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How to make a list of partial sums using forEach
Creating an array of cumulative sum in javascriptHow do JavaScript closures work?How do I check if an element is hidden in jQuery?jQuery get specific option tag textHow do I check if an array includes an object in JavaScript?How do I redirect to another webpage?How do I make the first letter of a string uppercase in JavaScript?How to check whether a string contains a substring in JavaScript?How do I remove a particular element from an array in JavaScript?Remove duplicate values from JS arrayHow does PHP 'foreach' actually work?
18
I have an array of arrays which looks like this:
changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
I want to get the next value in the array by adding the last value
values = [ [1, 2, 3, 2], [1, 0, -1], [1, 2] ];
so far I have tried to use a forEach:
changes.forEach(change =>
let i = changes.indexOf(change);
let newValue = change[i] + change[i + 1]
);
I think I am on the right lines but I cannot get this approach to work, or maybe there is a better way to do it.
javascript arrays ecmascript-6 foreach
edited Mar 20 at 14:51
Solomon Ucko
7752822
asked Mar 20 at 10:47
Team CafeTeam Cafe
1065
New contributor
Team Cafe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
|
show 3 more comments
18
I have an array of arrays which looks like this:
changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
I want to get the next value in the array by adding the last value
values = [ [1, 2, 3, 2], [1, 0, -1], [1, 2] ];
so far I have tried to use a forEach:
changes.forEach(change =>
let i = changes.indexOf(change);
let newValue = change[i] + change[i + 1]
);
I think I am on the right lines but I cannot get this approach to work, or maybe there is a better way to do it.
javascript arrays ecmascript-6 foreach
edited Mar 20 at 14:51
Solomon Ucko
7752822
asked Mar 20 at 10:47
Team CafeTeam Cafe
1065
New contributor
Team Cafe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Please elaborate this "I want to be able to increment the values based on the next value in the array to get this result:"
– Syed Mehtab Hassan
Mar 20 at 10:51
I want to add the numbers in the array that are next to each other together to make the next number in the array
– Team Cafe
Mar 20 at 10:52
1
@JollyJoker Thomas simulate.reducemethod, why not use .reduce directly ?
– R3tep
Mar 20 at 12:26
1
@Thomas Look my answer, you can use an array as accumulator.
– R3tep
Mar 20 at 12:52
1
Never useforEachif you want to produce a result.
– Bergi
Mar 20 at 13:55
|
show 3 more comments
18
18
18
1
I have an array of arrays which looks like this:
changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
I want to get the next value in the array by adding the last value
values = [ [1, 2, 3, 2], [1, 0, -1], [1, 2] ];
so far I have tried to use a forEach:
changes.forEach(change =>
let i = changes.indexOf(change);
let newValue = change[i] + change[i + 1]
);
I think I am on the right lines but I cannot get this approach to work, or maybe there is a better way to do it.
javascript arrays ecmascript-6 foreach
edited Mar 20 at 14:51
Solomon Ucko
7752822
asked Mar 20 at 10:47
Team CafeTeam Cafe
1065
New contributor
Team Cafe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I have an array of arrays which looks like this:
changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
I want to get the next value in the array by adding the last value
values = [ [1, 2, 3, 2], [1, 0, -1], [1, 2] ];
so far I have tried to use a forEach:
changes.forEach(change =>
let i = changes.indexOf(change);
let newValue = change[i] + change[i + 1]
);
I think I am on the right lines but I cannot get this approach to work, or maybe there is a better way to do it.
javascript arrays ecmascript-6 foreach
javascript arrays ecmascript-6 foreach
edited Mar 20 at 14:51
Solomon Ucko
7752822
asked Mar 20 at 10:47
Team CafeTeam Cafe
1065
New contributor
Team Cafe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 20 at 14:51
Solomon Ucko
7752822
asked Mar 20 at 10:47
Team CafeTeam Cafe
1065
New contributor
Team Cafe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 20 at 14:51
Solomon Ucko
7752822
edited Mar 20 at 14:51
Solomon Ucko
7752822
edited Mar 20 at 14:51
Solomon Ucko
7752822
7752822
asked Mar 20 at 10:47
Team CafeTeam Cafe
1065
New contributor
Team Cafe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 20 at 10:47
Team CafeTeam Cafe
1065
asked Mar 20 at 10:47
Team CafeTeam Cafe
1065
1065
New contributor
Team Cafe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Team Cafe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Team Cafe is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Please elaborate this "I want to be able to increment the values based on the next value in the array to get this result:"
– Syed Mehtab Hassan
Mar 20 at 10:51
I want to add the numbers in the array that are next to each other together to make the next number in the array
– Team Cafe
Mar 20 at 10:52
1
@JollyJoker Thomas simulate.reducemethod, why not use .reduce directly ?
