How to remove items from a slice while ranging over it? - go

What is the best way to remove items from a slice while ranging over it?
For example:
type MultiDataPoint []*DataPoint
func (m MultiDataPoint) Json() ([]byte, error) {
for i, d := range m {
err := d.clean()
if ( err != nil ) {
//Remove the DP from m
}
}
return json.Marshal(m)
}

As you have mentioned elsewhere, you can allocate new memory block and copy only valid elements to it. However, if you want to avoid the allocation, you can rewrite your slice in-place:
i := 0 // output index
for _, x := range s {
if isValid(x) {
// copy and increment index
s[i] = x
i++
}
}
// Prevent memory leak by erasing truncated values
// (not needed if values don't contain pointers, directly or indirectly)
for j := i; j < len(s); j++ {
s[j] = nil
}
s = s[:i]
Full example: http://play.golang.org/p/FNDFswPeDJ
Note this will leave old values after index i in the underlying array, so this will leak memory until the slice itself is garbage collected, if values are or contain pointers. You can solve this by setting all values to nil or the zero value from i until the end of the slice before truncating it.

I know its answered long time ago but i use something like this in other languages, but i don't know if it is the golang way.
Just iterate from back to front so you don't have to worry about indexes that are deleted. I am using the same example as Adam.
m = []int{3, 7, 2, 9, 4, 5}
for i := len(m)-1; i >= 0; i-- {
if m[i] < 5 {
m = append(m[:i], m[i+1:]...)
}
}

There might be better ways, but here's an example that deletes the even values from a slice:
m := []int{1,2,3,4,5,6}
deleted := 0
for i := range m {
j := i - deleted
if (m[j] & 1) == 0 {
m = m[:j+copy(m[j:], m[j+1:])]
deleted++
}
}
Note that I don't get the element using the i, d := range m syntax, since d would end up getting set to the wrong elements once you start deleting from the slice.

Here is a more idiomatic Go way to remove elements from slices.
temp := s[:0]
for _, x := range s {
if isValid(x) {
temp = append(temp, x)
}
}
s = temp
Playground link: https://play.golang.org/p/OH5Ymsat7s9
Note: The example and playground links are based upon #tomasz's answer https://stackoverflow.com/a/20551116/12003457

One other option is to use a normal for loop using the length of the slice and subtract 1 from the index each time a value is removed. See the following example:
m := []int{3, 7, 2, 9, 4, 5}
for i := 0; i < len(m); i++ {
if m[i] < 5 {
m = append(m[:i], m[i+1:]...)
i-- // -1 as the slice just got shorter
}
}
I don't know if len() uses enough resources to make any difference but you could also run it just once and subtract from the length value too:
m := []int{3, 7, 2, 9, 4, 5}
for i, s := 0, len(m); i < s; i++ {
if m[i] < 5 {
m = append(m[:i], m[i+1:]...)
s--
i--
}
}

Something like:
m = append(m[:i], m[i+1:]...)

You don't even need to count backwards but you do need to check that you're at the end of the array where the suggested append() will fail. Here's an example of removing duplicate positive integers from a sorted list:
// Remove repeating numbers
numbers := []int{1, 2, 3, 3, 4, 5, 5}
log.Println(numbers)
for i, numbersCount, prevNum := 0, len(numbers), -1; i < numbersCount; numbersCount = len(numbers) {
if numbers[i] == prevNum {
if i == numbersCount-1 {
numbers = numbers[:i]
} else {
numbers = append(numbers[:i], numbers[i+1:]...)
}
continue
}
prevNum = numbers[i]
i++
}
log.Println(numbers)
Playground: https://play.golang.org/p/v93MgtCQsaN

I just implement a method which removes all nil elements in slice.
And I used it to solve a leetcode problems, it works perfectly.
/**
* Definition for singly-linked list.
* type ListNode struct {
* Val int
* Next *ListNode
* }
*/
func removeNil(lists *[]*ListNode) {
for i := 0; i < len(*lists); i++ {
if (*lists)[i] == nil {
*lists = append((*lists)[:i], (*lists)[i+1:]...)
i--
}
}
}

