Delete element in a slice

Go

Go Problem Overview


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?

Go Solutions


Solution 1 - Go

Where a is the slice, and i is the index of the element you want to delete:

a = append(a[:i], a[i+1:]...)

... is syntax for variadic arguments in Go.

Basically, when defining a function it puts all the arguments that you pass into one slice of that type. By doing that, you can pass as many arguments as you want (for example, fmt.Println can take as many arguments as you want).

Now, when calling a function, ... does the opposite: it unpacks a slice and passes them as separate arguments to a variadic function.

So what this line does:

a = append(a[:0], a[1:]...)

is essentially:

a = append(a[:0], a[1], a[2])

Now, you may be wondering, why not just do

a = append(a[1:]...)

Well, the function definition of append is

func append(slice []Type, elems ...Type) []Type

So the first argument has to be a slice of the correct type, the second argument is the variadic, so we pass in an empty slice, and then unpack the rest of the slice to fill in the arguments.

Solution 2 - Go

There are two options:

A: You care about retaining array order:

a = append(a[:i], a[i+1:]...)
// or
a = a[:i+copy(a[i:], a[i+1:])]

B: You don't care about retaining order (this is probably faster):

a[i] = a[len(a)-1] // Replace it with the last one. CAREFUL only works if you have enough elements.
a = a[:len(a)-1]   // Chop off the last one.

See the link to see implications re memory leaks if your array is of pointers.

https://github.com/golang/go/wiki/SliceTricks

Solution 3 - Go

Rather than thinking of the indices in the [a:]-, [:b]- and [a:b]-notations as element indices, think of them as the indices of the gaps around and between the elements, starting with gap indexed 0 before the element indexed as 0.

enter image description here

Looking at just the blue numbers, it's much easier to see what is going on: [0:3] encloses everything, [3:3] is empty and [1:2] would yield {"B"}. Then [a:] is just the short version of [a:len(arrayOrSlice)], [:b] the short version of [0:b] and [:] the short version of [0:len(arrayOrSlice)]. The latter is commonly used to turn an array into a slice when needed.

Solution 4 - Go

... is syntax for variadic arguments.

I think it is implemented by the complier using slice ([]Type), just like the function append :

func append(slice []Type, elems ...Type) []Type

when you use "elems" in "append", actually it is a slice([]type). So "a = append(a[:0], a[1:]...)" means "a = append(a[0:0], a[1:])"

a[0:0] is a slice which has nothing

a[1:] is "Hello2 Hello3"

This is how it works

Solution 5 - Go

I'm getting an index out of range error with the accepted answer solution. Reason: When range start, it is not iterate value one by one, it is iterate by index. If you modified a slice while it is in range, it will induce some problem.

Old Answer:

chars := []string{"a", "a", "b"}

for i, v := range chars {
    fmt.Printf("%+v, %d, %s\n", chars, i, v)
    if v == "a" {
        chars = append(chars[:i], chars[i+1:]...)
    }
}
fmt.Printf("%+v", chars)

Expected :

[a a b], 0, a
[a b], 0, a
[b], 0, b
Result: [b]

Actual:

// Autual
[a a b], 0, a
[a b], 1, b
[a b], 2, b
Result: [a b]

Correct Way (Solution):

chars := []string{"a", "a", "b"}

for i := 0; i < len(chars); i++ {
    if chars[i] == "a" {
        chars = append(chars[:i], chars[i+1:]...)
        i-- // form the remove item index to start iterate next item
    }
}

fmt.Printf("%+v", chars)

Source: https://dinolai.com/notes/golang/golang-delete-slice-item-in-range-problem.html

Solution 6 - Go

Not an elegant approach but it works. (Using strings.Replace)

    var a = []string{"a", "b", "c", "d"}
	str := strings.Join(a, `,`)
	str2 := "a" // element to be removed

	arr := strings.Split(strings.Trim(strings.Replace(str, str2, "", -1), ","), ",")

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionJorge OliveroView Question on Stackoverflow
Solution 1 - GodaveView Answer on Stackoverflow
Solution 2 - GoChrisView Answer on Stackoverflow
Solution 3 - GoZylView Answer on Stackoverflow
Solution 4 - Gofrank.linView Answer on Stackoverflow
Solution 5 - GotimotewView Answer on Stackoverflow
Solution 6 - GobitsabhiView Answer on Stackoverflow