playground
package main
import (
"fmt"
"math/rand"
)
func randoms() *[]int {
var nums []int = make([]int, 5, 5) //Created slice with fixed Len, cap
fmt.Println(len(nums))
for i := range [5]int{} {//Added random numbers.
nums[i] = rand.Intn(10)
}
return &nums//Returning pointer to the slice
}
func main() {
fmt.Println("Hello, playground")
var nums []int = make([]int, 0, 25)
for _ = range [5]int{} {//Calling the functions 5 times
res := randoms()
fmt.Println(res)
//nums = append(nums, res)
for _, i := range *res {//Iterating and appending them
nums = append(nums, i)
}
}
fmt.Println(nums)
}
I am trying to mimic my problem. I have dynamic number of function calls i.e randoms and dynamic number of results. I need to append all of the results i.e numbers in this case.
I am able to do this with iteration and no issues with it. I am looking for a way to do something like nums = append(nums, res). Is there any way to do this/any built-in methods/did I misunderstand the pointers?
I think you're looking for append(nums, (*res)...):
nums = append(nums, (*res)...)
playground
See this answer for more about ..., but in short it expands the contents of a slice. Example:
x := []int{1, 2, 3}
y := []int{4, 5, 6}
x = append(x, y...) // Now x = []int{1, 2, 3, 4, 5, 6}
Further, since you have a pointer to a slice, you need to dereference the pointer with *.
x := []int{1, 2, 3}
y := &x
x = append(x, (*x)...) // x = []int{1, 2, 3, 1, 2, 3}
The result of s is [1, 2, 3], I thought that slices hold a reference to the underlying array. Isn't that the case?
package main
import (
"fmt"
)
func main() {
s := []int{1, 2, 3}
ss := s[1:]
ss = append(ss, 4)
for _, v := range ss {
v += 10
}
for i := range ss {
ss[i] += 10
}
fmt.Println(s)
}
I thought that slices hold a reference to the underlying array. Isn't that the case?
Yes it is. But you created an array with length 3 with this statement:
s := []int{1, 2, 3}
When you appended an element to ss, that required an allocation of a new, longer array. So you lost the link between ss and s with this statement:
ss = append(ss, 4)
You can verify that by running this example:
package main
import (
"fmt"
)
func main() {
s := []int{1, 2, 3}
ss := s[1:]
ss[0] += 5
ss = append(ss, 4)
ss[0] += 100
fmt.Println(s)
}
Which prints [1 7 3].
If you change your initialization of s to have a length greater than three, then no new array allocation will be required, and the link between s and ss will be maintained:
package main
import (
"fmt"
)
func main() {
s := make([]int, 3, 4)
s[0], s[1], s[2] = 1, 2, 3
ss := s[1:]
ss[0] += 5
ss = append(ss, 4)
ss[0] += 100
fmt.Println(s)
}
Output: [1 107 3]
The answer that theorizes that the problem is a range copy of the slice is incorrect, which can be shown with this example:
package main
import (
"fmt"
)
func main() {
s := make([]int, 3, 4)
s[0], s[1], s[2] = 1, 2, 3
ss := s[1:]
ss = append(ss, 4)
for i := range ss {
ss[i] += 10
}
fmt.Println(s)
}
Output: [1 12 13]
It seems like you've created a copy of slice s, and you've made changes to slice copy ss thinking that the changes would also be passed to the slice that the copy was made from.
The first for loop is also walking through the elements of slice ss, but is not actually doing anything with them, because range also creates a copy when providing you element values, so it doesn't actually do anything!
It does seem like you're trying to do the following:
Append the value 4 to slice s
Take each value from index 1 of slice to the end of the slice, and add 10
If that's the case, this should help you accomplish that:
package main
import (
"fmt"
)
func main() {
s := []int{1, 2, 3}
s = append(s, 4)
for i := range s {
if i == 0 {
continue
}
s[i] += 10
}
fmt.Println(s)
}
You can see this on the Go playground: Link
I am new to Go and doing a few exercises. One of them is to sort the numbers in an array by frequency, from most to least frequent.
Example:
Input: [2, 2, 5, 7, 4, 4, 4, 7, 2]
Output: [2, 4, 7, 5]
Note that [4, 2, 7, 5] would also be correct, since 4 and 2 have the same frequency.
For this purpose I am converting the array into a value value map, which here would look like this: [2:3][4:3][7:2][5:1] (2 and 3 have freq. of 3, 7 has the freq of 2,... )
Afterwards I would like to simply loop through the map and output the keys ordered by value. For that I use the following code, which apparently does not work. Why?
count := 0
max := -1
// break loop, if map is empty
for i := 0; i < 1; i-- {
if len(m) == 0 {
break
}
max = -1
// get key of biggest value
for k, v := range m {
if v > max {
max = k
}
}
// res (for result) is a slice of integers
res[count] = max
// remove key-value-pair from map
delete(m, max)
count++
}
return res
Please keep in mind that this is an exercise. I am very sure there are much better, build in ways to do this.