– R3tep
Mar 20 at 12:26
1
@Thomas Look my answer, you can use an array as accumulator.
– R3tep
Mar 20 at 12:52
1
Never useforEachif you want to produce a result.
– Bergi
Mar 20 at 13:55
|
show 3 more comments
1
Please elaborate this "I want to be able to increment the values based on the next value in the array to get this result:"
– Syed Mehtab Hassan
Mar 20 at 10:51
I want to add the numbers in the array that are next to each other together to make the next number in the array
– Team Cafe
Mar 20 at 10:52
1
@JollyJoker Thomas simulate.reducemethod, why not use .reduce directly ?
– R3tep
Mar 20 at 12:26
1
@Thomas Look my answer, you can use an array as accumulator.
– R3tep
Mar 20 at 12:52
1
Never useforEachif you want to produce a result.
– Bergi
Mar 20 at 13:55
1
1
Please elaborate this "I want to be able to increment the values based on the next value in the array to get this result:"
– Syed Mehtab Hassan
Mar 20 at 10:51
Please elaborate this "I want to be able to increment the values based on the next value in the array to get this result:"
– Syed Mehtab Hassan
Mar 20 at 10:51
I want to add the numbers in the array that are next to each other together to make the next number in the array
– Team Cafe
Mar 20 at 10:52
I want to add the numbers in the array that are next to each other together to make the next number in the array
– Team Cafe
Mar 20 at 10:52
1
1
@JollyJoker Thomas simulate
.reduce method, why not use .reduce directly ?– R3tep
Mar 20 at 12:26
@JollyJoker Thomas simulate
.reduce method, why not use .reduce directly ?– R3tep
Mar 20 at 12:26
1
1
@Thomas Look my answer, you can use an array as accumulator.
– R3tep
Mar 20 at 12:52
@Thomas Look my answer, you can use an array as accumulator.
– R3tep
Mar 20 at 12:52
1
1
Never use
forEach if you want to produce a result.– Bergi
Mar 20 at 13:55
Never use
forEach if you want to produce a result.– Bergi
Mar 20 at 13:55
|
show 3 more comments
7 Answers
7
active
oldest
votes
18
You could save a sum and add the values.
var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
result = array.map(a => a.map((s => v => s += v)(0)));
console.log(result);By using forEach, you need to take the object reference and the previous value or zero.
var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];
array.forEach(a => a.forEach((v, i, a) => a[i] = (a[i - 1] || 0) + v));
console.log(array);answered Mar 20 at 10:54
Nina ScholzNina Scholz
193k15107178
1
That first approach looks nice, but is really inefficient.
– T.J. Crowder
Mar 20 at 10:58
11
but man does it look nice
– Jeremy Thille
Mar 20 at 10:58
1
@T.J.Crowder could you explain or point me in the right direction, why the first approach is inefficient?
– Thomas
Mar 20 at 11:01
1
@Thomas - Look at all the functions it's creating and executing. Your solution is great -- as concise (way more so than mine) and still efficient.
– T.J. Crowder
Mar 20 at 11:01
2
really, for three items in the outer array? if you really want to speed up the execution, you never use some array methods, instead use nestedforloops and create new arrays.
– Nina Scholz
Mar 20 at 11:05
|
show 4 more comments
12
A version with map.
const changes = [
[1, 1, 1, -1],
[1, -1, -1],
[1, 1]
];
const values = changes.map(array =>
let acc = 0;
return array.map(v => acc += v);
);
console.log(values);.as-console-wrappertop:0;max-height:100%!importantAnd this doesn't change the source Array.
edited Mar 20 at 11:01
T.J. Crowder
696k12312401334
answered Mar 20 at 10:59
ThomasThomas
5,1411510
Note that the array of arrays structure is just a needless complication. The question is actually about the cumulative sum of an array. I'd split out the outer mapping function into something calledcumulativeSum
– JollyJoker
Mar 20 at 12:25
add a comment |
3
const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ]
let values = []
changes.forEach(arr =>
let accu = 0
let nestedArr = []
arr.forEach(n =>
accu += n
nestedArr.push(accu)
)
values.push(nestedArr)
)
console.log(values)answered Mar 20 at 10:52
holydragonholydragon
2,69021230
add a comment |
3
You may use map function of Array
const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
const result = changes.map((v) => v.slice(0).map((t, i, arr) => i === 0 ? t : (arr[i] += arr[i - 1])))
console.log(changes);
console.log(result);Update
Use slice to clone array. This will prevent changes to the original array.
answered Mar 20 at 10:51
AlexanderAlexander
816112
New contributor
Alexander is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
This modifies the source array, which is probably not a good idea.
