How To Sort Strings In Javascript Numerically
I would like to sort an array of strings (in javascript) such that groups of digits within the strings are compared as integers not strings. I am not worried about signed or float
Solution 1:
I think this does what you want
function sortArray(arr) {
var tempArr = [], n;
for (var i in arr) {
tempArr[i] = arr[i].match(/([^0-9]+)|([0-9]+)/g);
for (var j in tempArr[i]) {
if( ! isNaN(n = parseInt(tempArr[i][j])) ){
tempArr[i][j] = n;
}
}
}
tempArr.sort(function (x, y) {
for (var i in x) {
if (y.length < i || x[i] < y[i]) {
return -1; // x is longer
}
if (x[i] > y[i]) {
return 1;
}
}
return 0;
});
for (var i in tempArr) {
arr[i] = tempArr[i].join('');
}
return arr;
}
alert(
sortArray(["a1b3", "a10b11", "a10b2", "a9b2"]).join(",")
);
Solution 2:
Another variant is to use an instance of Intl.Collator
with numeric option:
var array = ["a100","a20","a3","a3b","a3b100","a3b20","a3b3","!!","~~","9","10","9.5"];
var collator = new Intl.Collator([], {numeric: true});
array.sort((a, b) => collator.compare(a, b));
console.log(array);
Solution 3:
Assuming what you want to do is just do a numeric sort by the digits in each array entry (ignoring the non-digits), you can use this:
function sortByDigits(array) {
var re = /\D/g;
array.sort(function(a, b) {
return(parseInt(a.replace(re, ""), 10) - parseInt(b.replace(re, ""), 10));
});
return(array);
}
It uses a custom sort function that removes the digits and converts to a number each time it's asked to do a comparison. You can see it work here: http://jsfiddle.net/jfriend00/t87m2/.
If this isn't what you want, then please clarify as your question is not very clear on how the sort should actually work.
Solution 4:
Use this compare function for sorting ..
function compareLists(a,b){
var alist = a.split(/(\d+)/), // split text on change from anything to digit and digit to anything
blist = b.split(/(\d+)/); // split text on change from anything to digit and digit to anything
alist.slice(-1) == '' ? alist.pop() : null; // remove the last element if empty
blist.slice(-1) == '' ? blist.pop() : null; // remove the last element if empty
for (var i = 0, len = alist.length; i < len;i++){
if (alist[i] != blist[i]){ // find the first non-equal part
if (alist[i].match(/\d/)) // if numeric
{
return +alist[i] - +blist[i]; // compare as number
} else {
return alist[i].localeCompare(blist[i]); // compare as string
}
}
}
return true;
}
Syntax
var data = ["a1b3","a10b11","b10b2","a9b2","a1b20","a1c4"];
data.sort( compareLists );
alert(data);
demo at http://jsfiddle.net/h9Rqr/7/
Solution 5:
Here's a more complete solution that sorts according to both letters and numbers in the strings
function sort(list) {
var i, l, mi, ml, x;
// copy the original array
list = list.slice(0);
// split the strings, converting numeric (integer) parts to integers
// and leaving letters as strings
for( i = 0, l = list.length; i < l; i++ ) {
list[i] = list[i].match(/(\d+|[a-z]+)/g);
for( mi = 0, ml = list[i].length; mi < ml ; mi++ ) {
x = parseInt(list[i][mi], 10);
list[i][mi] = !!x || x === 0 ? x : list[i][mi];
}
}
// sort deeply, without comparing integers as strings
list = list.sort(function(a, b) {
var i = 0, l = a.length, res = 0;
while( res === 0 && i < l) {
if( a[i] !== b[i] ) {
res = a[i] < b[i] ? -1 : 1;
break;
}
// If you want to ignore the letters, and only sort by numbers
// use this instead:
//
// if( typeof a[i] === "number" && a[i] !== b[i] ) {
// res = a[i] < b[i] ? -1 : 1;
// break;
// }
i++;
}
return res;
});
// glue it together again
for( i = 0, l = list.length; i < l; i++ ) {
list[i] = list[i].join("");
}
return list;
}
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