You can avoid memory leaks, as suggested in #tomasz's answer, controlling the capacity of the underlying array with a full slice expression. Look at the following function that remove duplicates from a slice of integers:
package main
import "fmt"
func removeDuplicates(a []int) []int {
for i, j := 0, 1; i < len(a) && j < len(a); i, j = i+1, j+1 {
if a[i] == a[j] {
copy(a[j:], a[j+1:])
// resize the capacity of the underlying array using the "full slice expression"
// a[low : high : max]
a = a[: len(a)-1 : len(a)-1]
i--
j--
}
}
return a
}
func main() {
a := []int{2, 3, 3, 3, 6, 9, 9}
fmt.Println(a)
a = removeDuplicates(a)
fmt.Println(a)
}
// [2 3 3 3 6 9 9]
// [2 3 6 9]

For reasons #tomasz has explained, there are issues with removing in place. That's why it is practice in golang not to do that, but to reconstruct the slice. So several answers go beyond the answer of #tomasz.
If elements should be unique, it's practice to use the keys of a map for this. I like to contribute an example of deletion by use of a map.
What's nice, the boolean values are available for a second purpose. In this example I calculate Set a minus Set b. As Golang doesn't have a real set, I make sure the output is unique. I use the boolean values as well for the algorithm.
The map gets close to O(n). I don't know the implementation. append() should be O(n). So the runtime is similar fast as deletion in place. Real deletion in place would cause a shifting of the upper end to clean up. If not done in batch, the runtime should be worse.
In this special case, I also use the map as a register, to avoid a nested loop over Set a and Set b to keep the runtime close to O(n).
type Set []int
func differenceOfSets(a, b Set) (difference Set) {
m := map[int]bool{}
for _, element := range a {
m[element] = true
}
for _, element := range b {
if _, registered := m[element]; registered {
m[element] = false
}
}
for element, present := range m {
if present {
difference = append(difference, element)
}
}
return difference
}

Try Sort and Binary search.
Example:
package main
import (
"fmt"
"sort"
)
func main() {
// Our slice.
s := []int{3, 7, 2, 9, 4, 5}
// 1. Iterate over it.
for i, v := range s {
func(i, v int) {}(i, v)
}
// 2. Sort it. (by whatever condition of yours)
sort.Slice(s, func(i, j int) bool {
return s[i] < s[j]
})
// 3. Cut it only once.
i := sort.Search(len(s), func(i int) bool { return s[i] >= 5 })
s = s[i:]
// That's it!
fmt.Println(s) // [5 7 9]
}
https://play.golang.org/p/LnF6o0yMJGT