– T.J. Crowder
Mar 20 at 10:56
Yes. Andslicewill help to prevent that. Thanks for the tip
– Alexander
Mar 20 at 11:02
add a comment |
3
New ESNext features of generators are nice for this.
Here I've created a simple sumpUp generator that you can re-use.
function* sumUp(a)
let sum = 0;
for (const v of a) yield sum += v;
const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
const values = changes.map(a => [...sumUp(a)]);
console.log(values);answered Mar 20 at 11:13
KeithKeith
9,0301821
add a comment |
2
Here is an easier to read way that iterates over the outer list of arrays. A copy of the inner array is made to keep the initial values (like [1, 1, 1, -1]). It then iterates over each value in the copied array and adds it to each index after it in the original array.
var changes = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];
changes.forEach(subArray =>
var subArrayCopy = subArray.slice(); // Create a copy of the current sub array (i.e. subArrayCopy = [1, 1, 1, -1];)
subArrayCopy.forEach((val, index) => // Iterate through each value in the copy
for (var i = subArray.length - 1; i > index; i--) // For each element from the end to the current index
subArray[i] += val; // Add the copy's current index value to the original array
);
)
console.log(changes);answered Mar 20 at 13:06
Nick GNick G
1,019614
add a comment |
1
Another way,
You can use .map to return your new array with the desired results. By using .reduce with an array as an accumulator, you can generate the subarray.
var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
result = array.map(a => a.reduce((ac, v, i) => 0;
return [...ac, lastVal + v];
, []));
console.log(result);
// shorter
result = array.map(a => a.reduce((ac, v, i) => [...ac, (ac[i-1] || 0) + v], []));
console.log(result);answered Mar 20 at 12:12
R3tepR3tep
8,00982962
add a comment |
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7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
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active
oldest
votes
18
You could save a sum and add the values.
var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
result = array.map(a => a.map((s => v => s += v)(0)));
console.log(result);By using forEach, you need to take the object reference and the previous value or zero.
var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];
array.forEach(a => a.forEach((v, i, a) => a[i] = (a[i - 1] || 0) + v));
console.log(array);answered Mar 20 at 10:54
Nina ScholzNina Scholz
193k15107178
1
That first approach looks nice, but is really inefficient.
– T.J. Crowder
Mar 20 at 10:58
11
but man does it look nice
– Jeremy Thille
Mar 20 at 10:58
1
@T.J.Crowder could you explain or point me in the right direction, why the first approach is inefficient?
– Thomas
Mar 20 at 11:01
1
@Thomas - Look at all the functions it's creating and executing. Your solution is great -- as concise (way more so than mine) and still efficient.
– T.J. Crowder
Mar 20 at 11:01
2
really, for three items in the outer array? if you really want to speed up the execution, you never use some array methods, instead use nestedforloops and create new arrays.
– Nina Scholz
Mar 20 at 11:05
|
show 4 more comments
18
You could save a sum and add the values.
var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
result = array.map(a => a.map((s => v => s += v)(0)));
console.log(result);By using forEach, you need to take the object reference and the previous value or zero.
var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];
array.forEach(a => a.forEach((v, i, a) => a[i] = (a[i - 1] || 0) + v));
console.log(array);answered Mar 20 at 10:54
Nina ScholzNina Scholz
193k15107178
1
That first approach looks nice, but is really inefficient.
– T.J. Crowder
Mar 20 at 10:58
11
but man does it look nice
– Jeremy Thille
Mar 20 at 10:58
1
@T.J.Crowder could you explain or point me in the right direction, why the first approach is inefficient?
– Thomas
Mar 20 at 11:01
1
@Thomas - Look at all the functions it's creating and executing. Your solution is great -- as concise (way more so than mine) and still efficient.
– T.J. Crowder
Mar 20 at 11:01
2
really, for three items in the outer array? if you really want to speed up the execution, you never use some array methods, instead use nestedforloops and create new arrays.
