Mixing
Can I increase the index of map() by 2?
I have a problem with converting the array to object.
const data = [0, 1, 2, 3, 4, 5];
const arr = data.map((element, idx) => {
return {
f: element,
s: arr[idx + 1],
};
});
Of course, arr is [{f: 0, s: 1}, {f: 1, s: 2}, ...], but I want to increase the index of map by 2. The result will like this:
arr = [{ f: 0, s: 1}, { f: 2, s: 3 }, ...]
Is there any way to make the result using method likes map?
javascript arrays functional-programming
asked Nov 20 '18 at 2:11
Caesium133Caesium133
6119
add a comment |
I have a problem with converting the array to object.
const data = [0, 1, 2, 3, 4, 5];
const arr = data.map((element, idx) => {
return {
f: element,
s: arr[idx + 1],
};
});
Of course, arr is [{f: 0, s: 1}, {f: 1, s: 2}, ...], but I want to increase the index of map by 2. The result will like this:
arr = [{ f: 0, s: 1}, { f: 2, s: 3 }, ...]
Is there any way to make the result using method likes map?
javascript arrays functional-programming
asked Nov 20 '18 at 2:11
Caesium133Caesium133
6119
1
I think you'll have to use another loop (forEach, or traditional for loop) and keep track of a secondary index yourself
– duxfox--
Nov 20 '18 at 2:14
Maybe use reduce instead.
– Dominique Fortin
Nov 20 '18 at 2:16
add a comment |
I have a problem with converting the array to object.
const data = [0, 1, 2, 3, 4, 5];
const arr = data.map((element, idx) => {
return {
f: element,
s: arr[idx + 1],
};
});
Of course, arr is [{f: 0, s: 1}, {f: 1, s: 2}, ...], but I want to increase the index of map by 2. The result will like this:
arr = [{ f: 0, s: 1}, { f: 2, s: 3 }, ...]
Is there any way to make the result using method likes map?
javascript arrays functional-programming
asked Nov 20 '18 at 2:11
Caesium133Caesium133
6119
I have a problem with converting the array to object.
const data = [0, 1, 2, 3, 4, 5];
const arr = data.map((element, idx) => {
return {
f: element,
s: arr[idx + 1],
};
});
Of course, arr is [{f: 0, s: 1}, {f: 1, s: 2}, ...], but I want to increase the index of map by 2. The result will like this:
arr = [{ f: 0, s: 1}, { f: 2, s: 3 }, ...]
Is there any way to make the result using method likes map?
javascript arrays functional-programming
javascript arrays functional-programming
asked Nov 20 '18 at 2:11
Caesium133Caesium133
6119
asked Nov 20 '18 at 2:11
Caesium133Caesium133
6119
asked Nov 20 '18 at 2:11
Caesium133Caesium133
6119
asked Nov 20 '18 at 2:11
Caesium133Caesium133
6119
asked Nov 20 '18 at 2:11
Caesium133Caesium133
6119
6119
1
I think you'll have to use another loop (forEach, or traditional for loop) and keep track of a secondary index yourself
– duxfox--
Nov 20 '18 at 2:14
Maybe use reduce instead.
– Dominique Fortin
Nov 20 '18 at 2:16
add a comment |
1
I think you'll have to use another loop (forEach, or traditional for loop) and keep track of a secondary index yourself
– duxfox--
Nov 20 '18 at 2:14
Maybe use reduce instead.
– Dominique Fortin
Nov 20 '18 at 2:16
1
1
I think you'll have to use another loop (forEach, or traditional for loop) and keep track of a secondary index yourself
– duxfox--
Nov 20 '18 at 2:14
I think you'll have to use another loop (forEach, or traditional for loop) and keep track of a secondary index yourself
– duxfox--
Nov 20 '18 at 2:14
Maybe use reduce instead.
– Dominique Fortin
Nov 20 '18 at 2:16
Maybe use reduce instead.
