从中间有空格的单词列表中查找字符串列中的确切单词

Question

如果列表中的任何单词与dataframe字符串列完全匹配，我希望创建一个带有1或0的新列。

列表中的单词在之间可以有多个空格，因此我无法使用str.split()进行精确匹配。

list_provided=["mul the","a b c"]
#how my dataframe looks
id  text
a    simultaneous there the
b    simultaneous there
c    mul why the
d    mul the
e    simul a b c
f    a c b

预期输出

id  text                      found
a    simultaneous there the    0
b    simultaneous there        0
c    mul why the               0
d    mul the                   1
e    simul a b c               1 
f    a c b                     0

列表元素中单词的排序也很重要！！

代码到现在为止尝试过

data=pd.DataFrame({"id":("a","b","c","d","e","f"), "text":("simultaneous there the","simultaneous there","mul why the","mul the","simul a b c","a c b")})
list_of_word=["mul the","a b c"]
pattern = '|'.join(list_of_word)
data['found'] = data['text'].apply(lambda x: sum(i in list_of_test_2 for i in x.split()))
data['found']=np.where(data['found']>0,1,0)
data
###Output generated###
id  text                   found
a   simultaneous there the  0
b   simultaneous there      0
c   mul why the             0
d   mul the                 0
e   simul a b c             0
f   a c b                   0

如何获得预期的输出，在其中我必须搜索与数据字符串列的列表中单词的准确匹配，其中有多个空格？

Answer 1

你快到了，你做了所有的基础工作，现在剩下的就是调用右边的函数，在这个例子中，是str.contains。

data['found'] = data.text.str.contains(pattern).astype(int)
data

  id                    text  found
0  a  simultaneous there the      0
1  b      simultaneous there      0
2  c             mul why the      0
3  d                 mul the      1
4  e             simul a b c      1
5  f                   a c b      0

如果您的模式本身包含regex或管道，那么首先尝试转义它们：

import re
pattern = '|'.join([re.escape(i) for i in list_of_word])