– Nina Scholz
Mar 20 at 11:05
|
show 4 more comments
18
18
18
You could save a sum and add the values.
var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
result = array.map(a => a.map((s => v => s += v)(0)));
console.log(result);By using forEach, you need to take the object reference and the previous value or zero.
var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];
array.forEach(a => a.forEach((v, i, a) => a[i] = (a[i - 1] || 0) + v));
console.log(array);answered Mar 20 at 10:54
Nina ScholzNina Scholz
193k15107178
You could save a sum and add the values.
var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
result = array.map(a => a.map((s => v => s += v)(0)));
console.log(result);By using forEach, you need to take the object reference and the previous value or zero.
var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];
array.forEach(a => a.forEach((v, i, a) => a[i] = (a[i - 1] || 0) + v));
console.log(array);var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
result = array.map(a => a.map((s => v => s += v)(0)));
console.log(result);var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
result = array.map(a => a.map((s => v => s += v)(0)));
console.log(result);var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];
array.forEach(a => a.forEach((v, i, a) => a[i] = (a[i - 1] || 0) + v));
console.log(array);var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];
array.forEach(a => a.forEach((v, i, a) => a[i] = (a[i - 1] || 0) + v));
console.log(array);answered Mar 20 at 10:54
Nina ScholzNina Scholz
193k15107178
answered Mar 20 at 10:54
Nina ScholzNina Scholz
193k15107178
answered Mar 20 at 10:54
Nina ScholzNina Scholz
193k15107178
answered Mar 20 at 10:54
Nina ScholzNina Scholz
193k15107178
193k15107178
1
That first approach looks nice, but is really inefficient.
– T.J. Crowder
Mar 20 at 10:58
11
but man does it look nice
– Jeremy Thille
Mar 20 at 10:58
1
@T.J.Crowder could you explain or point me in the right direction, why the first approach is inefficient?
– Thomas
Mar 20 at 11:01
1
@Thomas - Look at all the functions it's creating and executing. Your solution is great -- as concise (way more so than mine) and still efficient.
– T.J. Crowder
Mar 20 at 11:01
2
really, for three items in the outer array? if you really want to speed up the execution, you never use some array methods, instead use nestedforloops and create new arrays.
– Nina Scholz
Mar 20 at 11:05
|
show 4 more comments
1
That first approach looks nice, but is really inefficient.
– T.J. Crowder
Mar 20 at 10:58
11
but man does it look nice
– Jeremy Thille
Mar 20 at 10:58
1
@T.J.Crowder could you explain or point me in the right direction, why the first approach is inefficient?
– Thomas
Mar 20 at 11:01
1
@Thomas - Look at all the functions it's creating and executing. Your solution is great -- as concise (way more so than mine) and still efficient.
– T.J. Crowder
Mar 20 at 11:01
2
really, for three items in the outer array? if you really want to speed up the execution, you never use some array methods, instead use nestedforloops and create new arrays.
– Nina Scholz
Mar 20 at 11:05
1
1
That first approach looks nice, but is really inefficient.
– T.J. Crowder
Mar 20 at 10:58
That first approach looks nice, but is really inefficient.
– T.J. Crowder
Mar 20 at 10:58
11
11
but man does it look nice
– Jeremy Thille
Mar 20 at 10:58
but man does it look nice
– Jeremy Thille
Mar 20 at 10:58
1
1
@T.J.Crowder could you explain or point me in the right direction, why the first approach is inefficient?
– Thomas
Mar 20 at 11:01
@T.J.Crowder could you explain or point me in the right direction, why the first approach is inefficient?
– Thomas
Mar 20 at 11:01
1
1
@Thomas - Look at all the functions it's creating and executing. Your solution is great -- as concise (way more so than mine) and still efficient.
– T.J. Crowder
Mar 20 at 11:01
@Thomas - Look at all the functions it's creating and executing. Your solution is great -- as concise (way more so than mine) and still efficient.
– T.J. Crowder
Mar 20 at 11:01
2
2
really, for three items in the outer array? if you really want to speed up the execution, you never use some array methods, instead use nested
for loops and create new arrays.– Nina Scholz
Mar 20 at 11:05
really, for three items in the outer array? if you really want to speed up the execution, you never use some array methods, instead use nested
for loops and create new arrays.– Nina Scholz
Mar 20 at 11:05
|
show 4 more comments
12
A version with map.
const changes = [
[1, 1, 1, -1],
[1, -1, -1],
[1, 1]
];
const values = changes.map(array =>
let acc = 0;
return array.map(v => acc += v);
);
console.log(values);.as-console-wrappertop:0;max-height:100%!importantAnd this doesn't change the source Array.
edited Mar 20 at 11:01
T.J. Crowder
696k12312401334
answered Mar 20 at 10:59
ThomasThomas
5,1411510
Note that the array of arrays structure is just a needless complication. The question is actually about the cumulative sum of an array. I'd split out the outer mapping function into something calledcumulativeSum
– JollyJoker
Mar 20 at 12:25
add a comment |
12
A version with map.
