{{ message }}

# SliceTricks

Ian Lance Taylor edited this page Sep 13, 2021 · 75 revisions
##### Clone this wiki locally

Since the introduction of the `append` built-in, most of the functionality of the `container/vector` package, which was removed in Go 1, can be replicated using `append` and `copy`.

Here are the vector methods and their slice-manipulation analogues:

#### AppendVector

`a = append(a, b...)`

#### Copy

```b := make([]T, len(a))
copy(b, a)

// These two are often a little slower than the above one,
// but they would be more efficient if there are more
// elements to be appended to b after copying.
b = append([]T(nil), a...)
b = append(a[:0:0], a...)

// This one-line implementation is equivalent to the above
// two-line make+copy implementation logically. But it is
// actually a bit slower (as of Go toolchain v1.16).
b = append(make([]T, 0, len(a)), a...)```

#### Cut

`a = append(a[:i], a[j:]...)`

#### Delete

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

#### Delete without preserving order

```a[i] = a[len(a)-1]
a = a[:len(a)-1]```

NOTE If the type of the element is a pointer or a struct with pointer fields, which need to be garbage collected, the above implementations of `Cut` and `Delete` have a potential memory leak problem: some elements with values are still referenced by slice `a` and thus can not be collected. The following code can fix this problem:

Cut

```copy(a[i:], a[j:])
for k, n := len(a)-j+i, len(a); k < n; k++ {
a[k] = nil // or the zero value of T
}
a = a[:len(a)-j+i]```

Delete

```copy(a[i:], a[i+1:])
a[len(a)-1] = nil // or the zero value of T
a = a[:len(a)-1]```

Delete without preserving order

```a[i] = a[len(a)-1]
a[len(a)-1] = nil
a = a[:len(a)-1]```

#### Expand

`a = append(a[:i], append(make([]T, j), a[i:]...)...)`

#### Extend

`a = append(a, make([]T, j)...)`

#### Filter (in place)

```n := 0
for _, x := range a {
if keep(x) {
a[n] = x
n++
}
}
a = a[:n]```

#### Insert

`a = append(a[:i], append([]T{x}, a[i:]...)...)`

NOTE: The second `append` creates a new slice with its own underlying storage and copies elements in `a[i:]` to that slice, and these elements are then copied back to slice `a` (by the first `append`). The creation of the new slice (and thus memory garbage) and the second copy can be avoided by using an alternative way:

Insert

```s = append(s, 0 /* use the zero value of the element type */)
copy(s[i+1:], s[i:])
s[i] = x```

#### InsertVector

```a = append(a[:i], append(b, a[i:]...)...)

// The above one-line way copies a[i:] twice and
// allocates at least once.
// The following verbose way only copies elements
// in a[i:] once and allocates at most once.
// But, as of Go toolchain 1.16, due to lacking of
// optimizations to avoid elements clearing in the
// "make" call, the verbose way is not always faster.
//
// Future compiler optimizations might implement
// both in the most efficient ways.
//
// Assume element type is int.
func Insert(s []int, k int, vs ...int) []int {
if n := len(s) + len(vs); n <= cap(s) {
s2 := s[:n]
copy(s2[k+len(vs):], s[k:])
copy(s2[k:], vs)
return s2
}
s2 := make([]int, len(s) + len(vs))
copy(s2, s[:k])
copy(s2[k:], vs)
copy(s2[k+len(vs):], s[k:])
return s2
}

a = Insert(a, i, b...)```

#### Push

`a = append(a, x)`

#### Pop

`x, a = a[len(a)-1], a[:len(a)-1]`

#### Push Front/Unshift

`a = append([]T{x}, a...)`

#### Pop Front/Shift

`x, a = a[0], a[1:]`

### Filtering without allocating

This trick uses the fact that a slice shares the same backing array and capacity as the original, so the storage is reused for the filtered slice. Of course, the original contents are modified.

```b := a[:0]
for _, x := range a {
if f(x) {
b = append(b, x)
}
}```

For elements which must be garbage collected, the following code can be included afterwards:

```for i := len(b); i < len(a); i++ {
a[i] = nil // or the zero value of T
}```

### Reversing

To replace the contents of a slice with the same elements but in reverse order:

```for i := len(a)/2-1; i >= 0; i-- {
opp := len(a)-1-i
a[i], a[opp] = a[opp], a[i]
}```

The same thing, except with two indices:

```for left, right := 0, len(a)-1; left < right; left, right = left+1, right-1 {
a[left], a[right] = a[right], a[left]
}```

### Shuffling

Fisher–Yates algorithm:

Since go1.10, this is available at math/rand.Shuffle

```for i := len(a) - 1; i > 0; i-- {
j := rand.Intn(i + 1)
a[i], a[j] = a[j], a[i]
}```

### Batching with minimal allocation

Useful if you want to do batch processing on large slices.

```actions := []int{0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
batchSize := 3
batches := make([][]int, 0, (len(actions) + batchSize - 1) / batchSize)

for batchSize < len(actions) {
actions, batches = actions[batchSize:], append(batches, actions[0:batchSize:batchSize])
}
batches = append(batches, actions)```

Yields the following:

`[[0 1 2] [3 4 5] [6 7 8] [9]]`

### In-place deduplicate (comparable)

```import "sort"

in := []int{3,2,1,4,3,2,1,4,1} // any item can be sorted
sort.Ints(in)
j := 0
for i := 1; i < len(in); i++ {
if in[j] == in[i] {
continue
}
j++
// preserve the original data
// in[i], in[j] = in[j], in[i]
// only set what is required
in[j] = in[i]
}
result := in[:j+1]
fmt.Println(result) // [1 2 3 4]```

### Move to front, or prepend if not present, in place if possible.

```// moveToFront moves needle to the front of haystack, in place if possible.
func moveToFront(needle string, haystack []string) []string {
if len(haystack) != 0 && haystack[0] == needle {
return haystack
}
prev := needle
for i, elem := range haystack {
switch {
case i == 0:
haystack[0] = needle
prev = elem
case elem == needle:
haystack[i] = prev
return haystack
default:
haystack[i] = prev
prev = elem
}
}
return append(haystack, prev)
}

haystack := []string{"a", "b", "c", "d", "e"} // [a b c d e]
haystack = moveToFront("c", haystack)         // [c a b d e]
haystack = moveToFront("f", haystack)         // [f c a b d e]```

### Sliding Window

```func slidingWindow(size int, input []int) [][]int {
// returns the input slice as the first element
if len(input) <= size {
return [][]int{input}
}

// allocate slice at the precise size we need
r := make([][]int, 0, len(input)-size+1)

for i, j := 0, size; j <= len(input); i, j = i+1, j+1 {
r = append(r, input[i:j])
}

return r
}```