I have an alphanumeric array that I need to sort by alphabet ascending and then by number descending, the array is a as follows:
[A1, A4, A2, A3, C2, C4, C3, B3, B5, B4]
How do I sort the array using JavaScript, so it looks like:
[A4, A3, A2, A1, B5, B4, B3, C4, C3, C2]
I have tried:
arr.sort()
but that only sorts it alphabetically and numerically ascending, giving me:
[A1, A2, A3, A4, B3, B4, B5, C2, C3, C4]
If you have always a single letter, you could sort by the numerical rest and by the first letter.
const
array = ['A1', 'A4', 'A2', 'A3', 'C2', 'C4', 'C3', 'B3', 'B5', 'B4'];
array.sort((a, b) => a[0].localeCompare(b[0]) || b.slice(1) - a.slice(1));
console.log(...array);If you have more than one letter, you could take a regular expression and get only digits and non digits, separated into an object.
const
getValues = string => ({
letters: string.match(/\D+/)[0],
digits: string.match(/\d+/)[0]
}),
array = ['A1', 'A4', 'A2', 'A3', 'C2', 'C4', 'C3', 'B3', 'B5', 'B4'];
array.sort((a, b) => {
const [aa, bb] = [a, b].map(getValues);
return aa.letters.localeCompare(bb.letters)
|| bb.digits - aa.digits;
});
console.log(...array);array.sort take a callback where you can define the condition on how you compare two elements in your array
to perform your need an idea can be to return the comparison of second character if first one is identical
var data = ['A1', 'A4', 'A2', 'A3', 'C2', 'C4', 'C3', 'B3', 'B5', 'B4'];
var result = data.sort((a, b) => {
if (a === b) {
return 0;
}
if (a[0] === b[0]) {
return (a[1] > b[1]) ? -1 : 1;
}
return (a > b) ? 1 : -1;
});
console.log(result);
