英文:
Go: convert strings in array to integer
问题
如何在Go中将数组中的字符串转换为整数数组？
package mainimport ("fmt""strconv")func main() {strArray := []string{"1", "2", "3"}intArray := make([]int, len(strArray))for i, v := range strArray {num, err := strconv.Atoi(v)if err != nil {fmt.Printf("Error converting string to integer: %v\n", err)return}intArray[i] = num}fmt.Println(intArray)}
在上面的示例中，我们使用strconv.Atoi()函数将字符串转换为整数。我们首先创建一个与字符串数组相同长度的整数数组intArray。然后，我们使用range循环遍历字符串数组，并在每次迭代中将字符串转换为整数，并将其存储在整数数组中的相应位置。最后，我们打印整数数组。
希望这可以帮助到你！
英文:
How do I convert strings in an array to integers in an array in go?
["1", "2", "3"]
to
[1, 2, 3]
I've searched for some solutions online but couldn't find it. I've tried to loop through the array and did strconv.ParseFloat(v, 64) where v is the value but it didn't work.
答案1
得分: 40
你确实需要遍历切片。如果切片只包含整数，就不需要使用ParseFloat，只需要使用Atoi即可。
import "fmt"import "strconv"func main() {var t = []string{"1", "2", "3"}var t2 = []int{}for _, i := range t {j, err := strconv.Atoi(i)if err != nil {panic(err)}t2 = append(t2, j)}fmt.Println(t2)}
在Playground上可以运行。
英文:
You will have to loop through the slice indeed. If the slice only contains integers, no need of ParseFloat, Atoi is sufficient.
import "fmt"import "strconv"func main() {var t = []string{"1", "2", "3"}var t2 = []int{}for _, i := range t {j, err := strconv.Atoi(i)if err != nil {panic(err)}t2 = append(t2, j)}fmt.Println(t2)}
On Playground.
答案2
得分: 11
例如，
package mainimport ("fmt""strconv")func sliceAtoi(sa []string) ([]int, error) {si := make([]int, 0, len(sa))for _, a := range sa {i, err := strconv.Atoi(a)if err != nil {return si, err}si = append(si, i)}return si, nil}func main() {sa := []string{"1", "2", "3"}si, err := sliceAtoi(sa)if err != nil {fmt.Println(err)return}fmt.Printf("%q %v\n", sa, si)}
输出：
["1" "2" "3"] [1 2 3]
Playground:
http://play.golang.org/p/QwNO8R_f90
英文:
For example,
package mainimport ("fmt""strconv")func sliceAtoi(sa []string) ([]int, error) {si := make([]int, 0, len(sa))for _, a := range sa {i, err := strconv.Atoi(a)if err != nil {return si, err}si = append(si, i)}return si, nil}func main() {sa := []string{"1", "2", "3"}si, err := sliceAtoi(sa)if err != nil {fmt.Println(err)return}fmt.Printf("%q %v\n", sa, si)}
Output:
["1" "2" "3"] [1 2 3]
Playground:
答案3
得分: 4
这是一个古老的问题，但是所有的答案都忽略了输入长度事先已知的事实，因此可以通过预先分配目标切片来改进：
package mainimport "fmt"import "strconv"func main() {var t = []string{"1", "2", "3"}var t2 = make([]int, len(t))for idx, i := range t {j, err := strconv.Atoi(i)if err != nil {panic(err)}t2[idx] = j}fmt.Println(t2)}
Playground: https://play.golang.org/p/LBKnVdi_1Xz
英文:
This is an ancient question but all the answers ignore the fact that the input length is known in advance, so it can be improved by pre-allocating the destination slice:
package mainimport "fmt"import "strconv"func main() {var t = []string{"1", "2", "3"}var t2 = make([]int, len(t))for idx, i := range t {j, err := strconv.Atoi(i)if err != nil {panic(err)}t2[idx] = j}fmt.Println(t2)}
Playground: https://play.golang.org/p/LBKnVdi_1Xz
答案4
得分: 1
一个切片是数组段的描述符
它由以下部分组成：
- 数组的指针，
- 段的长度，
- 容量（段的最大长度）
下面，将字符串数组/切片转换为整数数组/切片：
package mainimport ("fmt""log""strconv""strings")func Slice_Atoi(strArr []string) ([]int, error) {// 注意：按您喜欢的方式将Arr读取为Slicevar str string // Ovar i int // Ovar err error // OiArr := make([]int, 0, len(strArr))for _, str = range strArr {i, err = strconv.Atoi(str)if err != nil {return nil, err // O}iArr = append(iArr, i)}return iArr, nil}func main() {strArr := []string{"0 0 24 3 15","0 0 2 5 1 5 11 13",}for i := 0; i < len(strArr); i++ {iArr, err := Slice_Atoi(strings.Split(strArr[i], " "))if err != nil {log.Print("Slice_Atoi failed: ", err)return}fmt.Println(iArr)}}
输出：
[0 0 24 3 15][0 0 2 5 1 5 11 13]
我在项目中使用了这段代码，并从其他回答中进行了一些小的优化，用 // O 标记了上面的部分，并对可读性进行了一些修复。
祝你好运！
英文:
A slice is a descriptor of an array segment
It consists of
- a pointer to the array,
- the length of the segment, and
- its capacity (the maximum length of the segment)
Below, string Array/Slice is converted to int Array/Slice:
package mainimport ("fmt""log""strconv""strings")func Slice_Atoi(strArr []string) ([]int, error) {// NOTE: Read Arr as Slice as you likevar str string // Ovar i int // Ovar err error // OiArr := make([]int, 0, len(strArr))for _, str = range strArr {i, err = strconv.Atoi(str)if err != nil {return nil, err // O}iArr = append(iArr, i)}return iArr, nil}func main() {strArr := []string{"0 0 24 3 15","0 0 2 5 1 5 11 13",}for i := 0; i < len(strArr); i++ {iArr, err := Slice_Atoi(strings.Split(strArr[i], " "))if err != nil {log.Print("Slice_Atoi failed: ", err)return}fmt.Println(iArr)}}
Output:
[0 0 24 3 15][0 0 2 5 1 5 11 13]
I used in a project, so did a small optimizations from other replies, marked as // O for above, also fixed a bit in readability for others
Best of luck
答案5
得分: 0
这是另一个示例，演示如何完成这个任务：
var t = []string{"1", "2", "3"}str := strings.Join(t, "")if _, err := strconv.Atoi(str); err != nil {// 如果str无法转换为整数，执行相应操作}var slice []int // 空切片for _, digit := range str {slice = append(slice, int(digit)-int('0')) // 构建切片}
我喜欢这个方法，因为你只需要检查一次t中的每个数字是否可以转换为整数。这样更有效率吗？我不知道。
为什么需要int('0')？因为int()会将字符转换为对应的ASCII码（ASCII码表在这里）。对于数字0，对应的ASCII码是48。所以你需要从你的数字在"ASCII十进制"中对应的值中减去48。
英文:
Here is another example how to do this:
var t = []string{"1", "2", "3"}str := strings.Join(t, "")if _, err := strconv.Atoi(str); err != nil {// do stuff, in case str can not be converted to an int}var slice []int // empty slicefor _, digit := range str {slice = append(slice, int(digit)-int('0')) // build up slice}
I like this because you have to check only once if each digit inside of t can be converted to an int. Is it more efficient? I don`t know.
Why do you need the int('0')? Because int() will convert the character to the corresponding ASCII code (ascii table here). For 0 that would be 48. So you will have to substract 48 from whatever your digit corresponds to in "ascii decimal".
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