[go] How to delete an element from a Slice in Golang

fmt.Println("Enter position to delete::")
fmt.Scanln(&pos)

new_arr := make([]int, (len(arr) - 1))
k := 0
for i := 0; i < (len(arr) - 1); {
    if i != pos {
        new_arr[i] = arr[k]
        k++
        i++
    } else {
        k++
    }
}

for i := 0; i < (len(arr) - 1); i++ {
    fmt.Println(new_arr[i])
}

I am using this command to delete an element from a Slice but it is not working, please suggest.

This question is related to go

The answer is


Find a way here without relocating.

  • changes order
a := []string{"A", "B", "C", "D", "E"}
i := 2

// Remove the element at index i from a.
a[i] = a[len(a)-1] // Copy last element to index i.
a[len(a)-1] = ""   // Erase last element (write zero value).
a = a[:len(a)-1]   // Truncate slice.

fmt.Println(a) // [A B E D]
  • keep order
a := []string{"A", "B", "C", "D", "E"}
i := 2

// Remove the element at index i from a.
copy(a[i:], a[i+1:]) // Shift a[i+1:] left one index.
a[len(a)-1] = ""     // Erase last element (write zero value).
a = a[:len(a)-1]     // Truncate slice.

fmt.Println(a) // [A B D E]

I take the below approach to remove the item in slice. This helps in readability for others. And also immutable.

func remove(items []string, item string) []string {
    newitems := []string{}

    for _, i := range items {
        if i != item {
            newitems = append(newitems, i)
        }
    }

    return newitems
}

This is how you Delete From a slice the idiomatic way. You don't need to build a function it is built into the append. Try it here https://play.golang.org/p/QMXn9-6gU5P

z := []int{9, 8, 7, 6, 5, 3, 2, 1, 0}
fmt.Println(z)  //will print Answer [9 8 7 6 5 3 2 1 0]

z = append(z[:2], z[4:]...)
fmt.Println(z)   //will print Answer [9 8 5 3 2 1 0]

From the book The Go Programming Language

To remove an element from the middle of a slice, preserving the order of the remaining elements, use copy to slide the higher-numbered elements down by one to fill the gap:

func remove(slice []int, i int) []int {
  copy(slice[i:], slice[i+1:])
  return slice[:len(slice)-1]
}

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.


Minor point (code golf), but in the case where order does not matter you don't need to swap the values. Just overwrite the array position being removed with a duplicate of the last position and then return a truncated array.

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

Same result.


Remove one element from the Slice (this is called 're-slicing'):

package main

import (
    "fmt"
)

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

func main() {
    all := []int{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
    fmt.Println(all) //[0 1 2 3 4 5 6 7 8 9]
    all = RemoveIndex(all, 5)
    fmt.Println(all) //[0 1 2 3 4 6 7 8 9]
}

Maybe you can try this method:

// DelEleInSlice delete an element from slice by index
//  - arr: the reference of slice
//  - index: the index of element will be deleted
func DelEleInSlice(arr interface{}, index int) {
    vField := reflect.ValueOf(arr)
    value := vField.Elem()
    if value.Kind() == reflect.Slice || value.Kind() == reflect.Array {
        result := reflect.AppendSlice(value.Slice(0, index), value.Slice(index+1, value.Len()))
        value.Set(result)
    }
}

Usage:

arrInt := []int{0, 1, 2, 3, 4, 5}
arrStr := []string{"0", "1", "2", "3", "4", "5"}
DelEleInSlice(&arrInt, 3)
DelEleInSlice(&arrStr, 4)
fmt.Println(arrInt)
fmt.Println(arrStr)

Result:

0, 1, 2, 4, 5
"0", "1", "2", "3", "5"

No need to check every single element unless you care contents and you can utilize slice append. try it out

pos := 0
arr := []int{1, 2, 3, 4, 5, 6, 7, 9}
fmt.Println("input your position")
fmt.Scanln(&pos)
/* you need to check if negative input as well */
if (pos < len(arr)){
    arr = append(arr[:pos], arr[pos+1:]...)
} else {
    fmt.Println("position invalid")
}

This is a little strange to see but most answers here are dangerous and gloss over what they are actually doing. Looking at the original question that was asked about removing an item from the slice a copy of the slice is being made and then it's being filled. This ensures that as the slices are passed around your program you don't introduce subtle bugs.

Here is some code comparing users answers in this thread and the original post. Here is a go playground to mess around with this code in.

Append based removal

package main

import (
    "fmt"
)

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

func main() {
    all := []int{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
    fmt.Println("all: ", all) //[0 1 2 3 4 5 6 7 8 9]
    removeIndex := RemoveIndex(all, 5)

    fmt.Println("all: ", all) //[0 1 2 3 4 6 7 8 9 9]
    fmt.Println("removeIndex: ", removeIndex) //[0 1 2 3 4 6 7 8 9]

    removeIndex[0] = 999
    fmt.Println("all: ", all) //[999 1 2 3 4 6 7 9 9]
    fmt.Println("removeIndex: ", removeIndex) //[999 1 2 3 4 6 7 8 9]
}

In the above example you can see me create a slice and fill it manually with numbers 0 to 9. We then remove index 5 from all and assign it to remove index. However when we go to print out all now we see that it has been modified as well. This is because slices are pointers to an underlying array. Writing it out to removeIndex causes all to be modified as well with the difference being all is longer by one element that is no longer reachable from removeIndex. Next we change a value in removeIndex and we can see all gets modified as well. Effective go goes into some more detail on this.