Related

Deleting multiple values from a map in Go at the same time within a loop

I'm trying to delete multiple values from my map[string][]interface{}
I am using the strings.Split function to separate each value i wish to delete, and then looping through them.
I have managed to get it so i can delete index values 0 and 1, however, 1,2 would delete index value 1, but error on index 2.
I have also managed to get it to delete a single value
My thought process was that if I can get it to delete just one value (any index i enter, inc first and last index), then i could use a loop to loop through, and delete the rest.
Everything is stored in the below:
package db
var DataStore map[string][]interface{}
The function
func HandleDelete(w http.ResponseWriter, k, v string) {
It takes the value you wish to delete in as a parameter (and the key, but that's fully functional)
The block of code the issue resides in
The loop starts at the end of the map slice, so when you remove index value 5 for example, 4 is still 4. Whereas if I go the other way, if i delete 5, 6 then becomes index 5. So 5,6 being deleted would effectively mean 5,7 is being deleted.
for i := len(db.DataStore) - 1; i >= 0; i-- {
for _, idxvalue := range values {
val, err := strconv.Atoi(idxvalue)
if err != nil {
log.Fatal(err)
return
}
dbval := db.DataStore[k][val]
if i == val {
if len(db.DataStore[k])-1 == i { //the length goes 1,2,3,4... the index goes 0,1,2,3 - therefore length -1, would be 3 - deletes the last index value
db.DataStore[k] = db.DataStore[k][:i]
} else { //delete everything else
db.DataStore[k] = append(db.DataStore[k][:i], db.DataStore[k][i+1:]...)
}
//when you delete the last value in that key, delete the key.
/*if len(db.DataStore[k]) == 0 {
delete(db.DataStore, k)
}*/
fmt.Fprintf(w, "Key: %v, Value: %v was deleted successfully", k, dbval)
}
}
}
I have tried both loops as below:
for i := len(db.DataStore) - 1; i >= 0; i-- {
Of course the reason the below didn't work, is because you're getting the length, before the loop (in the func body) which won't change after each iteration.
idx := len(db.DataStore) - 1
for i := idx; i >= 0; i-- {
The below code is to delete the index entered (this works with a single value)
if len(db.DataStore[k])-1 == i { //the length goes 1,2,3,4... the index goes 0,1,2,3 - therefore length -1, would be 3 - deletes the last index value
db.DataStore[k] = db.DataStore[k][:i]
} else { //delete everything else
db.DataStore[k] = append(db.DataStore[k][:i], db.DataStore[k][i+1:]...)
}
I expect the out put of '2,1' to delete index 1 and 2, but the actual input is that it just deletes index 1.
For example,
package main
import "fmt"
// Delete m k v elements indexed by d.
func deleteMKVD(m map[string][]interface{}, k string, d []int) {
v, ok := m[k]
if !ok {
return
}
for _, i := range d {
if 0 <= i && i < len(v) {
v[i] = nil
}
}
lw := 0
for i := range v {
if v[i] != nil {
lw++
}
}
if lw == 0 {
delete(m, k)
return
}
w := make([]interface{}, 0, lw)
for i := range v {
if v[i] != nil {
w = append(w, v[i])
}
}
m[k] = w
}
func main() {
m := map[string][]interface{}{
"k0": {"v0", "v1", "v2", "v3"},
}
fmt.Println(m)
deleteMKVD(m, "k0", []int{0, 3})
fmt.Println(m)
deleteMKVD(m, "k0", []int{1, 0})
fmt.Println(m)
}
Playground: https://play.golang.org/p/biEAxthTaj8
Output:
map[k0:[v0 v1 v2 v3]]
map[k0:[v1 v2]]
map[]
I think your problem is actually to remove elements from an array with an array of indices.
The easy fix here would be:
1) Find all the indices with certain k, make it an array(vals []int).
2) Sort this array int descendent. and iterate this array to delete
3) Then iterate this array to delete the elements.
In this way, every time you delete an element, it won't touch other elements' indices.
It may not be most efficient, but it would be a quick fix.
BTW, I think for i := len(db.DataStore) - 1; i >= 0; i--is not what you want.
If I understand correctly, this code here seems to make sure the val is the largest index in those indices.
So instead of write i:=len(db.DataStore) - 1, you actually need i:=len(db.DataStore[k])-1

Helper function to chunk any slice? [duplicate]

