更新时间:2023-11-28 18:34:04
使用 replace
由正则表达式创建,由列 word
的所有值创建,最后 strip
跟踪空格:
Use replace
by regex created by joined all values of column word
, last strip
traling whitespaces:
data.name = data.name.replace('|'.join(df['word']), '', regex=True).str.strip()
另一种解决方案是添加 \s*
以选择零个或多个空格:
Another solution is add \s*
for select zero or more whitespaces:
pat = '|'.join(['\s*{}\s*'.format(x) for x in df['word']])
print (pat)
\s*Clinton\s*|\s*James\s*|\s*Bill\s*|\s*Clark\s*
data.name = data.name.replace(pat, '', regex=True)
print (data)
name
0 Hayden
1 Rock
2 Gates
3 Vishal
4 Cameroon
5 Micky
6 Michael
7 Tony Waugh
8 Tom
9 Tom
10 Avinash
11 Shreyas
12 Ramesh
13 Adam