Search for the sum of subarrays

Question

I am working on a very large array of subarrays filled with numbers, which I want to reduce to a single sum for each subarray.

Example:

var arr = [[1,2,3], [4,5,6]];

arr.forEach(function(item) {
  item.reduce(function(a, b) {
    return a + b;
  });
});
console.log(arr);

//I want arr to equal [[6], [15]];

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My solution just returns the original element of the array

Answer 1

Try something like this:

var arr = [[1,2,3], [4,5,6]];
var newArr = [];

arr.forEach(function(item) {
  item = item.reduce(function(a, b) {
    return a + b;
  });
  newArr.push([item]);
});
console.log(newArr);

To be a little more concise, you can use the values ​​passed to the forEach callback: the element you are currently viewing, the index of that element, and the array itself:

arr.forEach(function(item, index, array) {
  array[index] = [array[index].reduce(function (a, b) {
    return a + b;
  })];
});

Answer 2

.reducedoes not change the original array. You can use .mapand return a new value, for example:

var arr = [[1,2,3], [4,5,6]];

var newArr = arr.map(function(item) {
  return [item.reduce(function(a, b) {
    return a + b;
  })];
});
console.log(newArr);

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Answer 3

:

var arr = [[1,2,3], [4,5,6]];
var result = arr.map(item => item.reduce((a, b) => a + b));
console.log(result); // [ 6, 15 ]

Answer 4

ES5

var arr = [[1,2,3], [4,5,6]];

arr.map(function (item) {
  return item.reduce(function (a, b) {
    return  a+b;
  });
})

ES6

arr.map(item => {
  return item.reduce((a,b) => a+b);
})

Answer 5

, , ?

var arr = [[1,2,3], [4,5,6]];

arr = arr.map(function(item,i) {
  return [item.reduce(function(a, b) {
    return a + b;
  })];
});
console.log(arr);

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Answer 6

If you want the final result to be on their own list, I would break the problem into two steps:

const arr = [[1,2,3], [4,5,6]];
const sumList = (xs) => [xs.reduce((x, y) => x + y, 0)];

alert(arr.map(sumList)); //=> [[6], [15]]

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