如何检测第一个单词并将其包含在 python 的字符串替换行中

一种可能过于复杂的方法是将正则表达式与以下捕获组一起使用:

res = df["Survey Name"].str.replace(r"(Q\d+)\s+(\w)\w+ (\w)\w+ - (\w)\w+ (\w)\w+", r"\1 \2\3 - \4\5", regex=True)
print(res)

Output

0    Q321 YV - IT
1    Q321 YV - IT
2    Q321 YV - IT
3    Q321 YV - IT
4    Q321 YV - IT
5    Q321 YV - BG
6    Q321 YV - BG
Name: Survey Name, dtype: object

注意，正则表达式模式捕获剩余每个单词的第一个单词和第一个字母.

另一种 Select 是使用apply，并具有替换功能:

def repl(x):
    head, tail = x.split("-")
    quarter, *chunk = head.split()

    head_initials = "".join(c[0] for c in chunk)
    tail_initials = "".join(c[0] for c in tail.split())

    return f"{quarter} {head_initials} - {tail_initials}"


res = df["Survey Name"].apply(repl)

Output

0    Q321 YV - IT
1    Q321 YV - IT
2    Q321 YV - IT
3    Q321 YV - IT
4    Q321 YV - IT
5    Q321 YV - BG
6    Q321 YV - BG
Name: Survey Name, dtype: object

UPDATE

更换零件的更通用方法是:

replacements = {
    "Your Voice - Information Tech": "YV - IT Group",
    "Your Voice - Business Group": "YV - BG",
    "Your Voice - Human Resources": "YV - LRECS"
}


def repl(match, repls=replacements):
    quarter = match.group(1)
    key = " ".join(match.group(2).strip().split())

    return f"{quarter} {replacements.get(key, '')}"


res = df["Survey Name"].str.replace(r"(Q\d+)\s+(.+)", repl, regex=True)
print(res)

Output

0    Q321 YV - IT Group
1    Q321 YV - IT Group
2    Q321 YV - IT Group
3    Q321 YV - IT Group
4    Q321 YV - IT Group
5          Q321 YV - BG
6       Q321 YV - LRECS
Name: Survey Name, dtype: object

注意，replacements是一个字典，其中键是预期要找到的字符串和相应替换的值.

上述输出数据来自:

{'Survey Name': {0: 'Q321 Your Voice - Information Tech',
                 1: 'Q321 Your Voice - Information Tech',
                 2: 'Q321 Your Voice - Information Tech',
                 3: 'Q321 Your Voice - Information Tech',
                 4: 'Q321 Your Voice - Information Tech',
                 5: 'Q321 Your Voice - Business Group',
                 6: 'Q321 Your Voice - Human Resources'}}