWith = function(str)
 return this.indexOf(str) === 0
 

share|improve this answer

edited Mar 25 at 15:19

answered Mar 25 at 7:04

Maheer AliMaheer Ali

10.6k825

const a = [
 id: '1-1-1', name: 'a111' ,
 id: '1-1-2', name: 'a112' ,
 id: '1-2-1', name: 'a121' ,
 id: '1-2-2', name: 'a122' ,
 id: '2-1-1', name: 'a211' ,
 id: '2-1-2', name: 'a212' 
]
const b = ['1-1', '1-2', '2-1']

let res = b.reduce((ac,b) => 
 
 ac[b] = a.filter(x => x.id.startsWith(b));
 return ac;

,)
console.log(res)

const a = [
 id: '1-1-1', name: 'a111' ,
 id: '1-1-2', name: 'a112' ,
 id: '1-2-1', name: 'a121' ,
 id: '1-2-2', name: 'a122' ,
 id: '2-1-1', name: 'a211' ,
 id: '2-1-2', name: 'a212' 
]
const b = ['1-1', '1-2', '2-1']


let res = a.reduce((ac,x) => 
 let temp = b.find(y => x.id.startsWith(y))
 if(!ac[temp]) ac[temp] = [];
 ac[temp].push(x);
 return ac;
,)

console.log(res)

if(!String.prototype.startWith)
 String.prototype.startsWith = function(str)
 return this.indexOf(str) === 0
 

const a = [
 id: '1-1-1', name: 'a111' ,
 id: '1-1-2', name: 'a112' ,
 id: '1-2-1', name: 'a121' ,
 id: '1-2-2', name: 'a122' ,
 id: '2-1-1', name: 'a211' ,
 id: '2-1-2', name: 'a212' 
]
const b = ['1-1', '1-2', '2-1']

let res = b.reduce((ac,b) => 
 
 ac[b] = a.filter(x => x.id.startsWith(b));
 return ac;

,)
console.log(res)

const a = [
 id: '1-1-1', name: 'a111' ,
 id: '1-1-2', name: 'a112' ,
 id: '1-2-1', name: 'a121' ,
 id: '1-2-2', name: 'a122' ,
 id: '2-1-1', name: 'a211' ,
 id: '2-1-2', name: 'a212' 
]
const b = ['1-1', '1-2', '2-1']

let res = b.reduce((ac,b) => 
 
 ac[b] = a.filter(x => x.id.startsWith(b));
 return ac;

,)
console.log(res)

const a = [
 id: '1-1-1', name: 'a111' ,
 id: '1-1-2', name: 'a112' ,
 id: '1-2-1', name: 'a121' ,
 id: '1-2-2', name: 'a122' ,
 id: '2-1-1', name: 'a211' ,
 id: '2-1-2', name: 'a212' 
]
const b = ['1-1', '1-2', '2-1']


let res = a.reduce((ac,x) => 
 let temp = b.find(y => x.id.startsWith(y))
 if(!ac[temp]) ac[temp] = [];
 ac[temp].push(x);
 return ac;
,)

console.log(res)

const a = [
 id: '1-1-1', name: 'a111' ,
 id: '1-1-2', name: 'a112' ,
 id: '1-2-1', name: 'a121' ,
 id: '1-2-2', name: 'a122' ,
 id: '2-1-1', name: 'a211' ,
 id: '2-1-2', name: 'a212' 
]
const b = ['1-1', '1-2', '2-1']


let res = a.reduce((ac,x) => 
 let temp = b.find(y => x.id.startsWith(y))
 if(!ac[temp]) ac[temp] = [];
 ac[temp].push(x);
 return ac;
,)

console.log(res)

if(!String.prototype.startWith)
 String.prototype.startsWith = function(str)
 return this.indexOf(str) === 0
 

if(!String.prototype.startWith)
 String.prototype.startsWith = function(str)
 return this.indexOf(str) === 0
 

share|improve this answer

edited Mar 25 at 15:19

answered Mar 25 at 7:04

Maheer AliMaheer Ali

10.6k825

answered Mar 25 at 7:04

Maheer AliMaheer Ali

10.6k825

answered Mar 25 at 7:04

Maheer AliMaheer Ali

10.6k825