How To Sort Array Based On The Numbers In String?

Given this array = [ '3ab1', '2a0', '1abc2' ] How do I sort it to [ '1abc2', '3ab1', '2a0' ] (descending order of the last number) and return [ 1,3,2 ]. (the first numbers of each

Solution 1:

You can use regular expression to get the numbers when using sort and reduce to get the result:

var array = [ "22x0", "3x9", "2x1" ];

var reS = /^\d+/,                          // regexp for getting all digits at the start of the string
    reE = /\d+$/;                          // regexp for getting all digits at the end of the stringvar result = array.sort(function(a, b) {   // First: sort the array
    a = reE.exec(a);                       // get the last number from the string a
    b = reE.exec(b);                       // get the last number from the string breturn b - a;                          // sort in a descending order
}).reduce(function(res, str, i) {          // Then: accumulate the result array var gap = reE.exec(array[i - 1]) - reE.exec(str); // calculate the gap between this string str and the last string array[i - 1] (gap = N_of_last_string - N_of_this_string)if(gap > 0)                            // if there is a gapwhile(--gap) res.push(0);          // then fill it with 0s
    res.push(+reS.exec(str));              // push this string numberreturn res;
}, []);

console.log("Sorted array:", array);       // array is now sortedconsole.log("Result:", result);            // result contain the numbers

In recent ECMAScript versions you can do it shortly using arrow functions like this:

let array = [ "22x0", "3x9", "2x1" ];

let reS = /^\d+/,                       
    reE = /\d+$/;                      
let result = array.sort((a, b) => reE.exec(b) - reE.exec(a))
                  .reduce((res, str, i) => { 
                      let gap = reE.exec(array[i - 1]) - reE.exec(str);
                      if(gap > 0)
                          while(--gap) res.push(0);
                      res.push(+reS.exec(str));
                      return res;
                  }, []);

console.log("Sorted array:", array); 
console.log("Result:", result);

Solution 2:

Your question is a bit odd, but you can achieve this using map and parseInt:

var arr = [ "1abc2", "3ab1", "2a0" ];

var res = arr.map(function (i) {
  returnparseInt(i);
});

console.log(res);

Solution 3:

A combination of sort and map should do the trick.

const ary = [ "1abc2", "3ab1", "2a0" ];
  const newarray = ary
      .sort((a, b) => {
        return a[a.length - 1] < b[b.length - 1];
      })
      .map((a) =>  {
        returnparseInt(a[0]);
      });
  console.log(newarray);

Solution 4:

The script below first sorts the array descending, based on the end-numbers, and then returns only the start-numbers of the sorted array.

(I changed your numbers in the array to show that they can be longer than one digit.)

var array = ["31ab12", "20a40", "11abc27"];

array.sort(function(a,b) {
  functiongetEndNum(str) {for (var i=str.length-1; i>=0; --i) {if (isNaN(str.charAt(i))) {return str.substring(i+1);}}} //this function returns the end-number of the supplied stringreturngetEndNum(b) - getEndNum(a); //sort array descendingly, based on the end-numbers
});

console.log(array.map(function(a){returnparseInt(a);})); //create a new array with only the start-numbers

jsfiddle: https://jsfiddle.net/nu8vf837/