Finding exact word in a column of strings from a list of words having spaces in between

Question

I want to create a new column with 1 or 0, if any of the words in a list is matched exaclty with the dataframe string column.

The words in the list can have multiple spaces in between , so I am not able to use str.split() for exact match.

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

Expected Output

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

Ordering of the words in the list element also matters!!

Code tried till now

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

How to obtain the expected output where I have to search for exact match of words from a list against a dataframe string column, having multiple spaces in between?

Answer 1

You were nearly there, you've done all the ground work, now all that's left is to call the right function, in this case, 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

If your patterns themselves contain the regex OR pipe, try escaping them first:

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

Answer 2

You can achieve this with the help of str.contains. This can take up regex too!

data['found'] = np.where(data['text'].str.contains(pattern),1,0)