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How to make a list of partial sums using forEach

The Next CEO of Stack OverflowCreating 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?

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.

edited Mar 20 at 14:51

Solomon Ucko

7902822

asked Mar 20 at 10:47

Team Cafe

1215

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

1

This is just a cumulative sum wrapped in .map. The answer by Thomas is better than anything you'll find there though.

– JollyJoker
Mar 20 at 12:21

2

@JollyJoker Thomas simulate .reduce method, 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 forEach if you want to produce a result.

– Bergi
Mar 20 at 13:55

|
show 4 more comments

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.

edited Mar 20 at 14:51

Solomon Ucko

7902822

asked Mar 20 at 10:47

Team Cafe

1215

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

1

This is just a cumulative sum wrapped in .map. The answer by Thomas is better than anything you'll find there though.

– JollyJoker
Mar 20 at 12:21

2

@JollyJoker Thomas simulate .reduce method, 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 forEach if you want to produce a result.

– Bergi
Mar 20 at 13:55

|
show 4 more comments

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.

edited Mar 20 at 14:51

Solomon Ucko

7902822

asked Mar 20 at 10:47

Team Cafe

1215

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

7902822

asked Mar 20 at 10:47

Team Cafe

1215

edited Mar 20 at 14:51

Solomon Ucko

7902822

asked Mar 20 at 10:47

Team Cafe

1215

edited Mar 20 at 14:51

Solomon Ucko

7902822

edited Mar 20 at 14:51

Solomon Ucko

7902822

edited Mar 20 at 14:51

Solomon Ucko

7902822

asked Mar 20 at 10:47

Team Cafe

1215

asked Mar 20 at 10:47

Team Cafe

1215

asked Mar 20 at 10:47

Team Cafe

1215

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

1

This is just a cumulative sum wrapped in .map. The answer by Thomas is better than anything you'll find there though.

– JollyJoker
Mar 20 at 12:21

2

@JollyJoker Thomas simulate .reduce method, 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 forEach if you want to produce a result.

– Bergi
Mar 20 at 13:55

|
show 4 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

1

This is just a cumulative sum wrapped in .map. The answer by Thomas is better than anything you'll find there though.

– JollyJoker
Mar 20 at 12:21

2

@JollyJoker Thomas simulate .reduce method, 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 forEach if you want to produce a result.

– Bergi
Mar 20 at 13:55

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

This is just a cumulative sum wrapped in .map. The answer by Thomas is better than anything you'll find there though.

– JollyJoker
Mar 20 at 12:21

@JollyJoker Thomas simulate .reduce method, why not use .reduce directly ?

– R3tep
Mar 20 at 12:26

@Thomas Look my answer, you can use an array as accumulator.

– R3tep
Mar 20 at 12:52

Never use forEach if you want to produce a result.

– Bergi
Mar 20 at 13:55

|
show 4 more comments

7 Answers
7

active

oldest

votes

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 Scholz

195k15107178

1

That first approach looks nice, but is really inefficient.

– T.J. Crowder
Mar 20 at 10:58

12

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 nested for loops and create new arrays.

– Nina Scholz
Mar 20 at 11:05

|
show 4 more comments

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%!important

And this doesn't change the source Array.

edited Mar 20 at 11:01

T.J. Crowder

697k12312401336

answered Mar 20 at 10:59

Thomas

5,2161510

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 |

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);

edited Mar 20 at 11:18

answered Mar 20 at 11:13

Keith

9,1031821

add a comment |

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

holydragon

2,74221230

add a comment |

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.

edited Mar 20 at 11:01

answered Mar 20 at 10:51

Alexander

1,024214

3

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

add a comment |

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 G

1,029614

add a comment |

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) => , []));

console.log(result);

// shorter
result = array.map(a => a.reduce((ac, v, i) => [...ac, (ac[i-1] || 0) + v], []));
console.log(result);

edited Mar 20 at 14:37

answered Mar 20 at 12:12

R3tep

8,14382962

add a comment |

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7 Answers
7

active

oldest

votes

7 Answers
7

active

oldest

votes

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 Scholz

195k15107178

1

That first approach looks nice, but is really inefficient.

– T.J. Crowder
Mar 20 at 10:58

12

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 nested for loops and create new arrays.

– Nina Scholz
Mar 20 at 11:05

|
show 4 more comments

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 Scholz

195k15107178

1

That first approach looks nice, but is really inefficient.

– T.J. Crowder
Mar 20 at 10:58

12

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 nested for loops and create new arrays.

– Nina Scholz
Mar 20 at 11:05

|
show 4 more comments

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 Scholz

195k15107178

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 Scholz

195k15107178

answered Mar 20 at 10:54

Nina Scholz

195k15107178

answered Mar 20 at 10:54

Nina Scholz

195k15107178

answered Mar 20 at 10:54

Nina Scholz

195k15107178

1

That first approach looks nice, but is really inefficient.

– T.J. Crowder
Mar 20 at 10:58

12

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 nested for loops 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

12

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 nested for loops and create new arrays.