This question already has answers here:
Slice chunking in Go
(8 answers)
Closed 8 months ago.
I created a small helper function to split a large array of items into smaller arrays with a maximum size of n.
func toPackages(e []int, n int) [][]int {
var p [][]int
packets := int(math.Ceil(float64(len(e)) / float64(n)))
for i := 0; i < packets; i++ {
start := i * n
end := n * (i + 1)
if len(e) < end {
end = len(e)
}
p = append(p, e[start:end])
}
return p
}
Working example at Golang Playground.
In the program I have several different types of arrays I would like to split. I have tried converting it to using interfaces with interface{}.
It is pretty hard to make a generic function to handle this well. You will often spend as much code converting []int to []interface{} and back, as it is to just copy the snippet. I do have a slightly nicer way to do it though:
playground link
func splitInts(src []int, n int) (p [][]int){
for len(src) > n{
p = append(p,src[:n])
src = src[n:]
}
if(len(src) > 0){
p = append(p,src)
}
return p
}
Nothing in the function changes because of types, it can easily be copied to:
func splitStrings(src []string, n int) (p [][]string){
for len(src) > n{
p = append(p,src[:n])
src = src[n:]
}
if(len(src) > 0){
p = append(p,src)
}
return p
}
By only changing the first line.
Generally it is a good idea to just write the function for every type you need it for. This is also a matter of speed: if you use reflect to write a generic function it will not be as fast.
If you still want a generic function here it is:
func genToPackages(e interface{}, n int) interface{} {
t := reflect.TypeOf(e)
if t.Kind() != reflect.Slice {
log.Println("e has to be a slice")
}
v := reflect.ValueOf(e)
packets := int(math.Ceil(float64(v.Len()) / float64(n)))
p := reflect.MakeSlice(reflect.SliceOf(t), packets, packets)
for i := 0; i < packets; i++ {
s := reflect.MakeSlice(t, n, n)
start := i * n
for j := 0; j < n; j++ {
s.Index(j).Set(v.Index(j+start))
}
p.Index(i).Set(s)
}
return p.Interface()
}
You will have to cast the result to the type you expect. For example:
res := genToPackages([]int{1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, 5).([][]int)

Golang: Find two number index where the sum of these two numbers equals to target number

