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?
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?
- Shouldn't the result be an object? – Jack Bashford Commented Mar 25, 2019 at 7:02
- @JackBashford Hey man, sry, u r right, I just updated it. – SPG Commented Mar 25, 2019 at 7:08
-
Do you really want
includes
? I'd remendstartsWith
– Bergi Commented Mar 25, 2019 at 8:26 -
@Bergi thx, I think
startWith
is better – SPG Commented Mar 26, 2019 at 0:30
1 Answer
Reset to default 13You can do that in following steps:
Apply
reduce()
on the arrayb
During each iteration use
filter()
on the the arraya
- Get all the items from
a
which starts with item ofb
usingString.prototype.startsWith()
- At last set it as property of the
ac
and returnac
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 ments 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
}
}