'how to use word_tokenize in data frame
I have recently started using the nltk module for text analysis. I am stuck at a point. I want to use word_tokenize on a dataframe, so as to obtain all the words used in a particular row of the dataframe.
data example:
text
1. This is a very good site. I will recommend it to others.
2. Can you please give me a call at 9983938428. have issues with the listings.
3. good work! keep it up
4. not a very helpful site in finding home decor.
expected output:
1. 'This','is','a','very','good','site','.','I','will','recommend','it','to','others','.'
2. 'Can','you','please','give','me','a','call','at','9983938428','.','have','issues','with','the','listings'
3. 'good','work','!','keep','it','up'
4. 'not','a','very','helpful','site','in','finding','home','decor'
Basically, i want to separate all the words and find the length of each text in the dataframe.
I know word_tokenize can for it for a string, but how to apply it onto the entire dataframe?
Please help!
Thanks in advance...
Solution 1:[1]
pandas.Series.apply is faster than pandas.DataFrame.apply
import pandas as pd
import nltk
df = pd.read_csv("/path/to/file.csv")
start = time.time()
df["unigrams"] = df["verbatim"].apply(nltk.word_tokenize)
print "series.apply", (time.time() - start)
start = time.time()
df["unigrams2"] = df.apply(lambda row: nltk.word_tokenize(row["verbatim"]), axis=1)
print "dataframe.apply", (time.time() - start)
On a sample 125 MB csv file,
series.apply 144.428858995
dataframe.apply 201.884778976
Edit: You could be thinking the Dataframe df after series.apply(nltk.word_tokenize) is larger in size, which might affect the runtime for the next operation dataframe.apply(nltk.word_tokenize).
Pandas optimizes under the hood for such a scenario. I got a similar runtime of 200s by only performing dataframe.apply(nltk.word_tokenize) separately.
Solution 2:[2]
I will show you an example. Suppose you have a data frame named twitter_df and you have stored sentiment and text within that. So, first I extract text data into a list as follows
tweetText = twitter_df['text']
then to tokenize
from nltk.tokenize import word_tokenize
tweetText = tweetText.apply(word_tokenize)
tweetText.head()
I think this will help you
Solution 3:[3]
May need to add str() to convert to pandas' object type to a string.
Keep in mind a faster way to count words is often to count spaces.
Interesting that tokenizer counts periods. May want to remove those first, maybe also remove numbers. Un-commenting the line below will result in equal counts, at least in this case.
import nltk
import pandas as pd
sentences = pd.Series([
'This is a very good site. I will recommend it to others.',
'Can you please give me a call at 9983938428. have issues with the listings.',
'good work! keep it up',
'not a very helpful site in finding home decor. '
])
# remove anything but characters and spaces
sentences = sentences.str.replace('[^A-z ]','').str.replace(' +',' ').str.strip()
splitwords = [ nltk.word_tokenize( str(sentence) ) for sentence in sentences ]
print(splitwords)
# output: [['This', 'is', 'a', 'very', 'good', 'site', 'I', 'will', 'recommend', 'it', 'to', 'others'], ['Can', 'you', 'please', 'give', 'me', 'a', 'call', 'at', 'have', 'issues', 'with', 'the', 'listings'], ['good', 'work', 'keep', 'it', 'up'], ['not', 'a', 'very', 'helpful', 'site', 'in', 'finding', 'home', 'decor']]
wordcounts = [ len(words) for words in splitwords ]
print(wordcounts)
# output: [12, 13, 5, 9]
wordcounts2 = [ sentence.count(' ') + 1 for sentence in sentences ]
print(wordcounts2)
# output: [12, 13, 5, 9]
If you aren't using Pandas, you might not need str()
Solution 4:[4]
Make it faster using pandarallel
Using Spacy
import spacy from pandarallel import pandarallel pandarallel.initialize(progress_bar=True) nlp = spacy.load("en_core_web_sm") df['new_col'] = df['text'].parallel_apply(lambda x: nlp(x))Using NLTK
import nltk from pandarallel import pandarallel pandarallel.initialize(progress_bar=True) df['new_col'] = df['text'].parallel_apply(lambda x: nltk.word_tokenize(x))
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | |
| Solution 2 | Yasuni Chamodya |
| Solution 3 | |
| Solution 4 | Ramkrishan Sahu |
