在Golang中按长度拆分字符串 [英] Split string by length in Golang

问题描述

有人知道如何在Golang中按长度分割字符串吗?

Does anyone know how to split a string in Golang by length?

例如，每隔3个字符分割一次"helloworld"，因此理想情况下应返回"hel"，"low"，"orl"，"d"的数组?

For example to split "helloworld" after every 3 characters, so it should ideally return an array of "hel" "low" "orl" "d"?

或者，一种可能的解决方案是在每3个字符后也添加一个换行符.

Alternatively a possible solution would be to also append a newline after every 3 characters..

所有想法都将不胜感激！

All ideas are greatly appreciated!

推荐答案

请确保 转换 ，将您的字符串切成一片符文:请参阅"将字符串切成字母".

Make sure to convert your string into a slice of rune: see "Slice string into letters".

for 自动将 string 转换为 rune ，因此在这种情况下，无需其他代码即可转换 string 首先是 rune .

for automatically converts string to rune so there is no additional code needed in this case to convert the string to rune first.

for i, r := range s {
    fmt.Printf("i%d r %c\n", i, r)
    // every 3 i, do something
}

r [n:n + 3] 最适合作为符文的一部分.

r[n:n+3] will work best with a being a slice of rune.

每个符文索引将增加一个，而在每个字节增加一个以上.golang.org/slices#TOC_12."rel ="nofollow noreferrer">字符串切片:世界": i 分别为0和3:一个字符(符文)可以由多个字节组成.

The index will increase by one every rune, while it might increase by more than one for every byte in a slice of string: "世界": i would be 0 and 3: a character (rune) can be formed of multiple bytes.

例如，考虑 s:=世间界bcd界efg世" :12个符文.(请参见 play.golang.org )

For instance, consider s := "世a界世bcd界efg世": 12 runes. (see play.golang.org)

如果您尝试逐字节地解析它，则会丢失(在每3个字符的天真的拆分中)一些索引模3"(等于2、5、8和11)，因为索引会增加超过这些值:

If you try to parse it byte by byte, you will miss (in a naive split every 3 chars implementation) some of the "index modulo 3" (equals to 2, 5, 8 and 11), because the index will increase past those values:

for i, r := range s {
    res = res + string(r)
    fmt.Printf("i %d r %c\n", i, r)
    if i > 0 && (i+1)%3 == 0 {
        fmt.Printf("=>(%d) '%v'\n", i, res)
        res = ""
    }
}

输出:

i  0 r 世
i  3 r a   <== miss i==2
i  4 r 界
i  7 r 世  <== miss i==5
i 10 r b  <== miss i==8
i 11 r c  ===============> would print '世a界世bc', not exactly '3 chars'!
i 12 r d
i 13 r 界
i 16 r e  <== miss i==14
i 17 r f  ===============> would print 'd界ef'
i 18 r g
i 19 r 世 <== miss the rest of the string

但是，如果您要遍历符文( a:= [] rune(s))，您将得到期望的结果，因为索引一次会增加一个符文，从而使其易于准确地合计3个字符:

But if you were to iterate on runes (a := []rune(s)), you would get what you expect, as the index would increase one rune at a time, making it easy to aggregate exactly 3 characters:

for i, r := range a {
    res = res + string(r)
    fmt.Printf("i%d r %c\n", i, r)
    if i > 0 && (i+1)%3 == 0 {
        fmt.Printf("=>(%d) '%v'\n", i, res)
        res = ""
    }
}

输出:

i 0 r 世
i 1 r a
i 2 r 界 ===============> would print '世a界'
i 3 r 世
i 4 r b
i 5 r c ===============> would print '世bc'
i 6 r d
i 7 r 界
i 8 r e ===============> would print 'd界e'
i 9 r f
i10 r g
i11 r 世 ===============> would print 'fg世'

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