– Dominique Fortin
Nov 20 '18 at 2:16
add a comment |
2 Answers
2
active
oldest
votes
You could do this with Array.reduce if I understood your question correctly:
const data = [0, 1, 2, 3, 4, 5];
const result = data.reduce((r,c,i,a) => {
if(i%2 == 0)
r.push({ f: i, s: a[i+1] })
return r
}, );
console.log(result)The idea is to use the % operator in combination with the reduce to push only the values you would want into the accumulator array.
Doing this with map would be trickier since map goes through every element and expects the same number of elements out as they ware in where the reduce can have any result type/length specified by the accumulator.
answered Nov 20 '18 at 2:17
AkrionAkrion
9,42711224
add a comment |
You can do pretty much anything with reduce, but it’s a bad fit for this:
arr = data.reduce( ( acc, item, i ) => {
if ( i % 2 === 0 ) {
acc[ acc.length - 1 ].s = item
} else {
acc.push({ f: item })
}
return acc
}, )
Better to split the array with a good old for loop, then map that:
var pairwise = arr => {
var pairs =
for ( var i = 0; i < arr.length; i++ ) {
pairs.push([ arr[ i ], arr[ i + 1 ] ])
}
return pairs
}
arr = pairwise( data ).map( ([ f, s ]) => ({ f, s }) )
edited Nov 20 '18 at 2:36
Mark Meyer
37.6k33159
answered Nov 20 '18 at 2:27
Ben WestBen West
2,6811814
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You could do this with Array.reduce if I understood your question correctly:
const data = [0, 1, 2, 3, 4, 5];
const result = data.reduce((r,c,i,a) => {
if(i%2 == 0)
r.push({ f: i, s: a[i+1] })
return r
}, );
console.log(result)The idea is to use the % operator in combination with the reduce to push only the values you would want into the accumulator array.
Doing this with map would be trickier since map goes through every element and expects the same number of elements out as they ware in where the reduce can have any result type/length specified by the accumulator.
answered Nov 20 '18 at 2:17
AkrionAkrion
9,42711224
add a comment |
You could do this with Array.reduce if I understood your question correctly:
const data = [0, 1, 2, 3, 4, 5];
const result = data.reduce((r,c,i,a) => {
if(i%2 == 0)
r.push({ f: i, s: a[i+1] })
return r
}, );
console.log(result)The idea is to use the % operator in combination with the reduce to push only the values you would want into the accumulator array.
Doing this with map would be trickier since map goes through every element and expects the same number of elements out as they ware in where the reduce can have any result type/length specified by the accumulator.
answered Nov 20 '18 at 2:17
AkrionAkrion
9,42711224
add a comment |
You could do this with Array.reduce if I understood your question correctly:
const data = [0, 1, 2, 3, 4, 5];
const result = data.reduce((r,c,i,a) => {
if(i%2 == 0)
r.push({ f: i, s: a[i+1] })
return r
}, );
console.log(result)The idea is to use the % operator in combination with the reduce to push only the values you would want into the accumulator array.
Doing this with map would be trickier since map goes through every element and expects the same number of elements out as they ware in where the reduce can have any result type/length specified by the accumulator.
answered Nov 20 '18 at 2:17
AkrionAkrion
9,42711224
You could do this with Array.reduce if I understood your question correctly:
const data = [0, 1, 2, 3, 4, 5];
const result = data.reduce((r,c,i,a) => {
if(i%2 == 0)
r.push({ f: i, s: a[i+1] })
return r
}, );
console.log(result)The idea is to use the % operator in combination with the reduce to push only the values you would want into the accumulator array.
Doing this with map would be trickier since map goes through every element and expects the same number of elements out as they ware in where the reduce can have any result type/length specified by the accumulator.
