How to add numbers in array using forEach

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 3 hours ago

asked 3 hours ago

Team Cafe

412

New contributor

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
3 hours ago

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
3 hours ago

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

– JollyJoker
1 hour ago

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

– R3tep
1 hour ago

@R3tep You may technically be right, but I think Thomas' version ends up easier to read.

– JollyJoker
1 hour ago

|
show 3 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 3 hours ago

asked 3 hours ago

Team Cafe

412

New contributor

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
3 hours ago

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
3 hours ago

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

– JollyJoker
1 hour ago

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

– R3tep
1 hour ago

@R3tep You may technically be right, but I think Thomas' version ends up easier to read.

– JollyJoker
1 hour ago

|
show 3 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 3 hours ago

asked 3 hours ago

Team Cafe

412

New contributor

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 3 hours ago

asked 3 hours ago

Team Cafe

412

New contributor

edited 3 hours ago

asked 3 hours ago

Team Cafe

412

New contributor

edited 3 hours ago

asked 3 hours ago

Team Cafe

412

New contributor

asked 3 hours ago

Team Cafe

412

asked 3 hours ago

Team Cafe

412

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.

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
3 hours ago

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
3 hours ago

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

– JollyJoker
1 hour ago

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

– R3tep
1 hour ago

@R3tep You may technically be right, but I think Thomas' version ends up easier to read.

– JollyJoker
1 hour ago

|
show 3 more comments

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
3 hours ago

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
3 hours ago

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

– JollyJoker
1 hour ago

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

– R3tep
1 hour ago

@R3tep You may technically be right, but I think Thomas' version ends up easier to read.

– JollyJoker
1 hour ago

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
3 hours ago

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
3 hours ago

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

– JollyJoker
1 hour ago

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

– R3tep
1 hour ago

@R3tep You may technically be right, but I think Thomas' version ends up easier to read.

– JollyJoker
1 hour ago

|
show 3 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 3 hours ago

Nina Scholz

192k15104176

That first approach looks nice, but is really inefficient.

– T.J. Crowder
2 hours ago

but man does it look nice

– Jeremy Thille
2 hours ago

1

@T.J.Crowder could you explain or point me in the right direction, why the first approach is inefficient?

– Thomas
2 hours ago

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
2 hours ago

1

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
2 hours ago

|
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-wrapper{top:0;max-height:100%!important}

And this doesn't change the source Array.

edited 2 hours ago

T.J. Crowder

694k12212401333

answered 2 hours ago

Thomas

5,016159

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
1 hour ago

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 3 hours ago

holydragon

2,64121230

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 2 hours ago

answered 3 hours ago

Alexander Yakushev

521111

New contributor

2

This modifies the source array, which is probably not a good idea.

– T.J. Crowder
2 hours ago

Yes. And slice will help to prevent that. Thanks for the tip

– Alexander Yakushev
2 hours ago

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 2 hours ago

answered 2 hours ago

Keith

8,9851821

add a comment |

Another way, you can use .map to loop your array and use reduce with an array accumulator

var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],

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



console.log(result);

edited 1 hour ago

answered 1 hour ago

R3tep

7,90182962

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 50 mins ago

Nick G

999614

add a comment |

Your Answer

StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});

}
});

Team Cafe is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55258925%2fhow-to-add-numbers-in-array-using-foreach%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

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 3 hours ago

Nina Scholz

192k15104176

That first approach looks nice, but is really inefficient.

– T.J. Crowder
2 hours ago

but man does it look nice

– Jeremy Thille
2 hours ago

1

@T.J.Crowder could you explain or point me in the right direction, why the first approach is inefficient?

– Thomas
2 hours ago

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
2 hours ago

1

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
2 hours ago

|
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 3 hours ago

Nina Scholz

192k15104176

That first approach looks nice, but is really inefficient.

– T.J. Crowder
2 hours ago

but man does it look nice

– Jeremy Thille
2 hours ago

1

@T.J.Crowder could you explain or point me in the right direction, why the first approach is inefficient?

– Thomas
2 hours ago

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
2 hours ago

1

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
2 hours ago

|
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 3 hours ago

Nina Scholz

192k15104176

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 3 hours ago

Nina Scholz

192k15104176

answered 3 hours ago

Nina Scholz

192k15104176

answered 3 hours ago

Nina Scholz

192k15104176

answered 3 hours ago

Nina Scholz

192k15104176

That first approach looks nice, but is really inefficient.

– T.J. Crowder
2 hours ago

but man does it look nice

– Jeremy Thille
2 hours ago

1

@T.J.Crowder could you explain or point me in the right direction, why the first approach is inefficient?

– Thomas
2 hours ago

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
2 hours ago

1

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
2 hours ago

|
show 4 more comments

That first approach looks nice, but is really inefficient.

– T.J. Crowder
2 hours ago

but man does it look nice

– Jeremy Thille
2 hours ago

1

@T.J.Crowder could you explain or point me in the right direction, why the first approach is inefficient?

– Thomas
2 hours ago

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
2 hours ago

1

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
2 hours ago

That first approach looks nice, but is really inefficient.

– T.J. Crowder
2 hours ago

but man does it look nice

– Jeremy Thille
2 hours ago

@T.J.Crowder could you explain or point me in the right direction, why the first approach is inefficient?

