如果数组的总和为0，如何在JavaScript中从数组中删除逻辑呢？

Question

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?

Answer 1

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) )) ); 

Answer 2

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) 

Answer 3

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)) 

Answer 4

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);

Answer 5

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)
    }
}