The problem is: find the index of two numbers that nums[index1] + nums[index2] == target. Here is my attempt in golang (index starts from 1):
package main
import (
"fmt"
)
var nums = []int{0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 25182, 25184, 25186, 25188, 25190, 25192, 25194, 25196} // The number list is too long, I put the whole numbers in a gist: https://gist.github.com/nickleeh/8eedb39e008da8b47864
var target int = 16021
func twoSum(nums []int, target int) (int, int) {
if len(nums) <= 1 {
return 0, 0
}
hdict := make(map[int]int)
for i := 1; i < len(nums); i++ {
if val, ok := hdict[nums[i+1]]; ok {
return val, i + 1
} else {
hdict[target-nums[i+1]] = i + 1
}
}
return 0, 0
}
func main() {
fmt.Println(twoSum(nums, target))
}
The nums list is too long, I put it into a gist:
https://gist.github.com/nickleeh/8eedb39e008da8b47864
This code works fine, but I find the return 0,0 part is ugly, and it runs ten times slower than the Julia translation. I would like to know is there any part that is written terrible and affect the performance?
Edit:
Julia's translation:
function two_sum(nums, target)
if length(nums) <= 1
return false
end
hdict = Dict()
for i in 1:length(nums)
if haskey(hdict, nums[i])
return [hdict[nums[i]], i]
else
hdict[target - nums[i]] = i
end
end
end
In my opinion if no elements found adding up to target, best would be to return values which are invalid indices, e.g. -1. Although returning 0, 0 would be enough as a valid index pair can't be 2 equal indices, this is more convenient (because if you forget to check the return values and you attempt to use the invalid indices, you will immediately get a run-time panic, alerting you not to forget checking the validity of the return values). As so, in my solutions I will get rid of that i + 1 shifts as it makes no sense.
Benchmarking of different solutions can be found at the end of the answer.
If sorting allowed:
If the slice is big and not changing, and you have to call this twoSum() function many times, the most efficient solution would be to sort the numbers simply using sort.Ints() in advance:
sort.Ints(nums)
And then you don't have to build a map, you can use binary search implemented in sort.SearchInts():
func twoSumSorted(nums []int, target int) (int, int) {
for i, v := range nums {
v2 := target - v
if j := sort.SearchInts(nums, v2); v2 == nums[j] {
return i, j
}
}
return -1, -1
}
Note: Note that after sorting, the indices returned will be indices of values in the sorted slice. This may differ from indices in the original (unsorted) slice (which may or may not be a problem). If you do need indices from the original order (original, unsorted slice), you may store sorted and unsorted index mapping so you can get what the original index is. For details see this question:
Get the indices of the array after sorting in golang
If sorting is not allowed:
Here is your solution getting rid of that i + 1 shifts as it makes no sense. Slice and array indices are zero based in all languages. Also utilizing for ... range:
func twoSum(nums []int, target int) (int, int) {
if len(nums) <= 1 {
return -1, -1
}
m := make(map[int]int)
for i, v := range nums {
if j, ok := m[v]; ok {
return j, i
}
m[target-v] = i
}
return -1, -1
}
If the nums slice is big and the solution is not found fast (meaning the i index grows big) that means a lot of elements will be added to the map. Maps start with small capacity, and they are internally grown if additional space is required to host many elements (key-value pairs). An internal growing requires rehashing and rebuilding with the already added elements. This is "very" expensive.
It does not seem significant but it really is. Since you know the max elements that will end up in the map (worst case is len(nums)), you can create a map with a big-enough capacity to hold all elements for the worst case. The gain will be that no internal growing and rehashing will be required. You can provide the initial capacity as the second argument to make() when creating the map. This speeds up twoSum2() big time if nums is big:
func twoSum2(nums []int, target int) (int, int) {
if len(nums) <= 1 {
return -1, -1
}
m := make(map[int]int, len(nums))
for i, v := range nums {
if j, ok := m[v]; ok {
return j, i
}
m[target-v] = i
}
return -1, -1
}
Benchmarking
Here's a little benchmarking code to test execution speed of the 3 solutions with the input nums and target you provided. Note that in order to test twoSumSorted(), you first have to sort the nums slice.
Save this into a file named xx_test.go and run it with go test -bench .:
package main
import (
"sort"
"testing"
)
func BenchmarkTwoSum(b *testing.B) {
for i := 0; i < b.N; i++ {
twoSum(nums, target)
}
}
func BenchmarkTwoSum2(b *testing.B) {
for i := 0; i < b.N; i++ {
twoSum2(nums, target)
}
}
func BenchmarkTwoSumSorted(b *testing.B) {
sort.Ints(nums)
b.ResetTimer()
for i := 0; i < b.N; i++ {
twoSumSorted(nums, target)
}
}
Output:
BenchmarkTwoSum-4 1000 1405542 ns/op
BenchmarkTwoSum2-4 2000 722661 ns/op
BenchmarkTwoSumSorted-4 10000000 133 ns/op
As you can see, making a map with big enough capacity speeds up: it runs twice as fast.
And as mentioned, if nums can be sorted in advance, that is ~10,000 times faster!
If nums is always sorted, you can do a binary search to see if the complement to whichever number you're on is also in the slice.
func binary(haystack []int, needle, startsAt int) int {
pivot := len(haystack) / 2
switch {
case haystack[pivot] == needle:
return pivot + startsAt
case len(haystack) <= 1:
return -1
case needle > haystack[pivot]:
return binary(haystack[pivot+1:], needle, startsAt+pivot+1)
case needle < haystack[pivot]:
return binary(haystack[:pivot], needle, startsAt)
}
return -1 // code can never fall off here, but the compiler complains
// if you don't have any returns out of conditionals.
}
func twoSum(nums []int, target int) (int, int) {
for i, num := range nums {
adjusted := target - num
if j := binary(nums, adjusted, 0); j != -1 {
return i, j
}
}
return 0, 0
}
playground example
Or you can use sort.SearchInts which implements binary searching.
func twoSum(nums []int, target int) (int, int) {
for i, num := range nums {
adjusted := target - num
if j := sort.SearchInts(nums, adjusted); nums[j] == adjusted {
// sort.SearchInts returns the index where the searched number
// would be if it was there. If it's not, then nums[j] != adjusted.
return i, j
}
}
return 0, 0
}

What is wrong with the following merge sort algorithm?

