CODEWITHSUNDEEP
javascriptarrays
Katsuya Obara
Katsuya Obara

Reputation: 963

How to implement logic if sum of array is 0,remove from array in JavaScript?

I have array of array like below. 

[["a",0,0,0],["b",30,20,10],["c",40,50,60],["d",0,0,0],...]

I want to remove some array if sum of number inside of each array is zero and get following return. 

[["b",30,20,10],["c",40,50,60],...]

Could anyone tell how I can implement logic to achieve this in JavaScript?

Upvotes: 1

Views: 441

Answers (5)

simsula
simsula

Reputation: 157

This is the dumb method:

const array = [["a",0,0,0],["b",30,20,10],["c",40,50,60],["d",0,0,0]]
var filteredArray = []

for (const subArray of array) {
    var subArrayJustNumbers = subArray.slice(0)   // copy the subArray
    subArrayJustNumbers.shift()                   // remove the the first object

    var sum = 0                                   // add all the numbers in the subarray
    for (const n of subArrayJustNumbers)
        sum += n

    if (sum != 0) {                               // if the sum of all numbers isn't zero, append it to the filteredArray
        filteredArray.push(subArray)
    }
}

Upvotes: 0

Kosh
Kosh

Reputation: 18423

Use filter and reduce to count sums of numbers:

const a = [
  ["a", 0, 0, 0],
  ["b", 30, 20, 10],
  ["c", 40, 50, 60],
  ["d", 0, 0, 0]
]

var filtered = a.filter(e => !!e.reduce((a, n) => isNaN(n) ? a : a+n, 0))

console.log(JSON.stringify(filtered))

Another approach:

const a = [
  ["a", 0, 0, 0],
  ["b", 30, 20, 10],
  ["c", 40, 50, 60],
  ["d", 0, 0, 0]
]

var f = []

while (e = a.pop())
  if(e.reduce((a, n) => isNaN(n) ? a : a+n, 0)) f.push(e)

console.log(JSON.stringify(f))

Upvotes: 1

garroad_ran
garroad_ran

Reputation: 174

Lots of ways to do this. For me, it's easy to split the problem up into two parts.

  1. Find out if the numbers add up to 0
  2. Remove arrays that have a sum of 0

Study up on the reduce and filter methods if you need to.

// Determine if a subarray adds up to 0.
function sumIsZero(arr) {
  // You could also do
  // const rest = arr.slice(1)
  // If you're not familiar with rest spread operators
  const [letter, ...rest] = arr;

  return rest.reduce((x, y) => x + y, 0) === 0;
}

// Create a new array that only includes the subarrays that add up to 0
const filtered = arrs.filter(sumIsZero);

Upvotes: 1

brk
brk

Reputation: 50326

Use filter method and return only those nestedarray where the sum is not 0

let x = [
  ["a", 0, 0, 0],
  ["b", 30, 20, 10],
  ["c", 40, 50, 60],
  ["d", 0, 0, 0]
];
let result = x.filter(function(item) {
  // a variable to hold the sum
  let sum = 0;
  // since the first value of inner array is string
  // start looping from index 1
  for (var i = 1; i < item.length; i++) {
    sum += item[i];
  }
  // return the nested array only if the sum is not 0
  if (sum !== 0) {
    return item;
  }

})

console.log(result)

Upvotes: 1

CertainPerformance
CertainPerformance

Reputation: 371049

filter the array of arrays, by iterating over each array and keeping track of the sum of the numbers in the array with reduce:

const input = [
  ["a", 0, 0, 0],
  ["b", 30, 20, 10],
  ["c", 40, 50, 60],
  ["d", 0, 0, 0],
  ['e', 30, -30]
];

console.log(
  input.filter((arr) => (
    arr.reduce((a, item) => (
      typeof item === 'number'
      ? a + item
      : a
    ), 0)
  ))
);

Upvotes: 3

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