How to split string in GO by array of runes?

huangapple
go评论116阅读模式
英文:

How to split string in GO by array of runes?

问题

如果你有一个由符文数组作为分隔符的字符串，有没有办法将它拆分成字符串数组？以下是我想要的一个示例：

separators = {' ',')','('}
SomeFunction("my string(qq bb)zz",separators) => {"my","string","qq","bb","zz"}

英文:

If there is any way to split string into array of strings, when you have as a separator an array of runes? There is an example what I want:

  1. seperators = {' ',')','('}
  2. SomeFunction("my string(qq bb)zz",seperators) => {"my","string","qq","bb","zz"}

答案1

得分: 13

例如，

  1. package main
  3. import (
  4.     "fmt"
  5.     "strings"
  6. )
  8. func split(s string, separators []rune) []string {
  9.     f := func(r rune) bool {
  10.         for _, s := range separators {
  11.             if r == s {
  12.                 return true
  13.             }
  14.         }
  15.         return false
  16.     }
  17.     return strings.FieldsFunc(s, f)
  19. }
  21. func main() {
  22.     separators := []rune{' ', ')', '('}
  23.     s := "my string(qq bb)zz"
  24.     ss := split(s, separators)
  25.     fmt.Printf("%q\n", s)
  26.     fmt.Printf("%q\n", ss)
  27. }

输出：

  1. "my string(qq bb)zz"
  2. ["my" "string" "qq" "bb" "zz"]
英文:

For example,

  1. package main
  3. import (
  4. 	"fmt"
  5. 	"strings"
  6. )
  8. func split(s string, separators []rune) []string {
  9. 	f := func(r rune) bool {
  10. 		for _, s := range separators {
  11. 			if r == s {
  12. 				return true
  13. 			}
  14. 		}
  15. 		return false
  16. 	}
  17. 	return strings.FieldsFunc(s, f)
  19. }
  21. func main() {
  22. 	separators := []rune{' ', ')', '('}
  23. 	s := "my string(qq bb)zz"
  24. 	ss := split(s, separators)
  25. 	fmt.Printf("%q\n", s)
  26. 	fmt.Printf("%q\n", ss)
  27. }

Output:

  1. "my string(qq bb)zz"
  2. ["my" "string" "qq" "bb" "zz"]

答案2

得分: 2

使用正则表达式：

  1. package main
  3. import (
  4. 	"fmt"
  5. 	"regexp"
  6. )
  8. var re = regexp.MustCompile(`[() ]`)
  10. func main() {
  11. 	text := "my string(qq bb)zz"
  12. 	splinedText := re.Split(text, -1)
  13. 	fmt.Printf("%q\n", text)
  14. 	fmt.Printf("%q\n", splinedText)
  15. }

输出：

  1. "my string(qq bb)zz"
  2. ["my" "string" "qq" "bb" "zz"]
英文:

with regexp:

  1. package main
  3. import (
  4. 	"fmt"
  5. 	"regexp"
  6. )
  8. var re = regexp.MustCompile("[() ]")
  10. func main() {
  11. 	text := "my string(qq bb)zz"
  12. 	splinedText := re.Split(text, -1)
  13. 	fmt.Printf("%q\n", text)
  14. 	fmt.Printf("%q\n", splinedText)
  15. }

output:

  1. "my string(qq bb)zz"
  2. ["my" "string" "qq" "bb" "zz"]

答案3

得分: 0

我相信一个更简单的方法是使用函数FieldsFunc。以下是它的实现示例：

  1. package main
  3. import (
  4. 	"fmt"
  5. 	"strings"
  6. 	"unicode"
  7. )
  9. func main() {
  10. 	f := func(c rune) bool {
  11. 		return !unicode.IsLetter(c) && !unicode.IsNumber(c)
  12. 	}
  13. 	fmt.Printf("Fields are: %q", strings.FieldsFunc("  foo1;bar2,baz3...", f))
  14. }

输出结果：

Fields are: ["foo1" "bar2" "baz3"]

英文:

I believe that a simpler approach would be to use the function FieldsFunc. Here is an example of it's implementation:

  1. package main
  3. import (
  4. 	"fmt"
  5. 	"strings"
  6. 	"unicode"
  7. )
  9. func main() {
  10. 	f := func(c rune) bool {
  11. 		return !unicode.IsLetter(c) && !unicode.IsNumber(c)
  12. 	}
  13. 	fmt.Printf("Fields are: %q", strings.FieldsFunc("  foo1;bar2,baz3...", f))
  14. }

Output :
> Fields are: ["foo1" "bar2" "baz3"]

huangapple
  • 本文由 发表于 2014年3月4日 03:16:41
  • 转载请务必保留本文链接：https://go.coder-hub.com/22155313.html
  • go
  • split
  • string
匿名

发表评论

匿名网友

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen:

确定