Your 'max' variable is meant to keep track of the maximum frequency seen so far. However when you do 'max = k' you're assigning a key.
You need to keep track of the maximum frequency and the key associated with that frequency in separate variables.
...
for k, v := range m {
if v > maxFreq {
maxFreq = v
mostFrequentKey = k
}
}
// res (for result) is a slice of integers
res[count] = mostFrequentKey
// remove key-value-pair from map
delete(m, mostFrequentKey)
count++
...
For sorted frequencies, use a map then a slice. For example,
package main
import (
"fmt"
"sort"
)
func main() {
Input := []int{2, 2, 5, 7, 4, 4, 4, 7, 2}
fmt.Println("Input: ", Input)
mFreq := make(map[int]int, len(Input))
for _, n := range Input {
mFreq[n]++
}
sFreq := make([][2]int, 0, len(mFreq))
for n, f := range mFreq {
sFreq = append(sFreq, [2]int{n, f})
}
sort.Slice(sFreq, func(i, j int) bool {
if sFreq[i][1] <= sFreq[j][1] {
if sFreq[i][1] < sFreq[j][1] {
return false
}
if sFreq[i][0] >= sFreq[j][0] {
return false
}
}
return true
},
)
Output := []int{2, 4, 7, 5}
fmt.Println("Output: ", Output)
fmt.Println("Frequencies:", sFreq)
}
Playground: https://play.golang.org/p/8tiSksz3S76
Output:
Input: [2 2 5 7 4 4 4 7 2]
Output: [2 4 7 5]
Frequencies: [[2 3] [4 3] [7 2] [5 1]]
how array shift function works with slices?
package main
import "fmt"
func main() {
s := []int{2, 3, 5, 7, 11, 13}
for k, v := range s {
x, a := s[0], s[1:] // get and remove the 0 index element from slice
fmt.Println(a) // print 0 index element
}
}
I found an example from slice tricks but can't get it right.
https://github.com/golang/go/wiki/SliceTricks
x, a := a[0], a[1:]
Edit can you please explain why x is undefined here?
Building upon the answer and merging with SliceTricks
import "fmt"
func main() {
s := []int{2, 3, 5, 7, 11, 13}
fmt.Println(len(s), s)
for len(s) > 0 {
x, s = s[0], s[1:] // undefined: x
fmt.Println(x) // undefined: x
}
fmt.Println(len(s), s)
}
For example,
package main
import "fmt"
func main() {
s := []int{2, 3, 5, 7, 11, 13}
fmt.Println(len(s), s)
for len(s) > 0 {
x := s[0] // get the 0 index element from slice
s = s[1:] // remove the 0 index element from slice
fmt.Println(x) // print 0 index element
}
fmt.Println(len(s), s)
}
Output:
6 [2 3 5 7 11 13]
2
3
5
7
11
13
0 []
References:
The Go Programming Language Specification: For statements
Addendum to answer edit to question:
Declare x,
package main
import "fmt"
func main() {
s := []int{2, 3, 5, 7, 11, 13}
fmt.Println(len(s), s)
for len(s) > 0 {
var x int
x, s = s[0], s[1:]
fmt.Println(x)
}
fmt.Println(len(s), s)
}
Output:
6 [2 3 5 7 11 13]
2
3
5
7
11
13
0 []
You can copy and paste my code for any slice type; it infers the type for x. It doesn't have to be changed if the type of s changes.
for len(s) > 0 {
x := s[0] // get the 0 index element from slice
s = s[1:] // remove the 0 index element from slice
fmt.Println(x) // print 0 index element
}
For your version, the type for x is explicit and must be changed if the type of s is changed.
for len(s) > 0 {
var x int
x, s = s[0], s[1:]
fmt.Println(x)
}
Just a quick explanation on how we implement shift-like functionality Go. It's actually a very manual process. Take this example:
catSounds := []string{"meow", "purr", "schnurr"}
firstValue := stuff[0] // meow
catSounds = catSounds[1:]
On the first line, we create our slice.
On the second line we get the first element of the slice.
On the third line, we re-assign the value of catSounds to everything currently in catSounds after the first element (catSounds[1:]).
So given all that, we can condense the second and third lines with a comma for brevity:
catSounds := []string{"meow", "purr", "schnurr"}
firstValue, catSounds := catSounds[0], catSounds[1:]
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.