– Thomas
2 hours ago

@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
2 hours ago

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
2 hours ago

|
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-wrapper{top:0;max-height:100%!important}

And this doesn't change the source Array.

edited 2 hours ago

T.J. Crowder

694k12212401333

answered 2 hours ago

Thomas

5,016159

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
1 hour ago

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-wrapper{top:0;max-height:100%!important}

And this doesn't change the source Array.

edited 2 hours ago

T.J. Crowder

694k12212401333

answered 2 hours ago

Thomas

5,016159

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
1 hour ago

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-wrapper{top:0;max-height:100%!important}

And this doesn't change the source Array.

edited 2 hours ago

T.J. Crowder

694k12212401333

answered 2 hours ago

Thomas

5,016159

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-wrapper{top: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-wrapper{top: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-wrapper{top:0;max-height:100%!important}

edited 2 hours ago

T.J. Crowder

694k12212401333

answered 2 hours ago

Thomas

5,016159

edited 2 hours ago

T.J. Crowder

694k12212401333

edited 2 hours ago

T.J. Crowder

694k12212401333

edited 2 hours ago

T.J. Crowder

694k12212401333

answered 2 hours ago

Thomas

5,016159

answered 2 hours ago

Thomas

5,016159

answered 2 hours ago

Thomas

5,016159

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
1 hour ago

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
1 hour ago

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
1 hour ago

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 3 hours ago

holydragon

2,64121230

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 3 hours ago

holydragon

2,64121230

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 3 hours ago

holydragon

2,64121230

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 3 hours ago

holydragon

2,64121230

answered 3 hours ago

holydragon

2,64121230

answered 3 hours ago

holydragon

2,64121230

answered 3 hours ago

holydragon

2,64121230

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 2 hours ago

answered 3 hours ago

Alexander Yakushev

521111

New contributor

2

This modifies the source array, which is probably not a good idea.

– T.J. Crowder
2 hours ago

Yes. And slice will help to prevent that. Thanks for the tip

– Alexander Yakushev
2 hours ago

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 2 hours ago

answered 3 hours ago

Alexander Yakushev

521111

New contributor

2

This modifies the source array, which is probably not a good idea.

– T.J. Crowder
2 hours ago

Yes. And slice will help to prevent that. Thanks for the tip

– Alexander Yakushev
2 hours ago

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 2 hours ago

answered 3 hours ago

Alexander Yakushev

521111

New contributor

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 2 hours ago

answered 3 hours ago

Alexander Yakushev

521111

New contributor

edited 2 hours ago

answered 3 hours ago

Alexander Yakushev

521111

New contributor

answered 3 hours ago

Alexander Yakushev

521111

answered 3 hours ago

Alexander Yakushev

521111

New contributor

Alexander Yakushev 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
2 hours ago

Yes. And slice will help to prevent that. Thanks for the tip

– Alexander Yakushev
2 hours ago

add a comment |

2

This modifies the source array, which is probably not a good idea.

– T.J. Crowder
2 hours ago

Yes. And slice will help to prevent that. Thanks for the tip

– Alexander Yakushev
2 hours ago

This modifies the source array, which is probably not a good idea.

– T.J. Crowder
2 hours ago

Yes. And slice will help to prevent that. Thanks for the tip

– Alexander Yakushev
2 hours ago

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 2 hours ago

answered 2 hours ago

Keith

8,9851821

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 2 hours ago

answered 2 hours ago

Keith

8,9851821

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 2 hours ago

answered 2 hours ago

Keith

8,9851821

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 2 hours ago

answered 2 hours ago

Keith

8,9851821

edited 2 hours ago

answered 2 hours ago

Keith

8,9851821

answered 2 hours ago

Keith

8,9851821

answered 2 hours ago

Keith

8,9851821

add a comment |

Another way, you can use .map to loop your array and use reduce with an array accumulator

var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],

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



console.log(result);

edited 1 hour ago

answered 1 hour ago

R3tep

7,90182962

add a comment |

Another way, you can use .map to loop your array and use reduce with an array accumulator

var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],

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



console.log(result);

edited 1 hour ago

answered 1 hour ago

R3tep

7,90182962

add a comment |

Another way, you can use .map to loop your array and use reduce with an array accumulator

var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],

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



console.log(result);

edited 1 hour ago

answered 1 hour ago

R3tep

7,90182962

Another way, you can use .map to loop your array and use reduce with an array accumulator

var array = [[1, 1, 1, -1], [1, -1, -1], [1, 1]],

    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) => [...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) => [...ac, (ac[i-1] || 0) + v], ));



console.log(result);

edited 1 hour ago

answered 1 hour ago

R3tep

7,90182962

edited 1 hour ago

answered 1 hour ago

R3tep

7,90182962

answered 1 hour ago

R3tep

7,90182962

answered 1 hour ago

R3tep

7,90182962

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 50 mins ago

Nick G

999614

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 50 mins ago

Nick G

999614

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 50 mins ago

Nick G

999614

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 50 mins ago

Nick G

999614

answered 50 mins ago

Nick G

999614

answered 50 mins ago

Nick G

999614

answered 50 mins ago

Nick G

999614

add a comment |

Team Cafe is a new contributor. Be nice, and check out our Code of Conduct.

draft saved

draft discarded

Team Cafe is a new contributor. Be nice, and check out our Code of Conduct.

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Ytdyklly