As the question states, I'm having trouble finding where is the issue within the following algorithm. It is the aux function for mergesort, i.e. the one used for combining sorted arrays.
func Merge(toSort *[]int, p, q, r int) {
arr := *toSort
L := arr[p:q]
R := arr[q:r+1]
fmt.Println(L)
fmt.Println(R)
i := 0
j := 0
for index := p; index <= r; index++ {
if i >= len(L) {
arr[index] = R[j]
j += 1
continue
} else if j >= len(R) {
arr[index] = L[i]
i += 1
continue
}
if L[i] > R[j] {
fmt.Println("right smaller")
arr[index] = R[j]
j += 1
continue
}
if L[i] <= R[j] {
fmt.Println("left smaller")
arr[index] = L[i]
i += 1
continue
}
}
}
For arr := []int{1,7,14,15,44,65,79,2,3,6,55,70} it gives as output [1 2 2 2 2 2 2 2 3 6 55 70].
Golang Play link
The JavaScript equivalent for this function works as expected, but I don't know why it isn't working in Go
Thank you
Golang slices are passed by reference. So you don't need to pass a pointer into the function in the first place, but you do need to take explicit copies of L and R or else merge into a different slice entirely. You are currently writing into the same underlying memory from which you are getting your values.
Code like L := arr[p:q] does not create a copy. I suppose you are overwriting your L and R parts during the assignments to arr. Have a look at http://blog.golang.org/slices to understand how slices work. (E.g. you'll basically never write stuff like toSort *[]int as []int is almost kinda pointer)
This seems to work: http://play.golang.org/p/vPo2ZKXtI9
You don't need all the indexes: slices are already views into an array. Here's a complete example using purely slice manipulation:
package main
import "fmt"
// Merge takes two sorted, increasing slices of ints and
// returns a slice combining them into a single sorted, increasing
// slice.
func Merge(a, b []int) []int {
res := make([]int, 0, len(a)+len(b))
for len(a) > 0 || len(b) > 0 {
if len(b) == 0 || len(a) > 0 && a[0] <= b[0] {
res = append(res, a[0])
a = a[1:]
} else {
res = append(res, b[0])
b = b[1:]
}
}
return res
}
func main() {
a := []int{1, 2, 5, 6, 3, 4, 7, 9}
fmt.Println(Merge(a[:4], a[4:]))
}

How do I reverse an array in Go?

