Counting specific keywords in a dataframe

Question

I have a dataframe as such:

    A
0   Please wait outside of the house
1   A glittering gem is not enough.
2   The memory we used to share is no longer coher...
3   She only paints with bold colors; she does not...

I have a set of keywords:

keywords = ["of","is","she"]

How can I create a column for each keyword containing the number of occurrences of the keyword in each sentence of my dataframe? It would look something like:

                                                   A  of  is  she
0                   Please wait outside of the house   1   0    0
1                    A glittering gem is not enough.   0   1    0
2  The memory we used to share is no longer coher...   0   1    0
3  She only paints with bold colors; she does not...   0   0    2

Note: I look at how to count specific words from a pandas Series?, but it does not answer my question.

Ajay Dabas · Accepted Answer · 2019-12-25 04:54:51Z

2

I assumed that you're looking for case-insensitive matches.

import pandas as pd
df = pd.DataFrame({
    'A': [
        'Please wait outside of the house',
        'A glittering gem is not enough.',
        'The memory we used to share is no longer coher...',
        'She only paints with bold colors; she does not...'
    ]
})
keywords = ["of","is","she"]
for keyword in keywords:
    df[keyword] = df['A'].apply(lambda _str: _str.lower().count(keyword))
print(df)

Output

                                                   A  of  is  she
0                   Please wait outside of the house   1   0    0
1                    A glittering gem is not enough.   0   1    0
2  The memory we used to share is no longer coher...   0   1    0
3  She only paints with bold colors; she does not...   0   0    2

answered Dec 25, 2019 at 4:54

Ajay Dabas

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oppressionslayer · Accepted Answer · 2019-12-25 05:26:19Z

0

You can also do it this way:

df['is'] = df.A.str.count(r'is', flags=re.IGNORECASE)
df['of'] = df.A.str.count(r'of', flags=re.IGNORECASE)
df['she'] = df.A.str.count(r'she', flags=re.IGNORECASE)


                                                   A  of  is  she
0                   Please wait outside of the house   1   0    0
1                    A glittering gem is not enough.   0   1    0
2  The memory we used to share is no longer coher...   0   1    0
3  She only paints with bold colors; she does not...   0   0    2

answered Dec 25, 2019 at 5:26

oppressionslayer

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