const changes = [
[1, 1, 1, -1],
[1, -1, -1],
[1, 1]
];
const values = changes.map(array =>
let acc = 0;
return array.map(v => acc += v);
);
console.log(values);.as-console-wrappertop:0;max-height:100%!importantAnd this doesn't change the source Array.
edited Mar 20 at 11:01
T.J. Crowder
696k12312401334
answered Mar 20 at 10:59
ThomasThomas
5,1411510
Note that the array of arrays structure is just a needless complication. The question is actually about the cumulative sum of an array. I'd split out the outer mapping function into something calledcumulativeSum
– JollyJoker
Mar 20 at 12:25
add a comment |
12
12
12
A version with map.
const changes = [
[1, 1, 1, -1],
[1, -1, -1],
[1, 1]
];
const values = changes.map(array =>
let acc = 0;
return array.map(v => acc += v);
);
console.log(values);.as-console-wrappertop:0;max-height:100%!importantAnd this doesn't change the source Array.
edited Mar 20 at 11:01
T.J. Crowder
696k12312401334
answered Mar 20 at 10:59
ThomasThomas
5,1411510
A version with map.
const changes = [
[1, 1, 1, -1],
[1, -1, -1],
[1, 1]
];
const values = changes.map(array =>
let acc = 0;
return array.map(v => acc += v);
);
console.log(values);.as-console-wrappertop:0;max-height:100%!importantAnd this doesn't change the source Array.
const changes = [
[1, 1, 1, -1],
[1, -1, -1],
[1, 1]
];
const values = changes.map(array =>
let acc = 0;
return array.map(v => acc += v);
);
console.log(values);.as-console-wrappertop:0;max-height:100%!importantconst changes = [
[1, 1, 1, -1],
[1, -1, -1],
[1, 1]
];
const values = changes.map(array =>
let acc = 0;
return array.map(v => acc += v);
);
console.log(values);.as-console-wrappertop:0;max-height:100%!importantedited Mar 20 at 11:01
T.J. Crowder
696k12312401334
answered Mar 20 at 10:59
ThomasThomas
5,1411510
edited Mar 20 at 11:01
T.J. Crowder
696k12312401334
edited Mar 20 at 11:01
T.J. Crowder
696k12312401334
edited Mar 20 at 11:01
T.J. Crowder
696k12312401334
696k12312401334
answered Mar 20 at 10:59
ThomasThomas
5,1411510
answered Mar 20 at 10:59
ThomasThomas
5,1411510
answered Mar 20 at 10:59
ThomasThomas
5,1411510
5,1411510
Note that the array of arrays structure is just a needless complication. The question is actually about the cumulative sum of an array. I'd split out the outer mapping function into something calledcumulativeSum
– JollyJoker
Mar 20 at 12:25
add a comment |
Note that the array of arrays structure is just a needless complication. The question is actually about the cumulative sum of an array. I'd split out the outer mapping function into something calledcumulativeSum
– JollyJoker
Mar 20 at 12:25
Note that the array of arrays structure is just a needless complication. The question is actually about the cumulative sum of an array. I'd split out the outer mapping function into something called
cumulativeSum– JollyJoker
Mar 20 at 12:25
Note that the array of arrays structure is just a needless complication. The question is actually about the cumulative sum of an array. I'd split out the outer mapping function into something called
cumulativeSum– JollyJoker
Mar 20 at 12:25
add a comment |
3
const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ]
let values = []
changes.forEach(arr =>
let accu = 0
let nestedArr = []
arr.forEach(n =>
accu += n
nestedArr.push(accu)
)
values.push(nestedArr)
)
console.log(values)answered Mar 20 at 10:52
holydragonholydragon
2,69021230
add a comment |
3
const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ]
let values = []
changes.forEach(arr =>
let accu = 0
let nestedArr = []
arr.forEach(n =>
accu += n
nestedArr.push(accu)
)
values.push(nestedArr)
)
console.log(values)answered Mar 20 at 10:52
holydragonholydragon
2,69021230
add a comment |
3
3
3
const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ]
let values = []
changes.forEach(arr =>
let accu = 0
let nestedArr = []
arr.forEach(n =>
accu += n
nestedArr.push(accu)
)
values.push(nestedArr)
)
console.log(values)answered Mar 20 at 10:52
holydragonholydragon
2,69021230
const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ]
let values = []
changes.forEach(arr =>
let accu = 0
let nestedArr = []
arr.forEach(n =>
accu += n
nestedArr.push(accu)
)
values.push(nestedArr)
)
console.log(values)const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ]
let values = []
changes.forEach(arr =>
let accu = 0
let nestedArr = []
arr.forEach(n =>
accu += n
nestedArr.push(accu)
)
values.push(nestedArr)
)
console.log(values)const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ]
let values = []
changes.forEach(arr =>
let accu = 0
let nestedArr = []
arr.forEach(n =>
accu += n
nestedArr.push(accu)
)
values.push(nestedArr)
)
console.log(values)answered Mar 20 at 10:52
holydragonholydragon
2,69021230
answered Mar 20 at 10:52
holydragonholydragon
2,69021230
answered Mar 20 at 10:52
holydragonholydragon
2,69021230
answered Mar 20 at 10:52
holydragonholydragon
2,69021230
2,69021230
add a comment |
add a comment |
3
You may use map function of Array
const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
const result = changes.map((v) => v.slice(0).map((t, i, arr) => i === 0 ? t : (arr[i] += arr[i - 1])))
console.log(changes);
console.log(result);Update
Use slice to clone array. This will prevent changes to the original array.