http://play.golang.org/p/W70J4GU7nA
s := []int{5, 2, 6, 3, 1, 4}
sort.Reverse(sort.IntSlice(s))
fmt.Println(s)
// 5, 2, 6, 3, 1, 4
It is hard to understand what it means in func Reverse(data Interface) Interface .
How do I reverse an array? I do not need to sort.
Honestly this one is simple enough that I'd just write it out like this:
package main
import "fmt"
func main() {
s := []int{5, 2, 6, 3, 1, 4}
for i, j := 0, len(s)-1; i < j; i, j = i+1, j-1 {
s[i], s[j] = s[j], s[i]
}
fmt.Println(s)
}
http://play.golang.org/p/vkJg_D1yUb
(The other answers do a good job of explaining sort.Interface and how to use it; so I won't repeat that.)
Normally, to sort an array of integers you wrap them in an IntSlice, which defines the methods Len, Less, and Swap. These methods are in turn used by sort.Sort. What sort.Reverse does is that it takes an existing type that defines Len, Less, and Swap, but it replaces the Less method with a new one that is always the inverse of the underlying Less:
type reverse struct {
// This embedded Interface permits Reverse to use the methods of
// another Interface implementation.
Interface
}
// Less returns the opposite of the embedded implementation's Less method.
func (r reverse) Less(i, j int) bool {
return r.Interface.Less(j, i)
}
// Reverse returns the reverse order for data.
func Reverse(data Interface) Interface {
return &reverse{data}
}
So when you write sort.Reverse(sort.IntSlice(s)), whats happening is that you're getting this new, 'modified' IntSlice that has it's Less method replaced. So if you call sort.Sort on it, which calls Less, it will get sorted in decreasing order.
I'm 2 years late, but just for fun and interest I'd like to contribute an "oddball" solution.
Assuming the task really is to reverse a list, then for raw performance bgp's solution is probably unbeatable. It gets the job done simply and effectively by swapping array items front to back, an operation that's efficient in the random-access structure of arrays and slices.
In Functional Programming languages, the idiomatic approach would often involve recursion. This looks a bit strange in Go and will have atrocious performance. That said, here's a recursive array reversal function (in a little test program):
package main
import (
"fmt"
)
func main() {
myInts := []int{ 8, 6, 7, 5, 3, 0, 9 }
fmt.Printf("Ints %v reversed: %v\n", myInts, reverseInts(myInts))
}
func reverseInts(input []int) []int {
if len(input) == 0 {
return input
}
return append(reverseInts(input[1:]), input[0])
}
Output:
Ints [8 6 7 5 3 0 9] reversed: [9 0 3 5 7 6 8]
Again, this is for fun and not production. Not only is it slow, but it will overflow the stack if the list is too large. I just tested, and it will reverse a list of 1 million ints but crashes on 10 million.
First of all, if you want to reverse the array, do like this,
for i, j := 0, len(a)-1; i < j; i, j = i+1, j-1 {
a[i], a[j] = a[j], a[i]
}
Then, look at the usage of Reverse in golang.org
package main
import (
"fmt"
"sort"
)
func main() {
s := []int{5, 2, 6, 3, 1, 4} // unsorted
sort.Sort(sort.Reverse(sort.IntSlice(s)))
fmt.Println(s)
}
// output
// [6 5 4 3 2 1]
And look at the description of Reverse and Sort
func Reverse(data Interface) Interface
func Sort(data Interface)
Sort sorts data. It makes one call to data.Len to determine n, and O(n*log(n)) calls to data.Less and data.Swap. The sort is not guaranteed to be stable.
So, as you know, Sort is not just a sort algorithm, you can view it as a factory, when you use Reverse it just return a reversed sort algorithm, Sort is just doing the sorting.
This is a more generic slice reverse function. It will panic if input is not a slice.
//panic if s is not a slice
func ReverseSlice(s interface{}) {
size := reflect.ValueOf(s).Len()
swap := reflect.Swapper(s)
for i, j := 0, size-1; i < j; i, j = i+1, j-1 {
swap(i, j)
}
}
If you want to reverse the array, you can just go through it in reverse order. Since there is no "reverse range" primitive in the language (at least not yet), you must do something like this (http://play.golang.org/p/AhvAfMjs_7):
s := []int{5, 2, 6, 3, 1, 4}
for i := len(s) - 1; i >= 0; i-- {
fmt.Print(s[i])
if i > 0 {
fmt.Print(", ")
}
}
fmt.Println()
Regarding whether it is hard to understand what sort.Reverse(data Interface) Interface does, I thought the same until I saw the source code from "http://golang.org/src/pkg/sort/sort.go".
It just makes the comparisons required for the sorting to be made "the other way around".
Here is a simple Go solution that uses an efficient (no extra memory) approach to reverse an array:
