There are quite a few question related to the topic, but I couldn't find the right solution for my case.
var arr = [a, b, null, d, null]
and am sorting this Array with below logic
return function(a,b){
if(a === null){
return 1;
}
else if(b === null){
return -1;
}
else if(a === b){
return 0;
}
else if(ascending) {
return a < b ? -1 : 1;
}
else if(!ascending) {
return a < b ? 1 : -1;
}
};
I get the following outputs for
Ascending : [a, b, d, null,null]
Descending : [d, b, a, null,null]
Expected : [null, null,d, b, a]
What am I doing wrong?
There are quite a few question related to the topic, but I couldn't find the right solution for my case.
var arr = [a, b, null, d, null]
and am sorting this Array with below logic
return function(a,b){
if(a === null){
return 1;
}
else if(b === null){
return -1;
}
else if(a === b){
return 0;
}
else if(ascending) {
return a < b ? -1 : 1;
}
else if(!ascending) {
return a < b ? 1 : -1;
}
};
I get the following outputs for
Ascending : [a, b, d, null,null]
Descending : [d, b, a, null,null]
Expected : [null, null,d, b, a]
What am I doing wrong?
Share Improve this question edited Nov 13, 2015 at 2:53 royhowie 11.2k14 gold badges53 silver badges67 bronze badges asked Nov 13, 2015 at 2:23 PeruPeru 2,9696 gold badges41 silver badges67 bronze badges 1-
Your parison to
null
does not rely onascending
variable value. PS: it's ugly and confusing when a parison function is not pure. – zerkms Commented Nov 13, 2015 at 2:27
3 Answers
Reset to default 8
function getSort (ascending) {
// if ascending, `null` will be pushed towards the end of the array by returning 1
var nullPosition = ascending ? 1 : -1
return function (a, b) {
// if a is null, push it towards whichever end null elements should end up
if (a == null) return nullPosition
// Note: at this point, a is non-null (previous if statement handled that case).
//
// If b is null, it must therefore be placed closer to whichever end the null
// elements should end up on. If ascending, null elements are pulled towards
// the right end of the array. If descending, null elements are pulled towards
// the left.
//
// Therefore, we return -nullPosition. If ascending, this is -1, meaning a es
// before b; if descending, this is 1, meaning a es after b. This is
// clearly the correct behavior, since ascending will push b, which is null,
// towards the end of the array (with -1) and descending will push b towards
// the beginning of the array.
if (b == null) return -nullPosition
// OTHERWISE, both elements are non-null, so sort normally.
// if a < b AND
// if ascending, a es first, so return -1 == -nullPosition
// if descending, a es after, so return -nullPosition == -(-1) == 1
if (a < b) return -nullPosition
// return the opposite of the previous condition
if (a > b) return nullPosition
// return 0 if both elements are equal
return 0
}
}
function write (arr) { arr.forEach(function (d) { document.write(d + "<br>")})}
var toSort = ['a', 'b', null, 'd', null]
var sortA = getSort(true)
var sortD = getSort(false)
document.write("<br>ASCENDING<br>")
write(toSort.sort(sortA))
document.write("<br>DESCENDING<br>")
write(toSort.sort(sortD))
You could use a two pass approach by checking null
values first and then order by string.
To change the sort order, you could swap the parameters or use a negated result of the one of the function.
var data = ['a', 'b', null, 'd', null];
// ascending
data.sort(function (a, b) {
return (a === null) - (b === null) || ('' + a).localeCompare(b);
});
console.log(data);
// descending
data.sort(function (a, b) {
return (b === null) - (a === null) || ('' + b).localeCompare(a);
});
console.log(data);
.as-console-wrapper { max-height: 100% !important; top: 0; }
What about the below solution?
var arr = [null, 'e', 'a', 'b', null, 'd', null];
function sortBy(arr, ascending) {
return arr.sort((a, b) => {
if(!a) return ascending ? 1 : -1;
if(!b) return ascending ? -1 : 1;
if (ascending) return a > b ? 1 : -1;
return a > b ? -1 : 1;
})
}
const ascendingArr = sortBy(arr, true);
console.log(ascendingArr);
const decendingArr = sortBy(arr, false);
console.log(decendingArr);