How to keep array index>value after sorting

Question

in javascript I have next array:

var a = [0, 2, 1, 3];

Where this array index>value pair is:

0 = 0, 1 = 2, 2 = 1, 3 = 3

What is simplest and most elegant way of keeping array index numbers after sorting an array. After sort() index>value pair should be like this:

0 = 0, 2 = 1, 1 = 2, 3 = 3

.. but I should be able to display those sorted values. Problem is that array cannot list by jumping index position 0, 2, 1, 3 but only as 0, 1, 2, 3.

Can I somehow make a new array whose array values will be those new index positions, and then sort this new array but keep memorized former index>value pairs.

Although it sound simple I cannot find solution to this.

Thanks

P.S. I actually want to sort by number of spaces between words in phrases contained in array. Then I want to display sorted by the number of spaces (phrases with most words first).

var input = ["zero", "here two spaces", "none", "here four spaces yes"];
var resort = [];
for (i = 0; i < input.length; i++) {
  var spaces = (input[i].split(" ").length - 1);
  resort.push(spaces); // new array with number of spaces list
}

Nina Scholz · Accepted Answer · 2017-06-18 10:11:06Z

5

You could use Sorting with map with a new array which keeps the original indices and values.

// the array to be sorted
var list = [0, 2, 1, 3];

// temporary array holds objects with position and sort-value
var mapped = list.map(function(el, i) {
    return { index: i, value: el };
})

// sorting the mapped array containing the reduced values
mapped.sort(function(a, b) {
    return a.value - b.value;
});

// container for the resulting order
var result = mapped.map(function(el){
    return list[el.index];
});

console.log(result);
console.log(mapped);

.as-console-wrapper { max-height: 100% !important; top: 0; }´

answered Jun 18, 2017 at 10:11

Nina Scholz

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1 Comment

Dan Over a year ago

Thanks for your kind help.

Niet the Dark Absol · Accepted Answer · 2017-06-18 10:50:02Z

1

If you want to sort by something nontrivial, pass a callback to sort.

input.sort(function(a,b) {
    // b - a for descending order
    return b.split(" ").length - a.split(" ").length;
});

answered Jun 18, 2017 at 10:50

Niet the Dark Absol

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2 Comments

Dan Over a year ago

Hey @ Niet, this works perfectly! And it's simple too. Unfortunately I cannot give you my vote due to the my fresh reputation. Many thanks.

Daniel Danielecki Over a year ago

@Dan, you can try now:)

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