– Nina Scholz
Mar 20 at 11:05

That first approach looks nice, but is really inefficient.

– T.J. Crowder
Mar 20 at 10:58

but man does it look nice

– Jeremy Thille
Mar 20 at 10:58

@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

@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

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

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%!important

And this doesn't change the source Array.

edited Mar 20 at 11:01

T.J. Crowder

697k12312401336

answered Mar 20 at 10:59

Thomas

5,2161510

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 |

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%!important

And this doesn't change the source Array.

edited Mar 20 at 11:01

T.J. Crowder

697k12312401336

answered Mar 20 at 10:59

Thomas

5,2161510

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 |

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%!important

And this doesn't change the source Array.

edited Mar 20 at 11:01

T.J. Crowder

697k12312401336

answered Mar 20 at 10:59

Thomas

5,2161510

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%!important

And 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%!important

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%!important

edited Mar 20 at 11:01

T.J. Crowder

697k12312401336

answered Mar 20 at 10:59

Thomas

5,2161510

edited Mar 20 at 11:01

T.J. Crowder

697k12312401336

edited Mar 20 at 11:01

T.J. Crowder

697k12312401336

edited Mar 20 at 11:01

T.J. Crowder

697k12312401336

answered Mar 20 at 10:59

Thomas

5,2161510

answered Mar 20 at 10:59

Thomas

5,2161510

answered Mar 20 at 10:59

Thomas

5,2161510

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 |

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 |

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);

edited Mar 20 at 11:18

answered Mar 20 at 11:13

Keith

9,1031821

add a comment |

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);

edited Mar 20 at 11:18

answered Mar 20 at 11:13

Keith

9,1031821

add a comment |

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);

edited Mar 20 at 11:18

answered Mar 20 at 11:13

Keith

9,1031821

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);

edited Mar 20 at 11:18

answered Mar 20 at 11:13

Keith

9,1031821

edited Mar 20 at 11:18

answered Mar 20 at 11:13

Keith

9,1031821

answered Mar 20 at 11:13

Keith

9,1031821

answered Mar 20 at 11:13

Keith

9,1031821

add a comment |

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

holydragon

2,74221230

add a comment |

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

holydragon

2,74221230

add a comment |

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

holydragon

2,74221230

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

holydragon

2,74221230

answered Mar 20 at 10:52

holydragon

2,74221230

answered Mar 20 at 10:52

holydragon

2,74221230

answered Mar 20 at 10:52

holydragon

2,74221230

add a comment |

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.

edited Mar 20 at 11:01

answered Mar 20 at 10:51

Alexander

1,024214

3

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

add a comment |

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.

edited Mar 20 at 11:01

answered Mar 20 at 10:51

Alexander

1,024214

3

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

add a comment |

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.

edited Mar 20 at 11:01

answered Mar 20 at 10:51

Alexander

1,024214

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);

edited Mar 20 at 11:01

answered Mar 20 at 10:51

Alexander

1,024214

edited Mar 20 at 11:01

answered Mar 20 at 10:51

Alexander

1,024214

answered Mar 20 at 10:51

Alexander

1,024214

answered Mar 20 at 10:51

Alexander

1,024214

3

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

add a comment |

3

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

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

add a comment |

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 G

1,029614

add a comment |

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 G

1,029614

add a comment |

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 G

1,029614

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 G

1,029614

answered Mar 20 at 13:06

Nick G

1,029614

answered Mar 20 at 13:06

Nick G

1,029614

answered Mar 20 at 13:06

Nick G

1,029614

add a comment |

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) => , []));

console.log(result);

// shorter
result = array.map(a => a.reduce((ac, v, i) => [...ac, (ac[i-1] || 0) + v], []));
console.log(result);

edited Mar 20 at 14:37

answered Mar 20 at 12:12

R3tep

8,14382962

add a comment |

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) => , []));

console.log(result);

// shorter
result = array.map(a => a.reduce((ac, v, i) => [...ac, (ac[i-1] || 0) + v], []));
console.log(result);

edited Mar 20 at 14:37

answered Mar 20 at 12:12

R3tep

8,14382962

add a comment |

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) => , []));

console.log(result);

// shorter
result = array.map(a => a.reduce((ac, v, i) => [...ac, (ac[i-1] || 0) + v], []));
console.log(result);

edited Mar 20 at 14:37

answered Mar 20 at 12:12

R3tep

8,14382962

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) => , []));

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) => , []));

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) => , []));

console.log(result);

// shorter
result = array.map(a => a.reduce((ac, v, i) => [...ac, (ac[i-1] || 0) + v], []));
console.log(result);

edited Mar 20 at 14:37

answered Mar 20 at 12:12

R3tep

8,14382962

edited Mar 20 at 14:37

answered Mar 20 at 12:12

R3tep

8,14382962

answered Mar 20 at 12:12

R3tep

8,14382962

answered Mar 20 at 12:12

R3tep

8,14382962

add a comment |

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