How can I order a table with decimal numbers and letters

Question

Hello I'm facing a problem.

I want to order my table by decimal and after by letter.

For exemple I got this :

5.3 Choice 3
A-Choice 4
1.2 Choice 1
1.5 Choice 2
C-Choice 5

And I want it to be that way :

1.2 Choice 1
1.5 Choice 2
5.3 Choice 3
A-Choice 4
C-Choice 5

I tried something like that

const compare = (a, b) => {
        if (!isNaN(b.label.charAt(0)))
        {
            if (a.label === b.label) {
                return 0
             };
             const aArr = a.label.split("."), bArr = b.label.split(".");
             for (let i = 0; i < Math.min(aArr.length, bArr.length); i++) {
                if (parseInt(aArr[i]) < parseInt(bArr[i])) {
                   return -1
                };
                if (parseInt(aArr[i]) > parseInt(bArr[i])) {
                   return 1
                };
             }
             if (aArr.length < bArr.length) {
                return -1
             };
             if (aArr.length > bArr.length) {
                return 1
             };
             return 0;
        }
        else
        {
            return a.label > b.label;
        }
        
     };
     processus.sort(compare);

But it's not working.. Thanks.

Answer 1

Try the sort using string comparison (ic localeCompare ).

localeCompare is not sufficient if the values after 'Choice' may be > 9 (eg '1.2 Choice 23' . In that case you'll need to to sort twice, something like the snippet.

 const toSort = getToSort(); log(`\\n**Choice always [< 10], localeCompare sufficient`, toSort.simple .sort((a, b) => a.localeCompare(b)) .join(`\\n`) ); // this does not work for a choice values > 9 log(`\\n**Choice may be [> 9], localeCompare insufficient`, toSort.complex .sort((a, b) => a.localeCompare(b)) .join(`\\n`) ); // so, [Choice > 9] needs more work // The plan: // 1. Sort on string start (eg '1.2', 'C'), create // an Object from that with key = string start, // values = [next values] and convert it to an // Array of entries; // 2. Sort numeric within the entries Array on the // last value ('Choice [value]'), which is already // converted to number. Remap the result to sorted // strings. const linesSorted = Object.entries(toSort.complex /*1*/ .map(v => v.split(/[\\s-]/)) .sort(([a, , ], [b, , ]) => a.localeCompare(b)) .reduce( (acc, [v1, v2, v3]) => ({ ...acc, [v1]: [...(acc[v1] || []), [v2, +v3]]}), {} ) ) .reduce( (acc, [key, value]) => /*2*/ [...acc, [key, value.sort(([, a], [, b]) => a - b, [] ).map(v => `${key} ${v.join(` `)}`)]], []) .map(([, value]) => value.join(`\\n`)) .join(`\\n`); log(`\\n**Choice may be [> 9], need more work`, linesSorted); // helpers function getToSort() { // return strings to sort, already splitted to array return { simple: `5.3 Choice 3 A-Choice 4 1.2 Choice 1 1.5 Choice 2 C-Choice 5`.split(`\\n`).map(v => v.trim()), complex: `1.5 Choice 3 5.3 Choice 3 1.2 Choice 55 A-Choice 4 1.2 Choice 1 1.5 Choice 11 C-Choice 5 A-Choice 110 1.2 Choice 2`.split(`\\n`).map(v => v.trim()) }; } function log(...strs) { const pre = document.querySelector(`pre`); strs.forEach(str => pre.textContent += `${str}\\n`); }

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Answer 2

This can easily be achieved with localeCompare .

 const arr = [ '5.3 Choice 3', 'A-Choice 4', '1.2 Choice 1', '1.5 Choice 2', 'C-Choice 5', ]; const result = arr.sort((a, b) => a.localeCompare(b)); console.log(result);