从文件中删除停用词

Question

I want to remove stopwords from the Data column in my file. I filtered out the line for when the end-user is speaking. But it doesn't filter out the stopwords with the usertext.apply(lambda x: [word for word in x if word not in stop_words]) what am i doing wrong?

import pandas as pd
from stop_words  import get_stop_words
df = pd.read_csv("F:/textclustering/data/cleandata.csv", encoding="iso-8859-1")
usertext = df[df.Role.str.contains("End-user",na=False)][['Data','chatid']]
stop_words = get_stop_words('dutch')
clean = usertext.apply(lambda x: [word for word in x if word not in stop_words])
print(clean)

Answer 1

You can build a regex pattern of your stop words and call the vectorised str.replace to remove them:

In [124]:
stop_words = ['a','not','the']
stop_words_pat = '|'.join(['\\b' + stop +  '\\b' for stop in stop_words])
stop_words_pat

Out[124]:
'\\ba\\b|\\bnot\\b|\\bthe\\b'

In [125]:    
df = pd.DataFrame({'text':['a to the b', 'the knot ace a']})
df['text'].str.replace(stop_words_pat, '')

Out[125]:
0         to  b
1     knot ace 
Name: text, dtype: object

here we perform a list comprehension to build a pattern surrounding each stop word with '\\b' which is a break and then we or all words using '|'

Answer 2

Two issues:

First, you have a module called stop_words and you later create a variable named stop_words . This is bad form.

Second, you are passing a lambda-function to .apply that wants its x parameter to be a list, rather than a value within a list.

That is, instead of doing df.apply(sqrt) you are doing df.apply(lambda x: [sqrt(val) for val in x]) .

You should either do the list-processing yourself:

clean = [x for x in usertext if x not in stop_words]

Or you should do the apply, with a function that takes one word at a time:

clean = usertext.apply(lambda x: x if x not in stop_words else '')

As @Jean-François Fabre suggested in a comment, you can speed things up if your stop_words is a set rather than a list:

from stop_words import get_stop_words

nl_stop_words = set(get_stop_words('dutch'))    # NOTE: set

usertext = ...
clean = usertext.apply(lambda word: word if word not in nl_stop_words else '')

Answer 3

clean = usertext.apply(lambda x:  x if x not in stop_words else '')