answered Mar 20 at 10:51
AlexanderAlexander
816112
New contributor
Alexander is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
This modifies the source array, which is probably not a good idea.
– T.J. Crowder
Mar 20 at 10:56
Yes. Andslicewill help to prevent that. Thanks for the tip
– Alexander
Mar 20 at 11:02
add a comment |
3
You may use map function of Array
const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
const result = changes.map((v) => v.slice(0).map((t, i, arr) => i === 0 ? t : (arr[i] += arr[i - 1])))
console.log(changes);
console.log(result);Update
Use slice to clone array. This will prevent changes to the original array.
answered Mar 20 at 10:51
AlexanderAlexander
816112
New contributor
Alexander is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
This modifies the source array, which is probably not a good idea.
– T.J. Crowder
Mar 20 at 10:56
Yes. Andslicewill help to prevent that. Thanks for the tip
– Alexander
Mar 20 at 11:02
add a comment |
3
3
3
You may use map function of Array
const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
const result = changes.map((v) => v.slice(0).map((t, i, arr) => i === 0 ? t : (arr[i] += arr[i - 1])))
console.log(changes);
console.log(result);Update
Use slice to clone array. This will prevent changes to the original array.
answered Mar 20 at 10:51
AlexanderAlexander
816112
New contributor
Alexander is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
You may use map function of Array
const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
const result = changes.map((v) => v.slice(0).map((t, i, arr) => i === 0 ? t : (arr[i] += arr[i - 1])))
console.log(changes);
console.log(result);Update
Use slice to clone array. This will prevent changes to the original array.
const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
const result = changes.map((v) => v.slice(0).map((t, i, arr) => i === 0 ? t : (arr[i] += arr[i - 1])))
console.log(changes);
console.log(result);const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
const result = changes.map((v) => v.slice(0).map((t, i, arr) => i === 0 ? t : (arr[i] += arr[i - 1])))
console.log(changes);
console.log(result);answered Mar 20 at 10:51
AlexanderAlexander
816112
New contributor
Alexander is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 20 at 11:01
answered Mar 20 at 10:51
AlexanderAlexander
816112
New contributor
Alexander is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 20 at 10:51
AlexanderAlexander
816112
answered Mar 20 at 10:51
AlexanderAlexander
816112
816112
New contributor
Alexander is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Alexander is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Alexander is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
This modifies the source array, which is probably not a good idea.
– T.J. Crowder
Mar 20 at 10:56
Yes. Andslicewill help to prevent that. Thanks for the tip
– Alexander
Mar 20 at 11:02
add a comment |
2
This modifies the source array, which is probably not a good idea.
– T.J. Crowder
Mar 20 at 10:56
Yes. Andslicewill help to prevent that. Thanks for the tip
– Alexander
Mar 20 at 11:02
2
2
This modifies the source array, which is probably not a good idea.
– T.J. Crowder
Mar 20 at 10:56
This modifies the source array, which is probably not a good idea.
– T.J. Crowder
Mar 20 at 10:56
Yes. And
slice will help to prevent that. Thanks for the tip– Alexander
Mar 20 at 11:02
Yes. And
slice will help to prevent that. Thanks for the tip– Alexander
Mar 20 at 11:02
add a comment |
3
New ESNext features of generators are nice for this.
Here I've created a simple sumpUp generator that you can re-use.
function* sumUp(a)
let sum = 0;
for (const v of a) yield sum += v;
const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
const values = changes.map(a => [...sumUp(a)]);
console.log(values);answered Mar 20 at 11:13
KeithKeith
9,0301821
add a comment |
3
New ESNext features of generators are nice for this.
Here I've created a simple sumpUp generator that you can re-use.
function* sumUp(a)
let sum = 0;
for (const v of a) yield sum += v;
const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
const values = changes.map(a => [...sumUp(a)]);
console.log(values);answered Mar 20 at 11:13
KeithKeith
9,0301821
add a comment |
3
3
3
New ESNext features of generators are nice for this.
