我采用下面的方法切片删除项目。这有助于提高其他人的可读性。也是不可变的。

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

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

    return newitems
}

下面是带有指针的操场示例。 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)
}

这看起来有点奇怪，但这里的大多数答案都很危险，掩盖了他们实际在做的事情。看看最初关于从切片中移除一个项目的问题切片的副本正在制作，然后被填充。这可以确保当切片在程序中传递时，不会引入微妙的错误。

这里有一些代码比较用户在这个帖子和原始帖子中的回答。这是一个可以随意处理代码的游乐场。

基于追加的删除

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.

下面的例子我就不细讲了，但它和我们的目的是一样的。只是说明了使用copy没有什么不同。

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]
}

问题的原始答案

看看最初的问题，它没有修改它要从其中删除项的切片。让这个帖子中的原始答案成为到目前为止对大多数来到这个页面的人来说最好的答案。

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]
}

正如您所看到的，这个输出的行为是大多数人所期望的，可能是大多数人想要的。对originalRemove的修改并不会导致所有的变化，删除索引并为其赋值的操作也不会导致变化!太棒了!

这段代码有点长，所以上面的代码可以改成这样。

正确答案

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]
}

几乎与原来的删除索引解决方案相同，但是我们在返回之前添加了一个新的切片。

T. Claverie目前投票最多的答案是正确的，但我发现如果只在需要时执行交换，即对片的最后一个元素以外的所有元素执行交换，算法会更清晰。这可以通过一个简单的if守卫来实现。

秩序不重要/没有进行边界检查

func remove(s []int, i int) []int {
    // bring element to remove at the end if its not there yet
    if i != len(s)-1 {
        s[i] = s[len(s)-1]
    }
 
    // drop the last element
    return s[:len(s)-1]
}

也许你可以试试这个方法:

// 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)
    }
}

用法:

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)

结果:

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

摘自《Go编程语言》这本书

从切片的中间删除一个元素，保持顺序 对于其余的元素，使用copy来滑动编号较高的元素 减少1来填补空白的元素: 删除(slice []int, I int) []int { 复制(片[我],片[i + 1:]) 返回片(:len(片)1) ｝