Sorting multidimensional array based on multiple criteria

Question

I'm trying to figure out an efficient method to sort a multidimensional array based on how many values of another array exist.

Given the following array: [1,2,3,4,5,6,7,8,9,10]

I am trying to sort another array of arrays based on how many of those values are included.

[
  [1,3,5,7,9,22],
  [1,200,300,400,500,600],
  [1,2,3,4,5,6]
]

So the code I'm trying to get to would return:

[
  [1,2,3,4,5,6],
  [1,3,5,7,9,22],
  [1,200,300,400,500,600]
]

I think what I'm doing is very inefficient and could be written better or more succinctly with a method I'm not aware of?

https://jsfiddle.net/gb3fsLdv/

const compareNums = [1,2,3,4,5,6,7,8,9,10];
let ourData = [
  [1,2,3,100,200,300],
  [100,200,300,400,500,600],
  [1,2,3,5,6,9]
];

function sortArr(compare, data){
  let indexMatches = [];
  data.map(arr => {
    let count = 0;
    compare.map(num => {
      if(arr.includes(num)){ 
        count++ 
        }
    })
    indexMatches.push(count);
  })
  // So now I have indexMatches with a count of how many hits each array has in the correct index order as the original data
  // And I can use data to sort them based on these values...
  // Little stuck how to relate the two so the same sorting happens to both arrays
}

sortArr(compareNums, ourData);

Maheer Ali · Accepted Answer · 2019-10-20 15:31:10Z

2

First convert the given array to set. And then use filter() to get the count of elements included in other array

const data = [
  [1,3,5,7,9,22],
  [1,200,300,400,500,600],
  [1,2,3,4,5,6]
]

let arr = [1,2,3,4,5,6,7,8,9,10];

function getCount(arr, set){
  return arr.filter(x => set.has(x)).length
}
function sortOnCount(data, arr){
  let set = new Set(arr);
  return data.slice(0).sort((a, b) => getCount(b, set) - getCount(a, set))
}

console.log(sortOnCount(data, arr))

answered Oct 20, 2019 at 15:31

Maheer Ali

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1 Comment

Nicholas Hazel Over a year ago

Looks extremely effective and easily maintainable. Thanks. Selecting correct answer.

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