i := 0
j := len(nums) - 1
for i < j {
nums[i], nums[j] = nums[j], nums[i]
i++
j--
}
The idea is that reversing an array is equivalent to swapping each element with its mirror image across the center.
https://play.golang.org/p/kLFpom4LH0g
Here is another way to do it
func main() {
example := []int{1, 25, 3, 5, 4}
sort.SliceStable(example, func(i, j int) bool {
return true
})
fmt.Println(example)
}
https://play.golang.org/p/-tIzPX2Ds9z
func Reverse(data Interface) Interface
This means that it takes a sort.Interface and returns another sort.Interface -- it doesn't actually doing any sorting itself. For example, if you pass in sort.IntSlice (which is essentially a []int that can be passed to sort.Sort to sort it in ascending order) you'll get a new sort.Interface which sorts the ints in descending order instead.
By the way, if you click on the function name in the documentation, it links directly to the source for Reverse. As you can see, it just wraps the sort.Interface that you pass in, so the value returned from Reverse gets all the methods of the original sort.Interface. The only method that's different is the Less method which returns the opposite of the Less method on the embedded sort.Interface. See this part of the language spec for details on embedded fields.
From Golang wiki SliceTricks:
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]
}
To reverse an array in place, iterate to its mid-point, and swap each element with its "mirror element":
func main() {
xs := []int{1, 2, 3, 4, 5, 6, 7, 8, 9}
itemCount := len(xs)
for i := 0; i < itemCount/2; i++ {
mirrorIdx := itemCount - i -1
xs[i], xs[mirrorIdx] = xs[mirrorIdx], xs[i]
}
fmt.Printf("xs: %v\n", xs)
}
https://play.golang.org/p/JeSApt80_k
Here is a method using append:
package main
import "fmt"
func main() {
a := []int{10, 20, 30, 40, 50}
for n := len(a) - 2; n >= 0; n-- {
a = append(a[:n], append(a[n + 1:], a[n])...)
}
fmt.Println(a)
}
Drawing of the steps:
10 20 30 40 50
10 20 30 50 40
10 20 50 40 30
10 50 40 30 20
50 40 30 20 10
This answer is mainly for those beginners who wish to write this code using only one variable in the for loop instead of using two variables (like i & j).
package main
import "fmt"
func main() {
array := []int{45, 17, 43, 67, 21, 4, 97, 44, 54, 98, 665}
fmt.Println("initial array:", array)
loop_iteration := len(array)
if len(array)%2 == 0 {
loop_iteration = (len(array) / 2) - 1
} else {
loop_iteration = int(len(array) / 2) //This will give the lower integer value of that float number.
}
for i := 0; i <= loop_iteration; i++ {
array[i], array[(len(array)-1)-i] = array[(len(array)-1)-i], array[i]
}
fmt.Println("reverse array:", array)
}
https://go.dev/play/p/bVp0x7v6Kbs
package main
import (
"fmt"
)
func main() {
arr := []int{1, 2, 3, 4, 5}
fmt.Println(reverseArray(arr))
}
func reverseArray(arr []int) []int {
reversed := make([]int, len(arr))
j := 0
for i := len(arr) - 1; i >= 0; i-- {
reversed[j] = arr[i]
j++
}
return reversed
}
Simple solution without involving math. Like this solution, this is inefficient as it does too much allocation and garbage collection. Good for non-critical code where clarity is more important than performance. Playground: https://go.dev/play/p/dQGwrc0Q9ZA
arr := []int{1, 3, 4, 5, 6}
var rev []int
for _, n := range arr {
rev = append([]int{n}, rev...)
}
fmt.Println(arr)
fmt.Println(rev)
Its very simple if you want to print reverse array
Use Index from length doing i--
ex.
a := []int{5, 4, 12, 7, 15, 9}
for i := 0; i <= len(a)-1; i++ {
fmt.Println(a[len(a)-(i+1)])
}
https://go.dev/play/p/bmyFh7-_VCZ
Here is my solution.
package main
import (
"fmt"
)
func main() {
var numbers = [10]int {1,2,3,4,5,6,7,8,9,10}
var reverseNumbers [10]int
j:=0
for i:=len(numbers)-1; i>=0 ; i-- {
reverseNumbers[j]=numbers[i]
j++
}
fmt.Println(reverseNumbers)
}
Here is my solution to reversing an array:
func reverse_array(array []string) []string {
lenx := len(array) // lenx holds the original array length
reversed_array := make([]string, lenx) // creates a slice that refer to a new array of length lenx
for i := 0; i < lenx; i++ {
j := lenx - (i + 1) // j initially holds (lenx - 1) and decreases to 0 while i initially holds 0 and increase to (lenx - 1)
reversed_array[i] = array[j]
}
return reversed_array
}
You can try this solution on the go playground the go playground
package main
import "fmt"
func main() {
array := []string{"a", "b", "c", "d"}
fmt.Println(reverse_array(array)) // prints [d c b a]
}
Do not reverse it, leave it as now and then just iterate it backwards.

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