熊猫:选择与字符串匹配的行，并使用该单词创建一个新列。

Question

熊猫:选择与字符串匹配的行，用该词创建新列，选择与字符串匹配的行，用该单词创建新列(找到)

list_provided=["mul","the","have", "then"]

我的数据看起来如何

id  text
a    simultaneous there the
b    simultaneous there
c    mul why

预期输出

id  text                        found
1    simultaneous there the      the
2    simultaneous there         
3    mul why                     mul
4    have the                    have, the 
5    then the late               then,the

Answer 1

我认为像这样的事情应该有效：

df['text'].apply(lambda x: [i for i in x.split() if i in list_provided])

Answer 2

另一种使用regex模式的方法：

pat = r'\b' + r'\b|\b'.join(list_provided) + r'\b'

df['found'] = df.text.str.findall(pat)


  id                    text        found
0  a  simultaneous there the        [the]
1  b      simultaneous there           []
2  c                 mul why        [mul]
3  d                have the  [have, the]
4  e           then the late  [then, the]