如何在多个指定字符串出现计数后添加字符串

Question

我想添加一个字符串（不替换）的基础上，第n次出现的多个字符串

示例：

tablespecs=“l c d r c l”

现在假设我想在字符串“here:”中添加l、c和r的第三个外观，所需的输出应该是：

所需的\u tablespecs=“l c d here:r c l”

所以只考虑l，c和r，忽略d

我只是像下面一样接近，代码没有添加copde，而是替换匹配项，因此它提供了“lcd here:cll”

tablespecs = "l c d r c l"


def replacenth(string, sub, wanted, n):
    pattern = re.compile(sub)
    where = [m for m in pattern.finditer(string)][n-1]
    before = string[:where.start()]
    after = string[where.end():]
    newString = before + wanted + after

    return newString

#Source of code chunk https://stackoverflow.com/questions/35091557/replace-nth-occurrence-of-substring-in-string


replacenth(tablespecs, "[lcr]", "[here:]", 3) 

#wrong_output: 'l c d [here:] c l'

Answer 1

稍有不同的方法：

tablespecs = "l c d r c l"


def replacenth(string, sub, wanted, n):
    count = 0
    out = ""

    for c in string:
        if c in sub:
            count += 1

        if count == n:
            out += wanted
            count = 0
        out += c
    return out


res = replacenth(tablespecs, "lcr", "here: ", 3)
assert res == "l c d here: r c l", res

Answer 2

您可以将after设置为从where.start()开始，而不是从where.end()开始，以便它包含该字符

tablespecs = "l c d r c l"

def replacenth(string, sub, wanted, n):
    pattern = re.compile(sub)
    where = [m for m in pattern.finditer(string)][n-1]
    before = string[:where.start()]
    after = string[where.start():]
    newString = before + wanted + after
    return newString

replacenth(tablespecs, "[lcr]", "here: ", 3) 

输出'l c d here: r c l'