英文:
Capture letter group and repeat them certain times with the number in group
问题
我明白你的问题。你想将字符串转换为"RLFLFPAPAPA"，即去掉数字并重复字母组。你的想法是使用正则表达式来捕获字母组，但是你在下面的代码中无法得到结果。你想知道如何解决这个问题，或者是否需要尝试其他方法。
在你的代码中，你已经定义了一个正则表达式(\d)\((\w+)\)，用于匹配数字和括号中的字母组。要解决这个问题，你可以使用FindAllStringSubmatch函数来找到所有匹配的子字符串，并将其重复。
下面是修改后的代码：
package main
import (
"fmt"
"regexp"
"strconv"
"strings"
)
func main() {
fmt.Println(move("R2(LF)3(PA)"))
}
func move(s string) string {
rgx := regexp.MustCompile(`(\d)\((\w+)\)`)
matches := rgx.FindAllStringSubmatch(s, -1)
for _, match := range matches {
count, _ := strconv.Atoi(match[1])
repeated := strings.Repeat(match[2], count)
s = strings.Replace(s, match[0], repeated, 1)
}
return s
}
这样，你就可以得到期望的结果"RLFLFPAPAPA"。希望能对你有所帮助！
英文:
I have string like: "R2(LF)3(PA)", and want to convert it to RLFLFPAPAPA (get rid of number and repeact the group of letters)
The idea come out is using regexp to capture the group, but I am stuck to get the result with code below, how can I solve that? Or I need to try other methods?
package main
import (
"fmt"
"regexp"
)
func main() {
fmt.Println(move("R2(LF)3(PA)"))
}
func move(s string) string {
rgx := regexp.MustCompile(`(\d)\((\w+)\)`)
// code need here
return s
}
答案1
得分: 2
似乎Go没有一个带有回调函数的替换函数（这对我们来说是最实用的情况）。但我找到了这个实现方法。
所以思路是将包含数字的第一个组转换为整数值。然后我们可以重复第二个组中的字母这个数字次数：
package main
import (
"fmt"
"regexp"
"strconv"
"strings"
)
func main() {
fmt.Println(move("R2(LF)3(PA)"))
}
func ReplaceAllStringSubmatchFunc(re *regexp.Regexp, str string, repl func([]string) string) string {
result := ""
lastIndex := 0
for _, v := range re.FindAllSubmatchIndex([]byte(str), -1) {
groups := []string{}
for i := 0; i < len(v); i += 2 {
groups = append(groups, str[v[i]:v[i+1]])
}
result += str[lastIndex:v[0]] + repl(groups)
lastIndex = v[1]
}
return result + str[lastIndex:]
}
func move(s string) string {
rgx := regexp.MustCompile(`(\d)\((\w+)\)`)
return ReplaceAllStringSubmatchFunc(rgx, s, func(groups []string) string {
// 将第一个组中的数字转换为整数值。
nbrTimes, err := strconv.Atoi(groups[1])
if err != nil {
// 处理错误（如果数字很长可能会发生错误）。
}
// 返回第二个组中的字母重复n次。
return strings.Repeat(groups[2], nbrTimes)
})
}
你可以在这里在线运行它：https://go.dev/play/p/FKNFJSc_tdS
英文:
It seems that Go doesn't have a replace function with a callback function (which is the most praticle case for us here). But I found this implementation to do it.
So the idea is to convert the first group containing a string with the number to an integer value. Then we can repeat the group 2 containing the letters this number of times:
package main
import (
"fmt"
"regexp"
"strconv"
"strings"
)
func main() {
fmt.Println(move("R2(LF)3(PA)"))
}
func ReplaceAllStringSubmatchFunc(re *regexp.Regexp, str string, repl func([]string) string) string {
result := ""
lastIndex := 0
for _, v := range re.FindAllSubmatchIndex([]byte(str), -1) {
groups := []string{}
for i := 0; i < len(v); i += 2 {
groups = append(groups, str[v[i]:v[i+1]])
}
result += str[lastIndex:v[0]] + repl(groups)
lastIndex = v[1]
}
return result + str[lastIndex:]
}
func move(s string) string {
rgx := regexp.MustCompile(`(\d)\((\w+)\)`)
return ReplaceAllStringSubmatchFunc(rgx, s, func(groups []string) string {
// Convert the first group with the digits to an integer value.
nbrTimes, err := strconv.Atoi(groups[1])
if err != nil {
// Handle error (could happen if very long number).
}
// Return the group 2 with the letters repeated n times.
return strings.Repeat(groups[2], nbrTimes)
})
}
You can run it online here: https://go.dev/play/p/FKNFJSc_tdS
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