The following example I won't go into but it does the same thing for our purposes. And just illustrates that using copy is no different.

package main

import (
    "fmt"
)

func RemoveCopy(slice []int, i int) []int {
    copy(slice[i:], slice[i+1:])
    return slice[:len(slice)-1]
}

func main() {
    all := []int{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
    fmt.Println("all: ", all) //[0 1 2 3 4 5 6 7 8 9]
    removeCopy := RemoveCopy(all, 5)

    fmt.Println("all: ", all) //[0 1 2 3 4 6 7 8 9 9]
    fmt.Println("removeCopy: ", removeCopy) //[0 1 2 3 4 6 7 8 9]

    removeCopy[0] = 999
    fmt.Println("all: ", all) //[99 1 2 3 4 6 7 9 9]
    fmt.Println("removeCopy: ", removeCopy) //[999 1 2 3 4 6 7 8 9]
}

The questions original answer

Looking at the original question it does not modify the slice that it's removing an item from. Making the original answer in this thread the best so far for most people coming to this page.

package main

import (
    "fmt"
)

func OriginalRemoveIndex(arr []int, pos int) []int {
    new_arr := make([]int, (len(arr) - 1))
    k := 0
    for i := 0; i < (len(arr) - 1); {
        if i != pos {
            new_arr[i] = arr[k]
            k++
        } else {
            k++
        }
        i++
    }

    return new_arr
}

func main() {
    all := []int{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
    fmt.Println("all: ", all) //[0 1 2 3 4 5 6 7 8 9]
    originalRemove := OriginalRemoveIndex(all, 5)

    fmt.Println("all: ", all) //[0 1 2 3 4 5 6 7 8 9]
    fmt.Println("originalRemove: ", originalRemove) //[0 1 2 3 4 6 7 8 9]

    originalRemove[0] = 999
    fmt.Println("all: ", all) //[0 1 2 3 4 5 6 7 8 9]
    fmt.Println("originalRemove: ", originalRemove) //[999 1 2 3 4 6 7 8 9]
}

As you can see this output acts as most people would expect and likely what most people want. Modification of originalRemove doesn't cause changes in all and the operation of removing the index and assigning it doesn't cause changes as well! Fantastic!

This code is a little lengthy though so the above can be changed to this.

A correct answer

package main

import (
    "fmt"
)

func RemoveIndex(s []int, index int) []int {
    ret := make([]int, 0)
    ret = append(ret, s[:index]...)
    return append(ret, s[index+1:]...)
}

func main() {
    all := []int{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
    fmt.Println("all: ", all) //[0 1 2 3 4 5 6 7 8 9]
    removeIndex := RemoveIndex(all, 5)

    fmt.Println("all: ", all) //[0 1 2 3 4 5 6 7 8 9]
    fmt.Println("removeIndex: ", removeIndex) //[0 1 2 3 4 6 7 8 9]

    removeIndex[0] = 999
    fmt.Println("all: ", all) //[0 1 2 3 4 5 6 7 9 9]
    fmt.Println("removeIndex: ", removeIndex) //[999 1 2 3 4 6 7 8 9]
}

Almost identical to the original remove index solution however we make a new slice to append to before returning.


The best way to do it is to use the append function:

package main

import (
    "fmt"
)

func main() {
    x := []int{4, 5, 6, 7, 88}
    fmt.Println(x)
    x = append(x[:2], x[4:]...)//deletes 6 and 7
    fmt.Println(x)
}

https://play.golang.org/p/-EEFCsqse4u


To remove an element from the middle of a slice, preserving the order of the remaining elements, use copy to slide the higher-numbered elements down by one to fill the gap:

func remove(slice []int, i int) []int {
  copy(slice[i:], slice[i+1:])
  return slice[:len(slice)-1]
}

If it is not necessary to preserve the order, we can simply move the last element to the gap.

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

Maybe this code will help.

It deletes item with a given index.

Takes the array, and the index to delete and returns a new array pretty much like append function.

func deleteItem(arr []int, index int) []int{
  if index < 0 || index >= len(arr){
    return []int{-1}
  }

    for i := index; i < len(arr) -1; i++{
      arr[i] = arr[i + 1]

    }

    return arr[:len(arr)-1]
}

Here you can play with the code : https://play.golang.org/p/aX1Qj40uTVs


here is the playground example with pointers in it. https://play.golang.org/p/uNpTKeCt0sH

package main

import (
    "fmt"
)

type t struct {
    a int
    b string
}

func (tt *t) String() string{
    return fmt.Sprintf("[%d %s]", tt.a, tt.b)
}

func remove(slice []*t, i int) []*t {
  copy(slice[i:], slice[i+1:])
  return slice[:len(slice)-1]
}

func main() {
    a := []*t{&t{1, "a"}, &t{2, "b"}, &t{3, "c"}, &t{4, "d"}, &t{5, "e"}, &t{6, "f"}}
    k := a[3]
    a = remove(a, 3)
    fmt.Printf("%v  ||  %v", a, k)
}