Delete element in a slice

func main() {
    a := []string{"Hello1", "Hello2", "Hello3"}
    fmt.Println(a)
    // [Hello1 Hello2 Hello3]
    a = append(a[:0], a[1:]...)
    fmt.Println(a)
    // [Hello2 Hello3]
}

How does this delete trick with the append function work?

It would seem that it’s grabbing everything before the first element (empty array)

Then appending everything after the first element (position zero)

What does the … (dot dot dot) do?

Order matters

If you want to keep your array ordered, you have to shift all of the elements at the right of the deleting index by one to the left. Hopefully, this can be done easily in Golang:

func remove(slice []int, s int) []int {
    return append(slice[:s], slice[s+1:]...)
}

However, this is inefficient because you may end up with moving all of the elements, which is costy.

Order is not important

If you do not care about ordering, you have the much faster possibility to swap the element to delete with the one at the end of the slice and then return the n-1 first elements:

func remove(s []int, i int) []int {
    s[len(s)-1], s[i] = s[i], s[len(s)-1]
    return s[:len(s)-1]
}

With the reslicing method, emptying an array of 1 000 000 elements take 224s, with this one it takes only 0.06ns. I suspect that internally, go only changes the length of the slice, without modifying it.

Edit 1

Quick notes based on the comments below (thanks to them !).

As the purpose is to delete an element, when the order does not matter a single swap is needed, the second will be wasted :

func remove(s []int, i int) []int {
    s[i] = s[len(s)-1]
    // We do not need to put s[i] at the end, as it will be discarded anyway
    return s[:len(s)-1]
}

Also, this answer does not perform bounds-checking. It expects a valid index as input. This means that negative values or indices that are greater or equal to len(s) will cause Go to panic. Slices and arrays being 0-indexed, removing the n-th element of an array implies to provide input n-1. To remove the first element, call remove(s, 0), to remove the second, call remove(s, 1), and so on and so forth.