如何计算多个指定字符串的出现之后如何添加字符串

Question

我想根据多个字符串的第n次出现添加一个字符串（而不是替换字符串）。

示例：

tablespecs =“ l c d r c l”

现在假设我要在字符串“ here：”中添加l，c和r的第三个外观，所需的输出应为：

desired_tablespecs =“此处的lc：r c l”

因此仅考虑l，c和r，而忽略d。

我只是来得很近，如下所示，代码没有替换匹配项，而是替换了匹配项，因此它提供了“ l c d here：c l”。

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

您可以将after设置为从where.start()开始而不是where.end()，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) 

此输出'l c d here: r c l'。

Answer 2

略有不同的方法：

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