javascript sort sparse array keep indexes
47,568
Solution 1
Here's one approach. It copies the defined array elements to a new array and saves their indexes. It sorts the new array and then puts the sorted results back into the indexes that were previously used.
var a = [];
a[0] = 3;
a[1] = 2;
a[2] = 6;
a[7] = 4;
a[8] = 5;
// sortFn is optional array sort callback function,
// defaults to numeric sort if not passed
function sortSparseArray(arr, sortFn) {
var tempArr = [], indexes = [];
for (var i = 0; i < arr.length; i++) {
// find all array elements that are not undefined
if (arr[i] !== undefined) {
tempArr.push(arr[i]); // save value
indexes.push(i); // save index
}
}
// sort values (numeric sort by default)
if (!sortFn) {
sortFn = function(a,b) {
return(a - b);
}
}
tempArr.sort(sortFn);
// put sorted values back into the indexes in the original array that were used
for (var i = 0; i < indexes.length; i++) {
arr[indexes[i]] = tempArr[i];
}
return(arr);
}
Working demo: http://jsfiddle.net/jfriend00/3ank4/
Solution 2
You can
- Use
filter
orObject.values
to obtain an array with the values of your sparse array. - Then
sort
that array, from largest to smaller. Be aware it's not stable, which may be specially problematic if some values are not numeric. You can use your own sorting implementation. - Use
map
andpop
to obtain the desired array. Assign it toa
.
var b = a.filter(function(x) {
return true;
}).sort(function(x,y) {
return y - x;
});
a = a.map([].pop, b);
Or, in ECMAScript 2017,
a = a.map([].pop, Object.values(a).sort((x,y) => y-x));
Author by
TestersGonnaTest
Updated on January 24, 2020Comments
-
TestersGonnaTest over 4 years
What is the best method to sort a sparse array and keep the elements on the same indexes? For example:
a[0] = 3, a[1] = 2, a[2] = 6, a[7] = 4, a[8] = 5,
I would like after the sort to have
a[0] = 2, a[1] = 3, a[2] = 4, a[7] = 5, a[8] = 6.