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