Sort an array to have specific items first in the array - JavaScript

Suppose, we have an array of objects like this −

const arr = [
   {flag: true, other: 1},
   {flag: true, other: 2},
   {flag: false, other: 3},
   {flag: true, other: 4},
   {flag: true, other: 5},
   {flag: true, other: 6},
   {flag: false, other: 7}
];

We are required to write a JavaScript function that takes in one such array and sorts it based on the following conditions −

• If arr.flag === false, the matching element gets placed first in the array, but only after the previous matching elements.
• The elements that do not match, remain in the same order they were in originally.
• Order of appearance is important.

Therefore, for the above array, the output should be −

const output = [
   {flag: false, other: 3},
   {flag: false, other: 7},
   {flag: true, other: 1},
   {flag: true, other: 2},
   {flag: true, other: 4},
   {flag: true, other: 5},
   {flag: true, other: 6}
];

Example

Following is the code −

const arr = [
   {flag: true, other: 1},
   {flag: true, other: 2},
   {flag: false, other: 3},
   {flag: true, other: 4},
   {flag: true, other: 5},
   {flag: true, other: 6},
   {flag: false, other: 7}
];
const sortByFlag = arr => {
   const sorter = (a, b) => {
      if(!a['flag'] && b['flag']){
         return -1;
      };
      if(a['flag'] && !b['flag']){
         return 1;
      }
      return a['other'] - b['other'];
   }
   arr.sort(sorter);
};
sortByFlag(arr);
console.log(arr);

Output

This will produce the following output on console −

[
   { flag: false, other: 3 },
   { flag: false, other: 7 },
   { flag: true, other: 1 },
   { flag: true, other: 2 },
   { flag: true, other: 4 },
   { flag: true, other: 5 },
   { flag: true, other: 6 }
]

AmitDiwan

Published on 01-Oct-2020 10:49:03

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