const data = [0, 1, 2, 3, 4, 5];
const result = data.reduce((r,c,i,a) => {
if(i%2 == 0)
r.push({ f: i, s: a[i+1] })
return r
}, );
console.log(result)const data = [0, 1, 2, 3, 4, 5];
const result = data.reduce((r,c,i,a) => {
if(i%2 == 0)
r.push({ f: i, s: a[i+1] })
return r
}, );
console.log(result)answered Nov 20 '18 at 2:17
AkrionAkrion
9,42711224
edited Nov 20 '18 at 2:33
answered Nov 20 '18 at 2:17
AkrionAkrion
9,42711224
answered Nov 20 '18 at 2:17
AkrionAkrion
9,42711224
answered Nov 20 '18 at 2:17
AkrionAkrion
9,42711224
9,42711224
add a comment |
add a comment |
You can do pretty much anything with reduce, but it’s a bad fit for this:
arr = data.reduce( ( acc, item, i ) => {
if ( i % 2 === 0 ) {
acc[ acc.length - 1 ].s = item
} else {
acc.push({ f: item })
}
return acc
}, )
Better to split the array with a good old for loop, then map that:
var pairwise = arr => {
var pairs =
for ( var i = 0; i < arr.length; i++ ) {
pairs.push([ arr[ i ], arr[ i + 1 ] ])
}
return pairs
}
arr = pairwise( data ).map( ([ f, s ]) => ({ f, s }) )
edited Nov 20 '18 at 2:36
Mark Meyer
37.6k33159
answered Nov 20 '18 at 2:27
Ben WestBen West
2,6811814
add a comment |
You can do pretty much anything with reduce, but it’s a bad fit for this:
arr = data.reduce( ( acc, item, i ) => {
if ( i % 2 === 0 ) {
acc[ acc.length - 1 ].s = item
} else {
acc.push({ f: item })
}
return acc
}, )
Better to split the array with a good old for loop, then map that:
var pairwise = arr => {
var pairs =
for ( var i = 0; i < arr.length; i++ ) {
pairs.push([ arr[ i ], arr[ i + 1 ] ])
}
return pairs
}
arr = pairwise( data ).map( ([ f, s ]) => ({ f, s }) )
edited Nov 20 '18 at 2:36
Mark Meyer
37.6k33159
answered Nov 20 '18 at 2:27
Ben WestBen West
2,6811814
add a comment |
You can do pretty much anything with reduce, but it’s a bad fit for this:
arr = data.reduce( ( acc, item, i ) => {
if ( i % 2 === 0 ) {
acc[ acc.length - 1 ].s = item
} else {
acc.push({ f: item })
}
return acc
}, )
Better to split the array with a good old for loop, then map that:
var pairwise = arr => {
var pairs =
for ( var i = 0; i < arr.length; i++ ) {
pairs.push([ arr[ i ], arr[ i + 1 ] ])
}
return pairs
}
arr = pairwise( data ).map( ([ f, s ]) => ({ f, s }) )
edited Nov 20 '18 at 2:36
Mark Meyer
37.6k33159
answered Nov 20 '18 at 2:27
Ben WestBen West
2,6811814
You can do pretty much anything with reduce, but it’s a bad fit for this:
arr = data.reduce( ( acc, item, i ) => {
if ( i % 2 === 0 ) {
acc[ acc.length - 1 ].s = item
} else {
acc.push({ f: item })
}
return acc
}, )
Better to split the array with a good old for loop, then map that:
var pairwise = arr => {
var pairs =
for ( var i = 0; i < arr.length; i++ ) {
pairs.push([ arr[ i ], arr[ i + 1 ] ])
}
return pairs
}
arr = pairwise( data ).map( ([ f, s ]) => ({ f, s }) )
edited Nov 20 '18 at 2:36
Mark Meyer
37.6k33159
answered Nov 20 '18 at 2:27
Ben WestBen West
2,6811814
edited Nov 20 '18 at 2:36
Mark Meyer
37.6k33159
edited Nov 20 '18 at 2:36
Mark Meyer
37.6k33159
edited Nov 20 '18 at 2:36
Mark Meyer
37.6k33159
37.6k33159
answered Nov 20 '18 at 2:27
Ben WestBen West
2,6811814
answered Nov 20 '18 at 2:27
Ben WestBen West
2,6811814
answered Nov 20 '18 at 2:27
Ben WestBen West
2,6811814
2,6811814
add a comment |
add a comment |
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1
I think you'll have to use another loop (forEach, or traditional for loop) and keep track of a secondary index yourself
– duxfox--
Nov 20 '18 at 2:14
Maybe use reduce instead.
– Dominique Fortin
Nov 20 '18 at 2:16