How to sort array based on the numbers in string?

Question

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 term)

When the last number and the next last number is not consecutive, the value returned should be 0.

[ "2x2", "3x0", "2x1" ] ==> [ 2, 2, 3 ]

[ "22x0", "3x9", "2x1" ] ==> [ 3, 0, 0, 0, 0, 0, 0, 0, 2, 22 ]

[ "2x4", "3x0" ] ==> [ 2, 0, 0, 3 ]

[ "axn", "bx(n-2)" ] ==> [ "axn", "0x(n-1)", bx(n-2) ] ==> [ a, 0, b ]

I was thinking of converting to the array to string, replacing the number and letters in front and then sorting the array. But I do not know how put the part that was replaced back to its original number. This is my attempt on returning the final array once it is sorted.

 var ary = [ "1abc2", "3ab1", "2a0" ]; console.log(((ary.toString()).match(/\\d+(?!,)/g)).slice(0, -1)); 

I saw these questions on sorting arrays based on numbers but they do not seem to work for me.

How to sort an array of integers correctly

Sort Array Elements (string with numbers), natural sort

Answer 1

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) { return parseInt(i); }); console.log(res); 

Answer 2

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) => { return parseInt(a[0]); }); console.log(newarray); 

Answer 3

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) { function getEndNum(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 string return getEndNum(b) - getEndNum(a); //sort array descendingly, based on the end-numbers }); console.log(array.map(function(a){return parseInt(a);})); //create a new array with only the start-numbers 

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

Answer 4

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 string var 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 b return 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 gap while(--gap) res.push(0); // then fill it with 0s res.push(+reS.exec(str)); // push this string number return res; }, []); console.log("Sorted array:", array); // array is now sorted console.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);