CODEWITHSUNDEEP
pythonpandasdataframescikit-learndataset
Angus
Angus

Reputation: 3728

Get count of matching word in string of pandas column with a predefined list

I have a DataFrame contains index and text columns.

For example:

index | text
1     | "I have a pen, but I lost it today."
2     | "I have pineapple and pen, but I lost it today."

Now I have a long list, and I want to match each of the words in text with the list.

Let's say:

long_list = ['pen', 'pineapple']

I would want to create a FunctionTransformer to match words in the long_list with each word of the column value, if there is a match, return the count.

index | text                                             | count
1     | "I have a pen, but I lost it today."             | 1
2     | "I have pineapple and pen, but I lost it today." | 2

I did in this way:

def count_words(df):
    long_list = ['pen', 'pineapple']
    count = 0
    for c in df['tweet_text']:
        if c in long_list:
            count = count + 1
            
    df['count'] = count   
    return df

count_word = FunctionTransformer(count_words, validate=False)

An example of how I develop my other FunctionTransformer will be:

def convert_twitter_datetime(df):
    df['hour'] = pd.to_datetime(df['created_at'], format='%a %b %d %H:%M:%S +0000 %Y').dt.strftime('%H').astype(int)
    return df

convert_datetime = FunctionTransformer(convert_twitter_datetime, validate=False)

Upvotes: 0

Views: 1193

Answers (3)

pythonic833
pythonic833

Reputation: 3224

Inspired by @Quang Hoang's answer

import pandas as pd
import sklearn as sk

y=['pen', 'pineapple']

def count_strings(X, y):
    pattern = r'\b{}\b'.format('|'.join(y))
    return X['text'].str.count(pattern)

string_transformer = sk.preprocessing.FunctionTransformer(count_strings, kw_args={'y': y})
df['count'] = string_transformer.fit_transform(X=df)

results in

    text                                              count
1   "I have a pen, but I lost it today."                1
2   "I have pineapple and pen, but I lost it today.     2

And for the following df2:

#df2
      text
1     "I have a pen, but I lost it today. pen pen"
2     "I have pineapple and pen, but I lost it today."

We get

string_transformer.transform(X=df2)
#result
1    3
2    2
Name: text, dtype: int64

This shows, that we converted the function to an sklearn-style object. To abstact this even further we can hand over the column name as key-word argument to count_strings.

Upvotes: 0

wwnde
wwnde

Reputation: 26676

Join elements in a list with with |. Find matching elements with .str.findall() and apply .str.len() for count

 p='|'.join(long_list)
df=df.assign(count=(df.text.str.findall(p)).str.len())
                                             text   count
0              "I have a pen, but I lost it today."      1
1  "I have pineapple and pen, but I lost it today."      2

Upvotes: 0

Quang Hoang
Quang Hoang

Reputation: 150765

Pandas has str.count:

# matching any of the words
pattern = r'\b{}\b'.format('|'.join(long_list))

df['count'] = df.text.str.count(pattern)

Output:

   index                                              text  count
0      1              "I have a pen, but I lost it today."      1
1      2  "I have pineapple and pen, but I lost it today."      2

Upvotes: 2

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