6.8 KiB
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.
Since the introduction of generics, generic implementations of several of these functions are available in the golang.org/x/exp/slices package.
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
Insert n elements at position i:
a = append(a[:i], append(make([]T, n), a[i:]...)...)
Extend
Append n elements:
a = append(a, make([]T, n)...)
Extend Capacity
Make sure there is space to append n elements without re-allocating:
if cap(a)-len(a) < n {
a = append(make([]T, 0, len(a)+n), a...)
}
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:]
Additional Tricks
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
}