英文:
Counting the occurrence of one or more substrings in a string
问题
我知道要计算一个子字符串的出现次数，可以使用"strings.Count(
英文:
I know that for counting the occurrence of one substring I can use "strings.Count(<string>, <substring>)". What if I want to count the number of occurrences of substring1 OR substring2? Is there a more elegant way than writing another new line with strings.count()?
答案1
得分: 17
使用正则表达式（regular expression）：
https://golang.org/pkg/regexp/
aORb := regexp.MustCompile("A|B")
matches := aORb.FindAllStringIndex("A B C B A", -1)
fmt.Println(len(matches))
英文:
Use a regular expression:
https://play.golang.org/p/xMsHIYKtkQ
aORb := regexp.MustCompile("A|B")
matches := aORb.FindAllStringIndex("A B C B A", -1)
fmt.Println(len(matches))
答案2
得分: 2
另一种进行子字符串匹配的方法是使用suffixarray包。下面是一个匹配多个模式的示例：
package main
import (
"fmt"
"index/suffixarray"
"regexp"
)
func main() {
r := regexp.MustCompile("an")
index := suffixarray.New([]byte("banana"))
results := index.FindAllIndex(r, -1)
fmt.Println(len(results))
}
你也可以使用Lookup函数来匹配单个子字符串。
英文:
Another way to do substring matching is with the suffixarray package. Here is an example of matching multiple patterns:
package main
import (
"fmt"
"index/suffixarray"
"regexp"
)
func main() {
r := regexp.MustCompile("an")
index := suffixarray.New([]byte("banana"))
results := index.FindAllIndex(r, -1)
fmt.Println(len(results))
}
You can also match a single substring with the Lookup function.
答案3
得分: 0
如果你想在一个大字符串中计算匹配项的数量，而不需要为了获取长度而分配所有索引的空间，然后再将它们丢弃，你可以使用Regexp.FindStringIndex在循环中匹配连续的子字符串：
func countMatches(s string, re *regexp.Regexp) int {
total := 0
for start := 0; start < len(s); {
remaining := s[start:] // 切片操作是廉价的
loc := re.FindStringIndex(remaining)
if loc == nil {
break
}
// loc[0] 是匹配的起始索引，
// loc[1] 是匹配的结束索引（不包含）
start += loc[1]
total++
}
return total
}
func main() {
s := "abracadabra"
fmt.Println(countMatches(s, regexp.MustCompile(`a|b`)))
}
英文:
If you want to count the number of matches in a large string, without allocating space for all the indices just to get the length and then throwing them away, you can use Regexp.FindStringIndex in a loop to match against successive substrings:
func countMatches(s string, re *regexp.Regexp) int {
total := 0
for start := 0; start < len(s); {
remaining := s[start:] // slicing the string is cheap
loc := re.FindStringIndex(remaining)
if loc == nil {
break
}
// loc[0] is the start index of the match,
// loc[1] is the end index (exclusive)
start += loc[1]
total++
}
return total
}
func main() {
s := "abracadabra"
fmt.Println(countMatches(s, regexp.MustCompile(`a|b`)))
}
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