Iterating through two objects with different structures in parallel
1
$begingroup$
I have two objects, one input = Array<Type> and one stored = Array<{..., Type}>
What's the best (cleanest) way to put the elements in the first array into the sub-elements of the second? We can assume they're of the same size.
My current code below works, but looks bad due to the double iterator usage.
var count = 0;
stored.forEach((obj)=> {
obj.value = input[count];
count++;
});
Is there some sort of prototyping that I can use to shorten the code up?
javascript array iteration
asked 4 hours ago
bxk21bxk21
1063
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bxk21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
1
$begingroup$
I have two objects, one input = Array<Type> and one stored = Array<{..., Type}>
What's the best (cleanest) way to put the elements in the first array into the sub-elements of the second? We can assume they're of the same size.
My current code below works, but looks bad due to the double iterator usage.
var count = 0;
stored.forEach((obj)=> {
obj.value = input[count];
count++;
});
Is there some sort of prototyping that I can use to shorten the code up?
javascript array iteration
asked 4 hours ago
bxk21bxk21
1063
New contributor
bxk21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
don't know if this qualifies as an answer but you don't need thecountvariable to increment the index,Array.forEachaccepts a second ( optional ) parameter which is the current index that you can use :stored.forEach((obj, ndx) => { obj.value = input[ndx]; });jsfiddle.net/qm6yep5r
$endgroup$
– Taki
3 hours ago
$begingroup$
@Taki That's exactly it. Thanks a lot! If you want to put it as an answer, I'll mark it.
$endgroup$
– bxk21
3 hours ago
add a comment |
1
1
1
$begingroup$
I have two objects, one input = Array<Type> and one stored = Array<{..., Type}>
What's the best (cleanest) way to put the elements in the first array into the sub-elements of the second? We can assume they're of the same size.
My current code below works, but looks bad due to the double iterator usage.
var count = 0;
stored.forEach((obj)=> {
obj.value = input[count];
count++;
});
Is there some sort of prototyping that I can use to shorten the code up?
javascript array iteration
asked 4 hours ago
bxk21bxk21
1063
New contributor
bxk21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have two objects, one input = Array<Type> and one stored = Array<{..., Type}>
What's the best (cleanest) way to put the elements in the first array into the sub-elements of the second? We can assume they're of the same size.
My current code below works, but looks bad due to the double iterator usage.
var count = 0;
stored.forEach((obj)=> {
obj.value = input[count];
count++;
});
Is there some sort of prototyping that I can use to shorten the code up?
javascript array iteration
javascript array iteration
asked 4 hours ago
bxk21bxk21
1063
New contributor
bxk21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
bxk21bxk21
1063
New contributor
bxk21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
bxk21bxk21
1063
New contributor
bxk21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
bxk21bxk21
1063
asked 4 hours ago
bxk21bxk21
1063
1063
New contributor
bxk21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
bxk21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
bxk21 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
don't know if this qualifies as an answer but you don't need thecountvariable to increment the index,Array.forEachaccepts a second ( optional ) parameter which is the current index that you can use :stored.forEach((obj, ndx) => { obj.value = input[ndx]; });jsfiddle.net/qm6yep5r
$endgroup$
– Taki
3 hours ago
$begingroup$
@Taki That's exactly it. Thanks a lot! If you want to put it as an answer, I'll mark it.
$endgroup$
– bxk21
3 hours ago
add a comment |
1
$begingroup$
don't know if this qualifies as an answer but you don't need thecountvariable to increment the index,Array.forEachaccepts a second ( optional ) parameter which is the current index that you can use :stored.forEach((obj, ndx) => { obj.value = input[ndx]; });jsfiddle.net/qm6yep5r
$endgroup$
– Taki
3 hours ago
$begingroup$
@Taki That's exactly it. Thanks a lot! If you want to put it as an answer, I'll mark it.
$endgroup$
– bxk21
3 hours ago
1
1
$begingroup$
don't know if this qualifies as an answer but you don't need the
count variable to increment the index, Array.forEach accepts a second ( optional ) parameter which is the current index that you can use : stored.forEach((obj, ndx) => { obj.value = input[ndx]; }); jsfiddle.net/qm6yep5r$endgroup$
– Taki
3 hours ago
$begingroup$
don't know if this qualifies as an answer but you don't need the
count variable to increment the index, Array.forEach accepts a second ( optional ) parameter which is the current index that you can use : stored.forEach((obj, ndx) => { obj.value = input[ndx]; }); jsfiddle.net/qm6yep5r$endgroup$
– Taki
3 hours ago
$begingroup$
@Taki That's exactly it. Thanks a lot! If you want to put it as an answer, I'll mark it.
$endgroup$
– bxk21
3 hours ago
$begingroup$
@Taki That's exactly it. Thanks a lot! If you want to put it as an answer, I'll mark it.
