Understanding Array.sort compareFn
SeongKuk Han
Posted on February 4, 2023
Understanding Array.sort compareFn
With compareFn
which is an option of sort
, you can sort elements in an order that you want.
compareFn: Specifies a function that defines the sort order. If omitted, the array elements are converted to strings, then sorted according to each character's Unicode code point value.
In the documentation, in order to use compareFn
, you should read it thoroughly
- A comparator conforming to the constraints above will always be able to return all of 1, 0, and -1, or consistently return 0. For example, if a comparator only returns 1 and 0, or only returns 0 and -1, it will not be able to sort reliably because anti-symmetry is broken. A comparator that always returns 0 will cause the array to not be changed at all, but is reliable nonetheless.
Let's begin with a number array.
const arr = [4, 3, 1, 5, 2];
arr.sort((a, b) => a - b);
console.log(arr);
arr.sort((a, b) => b - a);
console.log(arr);
[ 1, 2, 3, 4, 5 ]
[ 5, 4, 3, 2, 1 ]
I sorted it in ascending and reversed.
What do value a
and b
have?
Let's print it out.
const arr = [4, 3, 1, 5, 2];
arr.sort((a, b) => {
console.log(a, b);
return a - b;
});
3 4
1 3
5 1
5 3
5 4
2 4
2 3
2 1
In the result, only one thing we should know that a
element is next element of b
element.
Let's say. We want to place a specific value 3
at the first position.
We can do like this.
const arr = [4, 3, 1, 5, 2];
arr.sort((a, b) => {
if (a === 3) return -1;
else if (b === 3) return 0;
return a - b;
});
console.log(arr);
[ 3, 1, 2, 4, 5 ]
We are also be able to sort an object array.
const objArr = [
{
name: "Bob",
age: 30,
},
{
name: "Delta",
age: 23,
},
{
name: "Charlie",
age: 25,
},
{
name: "Echo",
age: 29,
},
{
name: "Charlie",
age: 25,
},
{
name: "Alice",
age: 20,
},
];
Let's try to sort it by age in ascending.
const objArr = [
{
name: "Bob",
age: 30,
},
{
name: "Delta",
age: 23,
},
{
name: "Charlie",
age: 25,
},
{
name: "Echo",
age: 29,
},
{
name: "Alice",
age: 20,
},
];
objArr.sort((a, b) => {
return a.age - b.age;
});
console.log(objArr);
[
{ name: 'Alice', age: 20 },
{ name: 'Delta', age: 23 },
{ name: 'Charlie', age: 25 },
{ name: 'Echo', age: 29 },
{ name: 'Bob', age: 30 }
]
We can sort it by string using String.localeCompare.
- The localeCompare() method returns a number indicating whether a reference string comes before, or after, or is the same as the given string in sort order.
You can check out more details about the function in the document.
For the first, let's see results of the function.
console.log("a".localeCompare("a"));
console.log("a".localeCompare("b"));
console.log("b".localeCompare("a"));
0
-1
1
When the parameter comes after the string, it returns -1
.
In other words, when the string comes before the parameter, it returns -1
.
Just pick two names that has to be switched.
I picked Charlie
and Delta
.
To switch these, we have to return -1.
.
The element that has Charlie
comes after the element that has Delta
, right?
So, a
will be an element that has Charlie
and b
will be an element that has Delta
.
"Charlie".localeCompare("Delta");
It will return -1
.
It means
a.name.localeCompare(b.name);
This will be our compareFn
function.
Okay, let's sort the object array by name
.
const objArr = [
{
name: "Bob",
age: 30,
},
{
name: "Delta",
age: 23,
},
{
name: "Charlie",
age: 25,
},
{
name: "Echo",
age: 29,
},
{
name: "Alice",
age: 20,
},
];
objArr.sort((a, b) => {
return a.name.localeCompare(b.name);
});
console.log(objArr);
[
{ name: 'Alice', age: 20 },
{ name: 'Bob', age: 30 },
{ name: 'Charlie', age: 25 },
{ name: 'Delta', age: 23 },
{ name: 'Echo', age: 29 }
]
It works well.
Let's say we have to place an element that has a name Charlie
at the first.
When a
is Charlie
, we should return -1
, and when b
is Charlie
, we should keep it as consistency.
const objArr = [
{
name: "Bob",
age: 30,
},
{
name: "Delta",
age: 23,
},
{
name: "Charlie",
age: 25,
},
{
name: "Echo",
age: 29,
},
{
name: "Alice",
age: 20,
},
];
objArr.sort((a, b) => {
if (a.name === "Charlie") return -1;
if (b.name === "Charlie") return 0;
return a.name.localeCompare(b.name);
});
console.log(objArr);
[
{ name: 'Charlie', age: 25 },
{ name: 'Alice', age: 20 },
{ name: 'Bob', age: 30 },
{ name: 'Delta', age: 23 },
{ name: 'Echo', age: 29 }
]
That's it!
I hope you found it useful.
Happy Coding!
Posted on February 4, 2023
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