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SliceTricks
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:
a = append(a, b...)
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
// tw0-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...)
a = append(a[:i], a[j:]...)
a = append(a[:i], a[i+1:]...)
// or
a = a[:i+copy(a[i:], a[i+1:])]
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]
a = append(a[:i], append(make([]T, j), a[i:]...)...)
a = append(a, make([]T, j)...)
n := 0
for _, x := range a {
if keep(x) {
a[n] = x
n++
}
}
a = a[:n]
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
a = append(a[:i], append(b, a[i:]...)...)
NOTE: To get the best efficiency, it is best to do the insertion without using append
, in particular when the number of the inserted elements is known:
// 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...)
a = append(a, x)
x, a = a[len(a)-1], a[:len(a)-1]
a = append([]T{x}, a...)
x, a = a[0], a[1:]
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
}
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]
}
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]
}
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]]
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]
// 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]
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
}