对具有负数和正数的字符串数组进行排序

Question

Firstly I could not find a question addressing the whole issue.

I used to compare arrays like this:

array.sort((a, b) => {
    return a.localeCompare(b, undefined, {numeric: true, sensitivity: 'base'})
})

But I realized it does not work on an array like ['a', 'A', 'B', -1.50', '0', '1.50', '-2', '2'] .

The expected output would be: ['-2', '-1.50', '0', '1.50', '2', 'A', 'a', 'B'] .

I have some dirty ideas to achieve it. But maybe there is a clean and easy way.

Answer 1

You could prepend the comparison by taking the delta of the wanted properties. This saves the order for numerical values.

 console.log( ['a', 'A', 'B', '-1.50', '0', '1.50', '-2', '2', 'D'] .sort((a, b) => a - b || a.localeCompare(b, undefined, {sensitivity: 'base'})) ); 

numeric: true option can be omitted for there won't be two numbers to compare at the left hand side of the expression.

Answer 2

You should use isNaN method in order to check if you compare numerical strings. If yes, localeCompare is not recommended , you need to use Number(a) - Number(b)

 array = ['a', 'A', 'B','-1.50', '0', '1.50', '-2', '2'] array.sort((a, b) => { return !isNaN(a) && !isNaN(b) ? Number(a)-Number(b) : a.localeCompare(b, undefined, { sensitivity: 'base'}); }) console.log(array); 

Since in Javascript the - operator is defined only for numeric subtraction, the operands will be coerced to Number . So you could simply use:

array.sort((a, b) => {
  return a - b || a.localeCompare(b, undefined, { sensitivity: 'base'});
})

Answer 3

const customSort = (array) => {
    return array.sort((a, b) => {

        let r = /^-?\d+(?:\.\d+)?$/;

        if (r.test(a) && r.test(b))
        {
            return a - b;
        }

        return a.localeCompare(b, undefined, {numeric: true, sensitivity: 'base'})
    })
}

console.log(customSort(['a', 'A', 'B', '-1.50', '0', '1.50', '-2', '2']));
// => [ '-2', '-1.50', '0', '1.50', '2', 'a', 'A', 'B' ]