$endgroup$
– bxk21
3 hours ago
add a comment |
1 Answer
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0
$begingroup$
you don't need the count variable to increment the index, Array.forEach accepts a second ( optional ) parameter which is the current index that you can use :
stored.forEach((obj, ndx) => { obj.value = input[ndx]; });
OR
stored = stored.map((e, ndx) => ({...e, value: input[ndx]}))
example :
let stored = [{ name : 'a', value: 2 }, {name : 'b', value: 4 }, { name : 'c', value: 6 }];
let input = [10, 11, 12];
stored.forEach((obj, ndx) => { obj.value = input[ndx]; });
// OR
// stored = stored.map((e, ndx) => ({...e, value: input[ndx]}))
console.log(stored)answered 2 hours ago
TakiTaki
3087
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0
$begingroup$
you don't need the count variable to increment the index, Array.forEach accepts a second ( optional ) parameter which is the current index that you can use :
stored.forEach((obj, ndx) => { obj.value = input[ndx]; });
OR
stored = stored.map((e, ndx) => ({...e, value: input[ndx]}))
example :
let stored = [{ name : 'a', value: 2 }, {name : 'b', value: 4 }, { name : 'c', value: 6 }];
let input = [10, 11, 12];
stored.forEach((obj, ndx) => { obj.value = input[ndx]; });
// OR
// stored = stored.map((e, ndx) => ({...e, value: input[ndx]}))
console.log(stored)answered 2 hours ago
TakiTaki
3087
$endgroup$
add a comment |
0
$begingroup$
you don't need the count variable to increment the index, Array.forEach accepts a second ( optional ) parameter which is the current index that you can use :
stored.forEach((obj, ndx) => { obj.value = input[ndx]; });
OR
stored = stored.map((e, ndx) => ({...e, value: input[ndx]}))
example :
let stored = [{ name : 'a', value: 2 }, {name : 'b', value: 4 }, { name : 'c', value: 6 }];
let input = [10, 11, 12];
stored.forEach((obj, ndx) => { obj.value = input[ndx]; });
// OR
// stored = stored.map((e, ndx) => ({...e, value: input[ndx]}))
console.log(stored)answered 2 hours ago
TakiTaki
3087
$endgroup$
add a comment |
0
0
0
$begingroup$
you don't need the count variable to increment the index, Array.forEach accepts a second ( optional ) parameter which is the current index that you can use :
stored.forEach((obj, ndx) => { obj.value = input[ndx]; });
OR
stored = stored.map((e, ndx) => ({...e, value: input[ndx]}))
example :
let stored = [{ name : 'a', value: 2 }, {name : 'b', value: 4 }, { name : 'c', value: 6 }];
let input = [10, 11, 12];
stored.forEach((obj, ndx) => { obj.value = input[ndx]; });
// OR
// stored = stored.map((e, ndx) => ({...e, value: input[ndx]}))
console.log(stored)answered 2 hours ago
TakiTaki
3087
$endgroup$
you don't need the count variable to increment the index, Array.forEach accepts a second ( optional ) parameter which is the current index that you can use :
stored.forEach((obj, ndx) => { obj.value = input[ndx]; });
OR
stored = stored.map((e, ndx) => ({...e, value: input[ndx]}))
example :
let stored = [{ name : 'a', value: 2 }, {name : 'b', value: 4 }, { name : 'c', value: 6 }];
let input = [10, 11, 12];
stored.forEach((obj, ndx) => { obj.value = input[ndx]; });
// OR
// stored = stored.map((e, ndx) => ({...e, value: input[ndx]}))
console.log(stored)let stored = [{ name : 'a', value: 2 }, {name : 'b', value: 4 }, { name : 'c', value: 6 }];
let input = [10, 11, 12];
stored.forEach((obj, ndx) => { obj.value = input[ndx]; });
// OR
// stored = stored.map((e, ndx) => ({...e, value: input[ndx]}))
console.log(stored)let stored = [{ name : 'a', value: 2 }, {name : 'b', value: 4 }, { name : 'c', value: 6 }];
let input = [10, 11, 12];
stored.forEach((obj, ndx) => { obj.value = input[ndx]; });
// OR
// stored = stored.map((e, ndx) => ({...e, value: input[ndx]}))
console.log(stored)answered 2 hours ago
TakiTaki
3087
answered 2 hours ago
TakiTaki
3087
answered 2 hours ago
TakiTaki
3087
answered 2 hours ago
TakiTaki
3087
3087
add a comment |
add a comment |
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bxk21 is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
don't know if this qualifies as an answer but you don't need the
countvariable to increment the index,Array.forEachaccepts a second ( optional ) parameter which is the current index that you can use :stored.forEach((obj, ndx) => { obj.value = input[ndx]; });jsfiddle.net/qm6yep5r$endgroup$
– Taki
3 hours ago
$begingroup$
@Taki That's exactly it. Thanks a lot! If you want to put it as an answer, I'll mark it.
$endgroup$
– bxk21
3 hours ago