Here I've created a simple sumpUp generator that you can re-use.
function* sumUp(a)
let sum = 0;
for (const v of a) yield sum += v;
const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
const values = changes.map(a => [...sumUp(a)]);
console.log(values);answered Mar 20 at 11:13
KeithKeith
9,0301821
New ESNext features of generators are nice for this.
Here I've created a simple sumpUp generator that you can re-use.
function* sumUp(a)
let sum = 0;
for (const v of a) yield sum += v;
const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
const values = changes.map(a => [...sumUp(a)]);
console.log(values);function* sumUp(a)
let sum = 0;
for (const v of a) yield sum += v;
const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
const values = changes.map(a => [...sumUp(a)]);
console.log(values);function* sumUp(a)
let sum = 0;
for (const v of a) yield sum += v;
const changes = [ [1, 1, 1, -1], [1, -1, -1], [1, 1] ];
const values = changes.map(a => [...sumUp(a)]);
console.log(values);answered Mar 20 at 11:13
KeithKeith
9,0301821
edited Mar 20 at 11:18
answered Mar 20 at 11:13
KeithKeith
9,0301821
answered Mar 20 at 11:13
KeithKeith
9,0301821
answered Mar 20 at 11:13
KeithKeith
9,0301821
9,0301821
add a comment |
add a comment |
2
Here is an easier to read way that iterates over the outer list of arrays. A copy of the inner array is made to keep the initial values (like [1, 1, 1, -1]). It then iterates over each value in the copied array and adds it to each index after it in the original array.
var changes = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];
changes.forEach(subArray =>
var subArrayCopy = subArray.slice(); // Create a copy of the current sub array (i.e. subArrayCopy = [1, 1, 1, -1];)
subArrayCopy.forEach((val, index) => // Iterate through each value in the copy
for (var i = subArray.length - 1; i > index; i--) // For each element from the end to the current index
subArray[i] += val; // Add the copy's current index value to the original array
);
)
console.log(changes);answered Mar 20 at 13:06
Nick GNick G
1,019614
add a comment |
2
Here is an easier to read way that iterates over the outer list of arrays. A copy of the inner array is made to keep the initial values (like [1, 1, 1, -1]). It then iterates over each value in the copied array and adds it to each index after it in the original array.
var changes = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];
changes.forEach(subArray =>
var subArrayCopy = subArray.slice(); // Create a copy of the current sub array (i.e. subArrayCopy = [1, 1, 1, -1];)
subArrayCopy.forEach((val, index) => // Iterate through each value in the copy
for (var i = subArray.length - 1; i > index; i--) // For each element from the end to the current index
subArray[i] += val; // Add the copy's current index value to the original array
);
)
console.log(changes);answered Mar 20 at 13:06
Nick GNick G
1,019614
add a comment |
2
2
2
Here is an easier to read way that iterates over the outer list of arrays. A copy of the inner array is made to keep the initial values (like [1, 1, 1, -1]). It then iterates over each value in the copied array and adds it to each index after it in the original array.
var changes = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];
changes.forEach(subArray =>
var subArrayCopy = subArray.slice(); // Create a copy of the current sub array (i.e. subArrayCopy = [1, 1, 1, -1];)
subArrayCopy.forEach((val, index) => // Iterate through each value in the copy
for (var i = subArray.length - 1; i > index; i--) // For each element from the end to the current index
subArray[i] += val; // Add the copy's current index value to the original array
);
)
console.log(changes);answered Mar 20 at 13:06
Nick GNick G
1,019614
Here is an easier to read way that iterates over the outer list of arrays. A copy of the inner array is made to keep the initial values (like [1, 1, 1, -1]). It then iterates over each value in the copied array and adds it to each index after it in the original array.
