Javascript Multiplying Within an Array

Question

I have an array with multiple arrays inside. I would like to multiply the contents of each array.

Here is my code--currently not getting anything in console. The desired output is 3 numbers-- the result of multiplication of the three arrays

var arr = [
  [1, 2, 3],
  [3, 4],
  [7, 8, 9]
]
var mult = 1
for (i = 0; i < arr.length; i++) {
  for (j = 0; j < arr.length; j++) {
    mult *= arr[i][j]
  }
  console.log(mult)
}

It seems that the inner j loop should range from 0 to arr[i].length. — showdev
– showdev, Commented Feb 12, 2022 at 18:59

Nina Scholz · Accepted Answer · 2022-02-12 19:03:04Z

2

You could map the result of the nested multiplying.

const
    multiply = (a, b) => a * b,
    array = [[1, 2, 3], [3, 4], [7, 8, 9]],
    result = array.map(a => a.reduce(multiply, 1));

console.log(result);

answered Feb 12, 2022 at 19:03

Nina Scholz

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2 Comments

Peter Seliger Over a year ago

The reduce method's 2nd parameter, here the initial value of 1, is not needed.

Terry Over a year ago

Not necessarily: having a seed value to start with in a reducer is always a good and safe idea… especially when you start using TypeScript and it actually uses the seed value to infer the correct return type.

Hector Trivino · Accepted Answer · 2022-02-12 19:06:55Z

1

You can try this, it is easier using map and reduce functions. arr.map(array => array.reduce(arrayMult, 1)) is the multiplication of each inner array, like [ 6, 12, 504 ], and the last reduce is to multiply these values.

var arr = [[1, 2, 3],[3, 4],[7, 8, 9]];
const arrayMult = (prev, curr) => prev * curr;
const total = arr.map(array => array.reduce(arrayMult, 1)).reduce(arrayMult, 1);

answered Feb 12, 2022 at 19:06

Hector Trivino

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1 Comment

Peter Seliger Over a year ago

The reduce method's 2nd parameter, here the initial value of 1, is not needed.

tmsbrndz · Accepted Answer · 2022-02-12 19:12:48Z

1

You should iterate over rows. So you need to change inner loop on this :

for (j = 0; j < arr[i].length; j++)

const array = [
  [1, 2, 3],
  [3, 4],
  [7, 8, 9]
];

for (i = 0; i < array.length; i++) {
let result = 1;
  for (j = 0; j < array[i].length; j++) {
    result *= array[i][j]
  }
  console.log(result);
}

edited Feb 12, 2022 at 19:12

answered Feb 12, 2022 at 19:00

tmsbrndz

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5 Comments

ramango Over a year ago

Thanks yeah I see what I should add to the inner loop. For some reason still not seeing anything in the console with what you provided

tmsbrndz Over a year ago

@ramango Whats your expected result?

ramango Over a year ago

expecting three numbers printed, each the result of multiplying the contents of each array

tmsbrndz Over a year ago

@ramango this is what you get if you click on Run code snippet button

ramango Over a year ago

True--i'll check what's goin on Thx!

Daniel Beck · Accepted Answer · 2022-02-12 19:19:31Z

1

Nothing wrong with good old iterating; your existing approach only had one bug, on the inner loop you were still checking the length of the top-level array instead of the inner array; j should range from 0 to arr[i].length, not arr.length.

An alternative approach uses map and reduce; comments below explain how it works:

var arr = [
  [1, 2, 3],
  [3, 4],
  [7, 8, 9]
]

console.log(
  arr.map(a => {  // <-- calls a function on each element of arr, 
                  // leaving the results in an array
    return a.reduce( // <-- reducer is like map, calling a function 
                     // on each element of a,  but collects its results
                     // in a "collector" variable passed to each step
      (val, sum) => {return val * sum}, // <-- multiply this value with 
                                        // the ones so far
      1 // <-- initial value for the collector
    )
  })
)

edited Feb 12, 2022 at 19:19

answered Feb 12, 2022 at 19:07

Daniel Beck

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2 Comments

Peter Seliger Over a year ago

The reduce method's 2nd parameter, here the initial value of 1, is not needed.

Daniel Beck Over a year ago

Strictly speaking you can get away with leaving it out in this case -- but I think it's safer and easier to reason about to explicitly init the collector.

Peter Seliger · Accepted Answer · 2022-02-12 19:16:48Z

0

One approach was to iterate the array of arrays via map which for each array puts the product of all of its items where the latter gets achieved via a multiplying reduce task.

console.log([

  [1, 2, 3],
  [3, 4],
  [7, 8, 9],

].map(arr =>
  arr.reduce((x, y) => x * y)
));

answered Feb 12, 2022 at 19:16

Peter Seliger

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