var changes = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];
changes.forEach(subArray =>
var subArrayCopy = subArray.slice(); // Create a copy of the current sub array (i.e. subArrayCopy = [1, 1, 1, -1];)
subArrayCopy.forEach((val, index) => // Iterate through each value in the copy
for (var i = subArray.length - 1; i > index; i--) // For each element from the end to the current index
subArray[i] += val; // Add the copy's current index value to the original array
);
)
console.log(changes);var changes = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];
changes.forEach(subArray =>
var subArrayCopy = subArray.slice(); // Create a copy of the current sub array (i.e. subArrayCopy = [1, 1, 1, -1];)
subArrayCopy.forEach((val, index) => // Iterate through each value in the copy
for (var i = subArray.length - 1; i > index; i--) // For each element from the end to the current index
subArray[i] += val; // Add the copy's current index value to the original array
);
)
console.log(changes);var changes = [[1, 1, 1, -1], [1, -1, -1], [1, 1]];
changes.forEach(subArray =>
var subArrayCopy = subArray.slice(); // Create a copy of the current sub array (i.e. subArrayCopy = [1, 1, 1, -1];)
subArrayCopy.forEach((val, index) => // Iterate through each value in the copy
for (var i = subArray.length - 1; i > index; i--) // For each element from the end to the current index
subArray[i] += val; // Add the copy's current index value to the original array
);
)
console.log(changes);answered Mar 20 at 13:06
Nick GNick G
1,019614
answered Mar 20 at 13:06
Nick GNick G
1,019614
answered Mar 20 at 13:06
Nick GNick G
1,019614
answered Mar 20 at 13:06
Nick GNick G
1,019614
1,019614
add a comment |
add a comment |
1
Another way,
You can use .map to return your new array with the desired results. By using .reduce with an array as an accumulator, you can generate the subarray.
var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
result = array.map(a => a.reduce((ac, v, i) => 0;
return [...ac, lastVal + v];
, []));
console.log(result);
// shorter
result = array.map(a => a.reduce((ac, v, i) => [...ac, (ac[i-1] || 0) + v], []));
console.log(result);answered Mar 20 at 12:12
R3tepR3tep
8,00982962
add a comment |
1
Another way,
You can use .map to return your new array with the desired results. By using .reduce with an array as an accumulator, you can generate the subarray.
var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
result = array.map(a => a.reduce((ac, v, i) => 0;
return [...ac, lastVal + v];
, []));
console.log(result);
// shorter
result = array.map(a => a.reduce((ac, v, i) => [...ac, (ac[i-1] || 0) + v], []));
console.log(result);answered Mar 20 at 12:12
R3tepR3tep
8,00982962
add a comment |
1
1
1
Another way,
You can use .map to return your new array with the desired results. By using .reduce with an array as an accumulator, you can generate the subarray.
var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
result = array.map(a => a.reduce((ac, v, i) => 0;
return [...ac, lastVal + v];
, []));
console.log(result);
// shorter
result = array.map(a => a.reduce((ac, v, i) => [...ac, (ac[i-1] || 0) + v], []));
console.log(result);answered Mar 20 at 12:12
R3tepR3tep
8,00982962
Another way,
You can use .map to return your new array with the desired results. By using .reduce with an array as an accumulator, you can generate the subarray.
var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
result = array.map(a => a.reduce((ac, v, i) => 0;
return [...ac, lastVal + v];
, []));
console.log(result);
// shorter
result = array.map(a => a.reduce((ac, v, i) => [...ac, (ac[i-1] || 0) + v], []));
console.log(result);var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
result = array.map(a => a.reduce((ac, v, i) => 0;
return [...ac, lastVal + v];
, []));
console.log(result);
// shorter
result = array.map(a => a.reduce((ac, v, i) => [...ac, (ac[i-1] || 0) + v], []));
console.log(result);var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],
result = array.map(a => a.reduce((ac, v, i) => 0;
return [...ac, lastVal + v];
, []));
console.log(result);
// shorter
result = array.map(a => a.reduce((ac, v, i) => [...ac, (ac[i-1] || 0) + v], []));
console.log(result);answered Mar 20 at 12:12
R3tepR3tep
8,00982962
edited Mar 20 at 14:37
answered Mar 20 at 12:12
R3tepR3tep
8,00982962
answered Mar 20 at 12:12
R3tepR3tep
8,00982962
answered Mar 20 at 12:12
R3tepR3tep
8,00982962
8,00982962
add a comment |
add a comment |
Team Cafe is a new contributor. Be nice, and check out our Code of Conduct.
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Team Cafe is a new contributor. Be nice, and check out our Code of Conduct.
Team Cafe is a new contributor. Be nice, and check out our Code of Conduct.
Team Cafe is a new contributor. Be nice, and check out our Code of Conduct.
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1
Please elaborate this "I want to be able to increment the values based on the next value in the array to get this result:"
– Syed Mehtab Hassan
Mar 20 at 10:51
I want to add the numbers in the array that are next to each other together to make the next number in the array
– Team Cafe
Mar 20 at 10:52
1
@JollyJoker Thomas simulate
.reducemethod, why not use .reduce directly ?– R3tep
Mar 20 at 12:26
1
@Thomas Look my answer, you can use an array as accumulator.
– R3tep
Mar 20 at 12:52
1
Never use
forEachif you want to produce a result.– Bergi
Mar 20 at 13:55