Write lambda function to function

Question

I have a lambda function like this:

from Sastrawi.Stemmer.StemmerFactory import StemmerFactory

factory = StemmerFactory()
stemmer = factory.create_stemmer()

df['tweets_stemming'] = df['tweets_stopwords'].apply(lambda x: [stemmer.stem(stem) for stem in x])

And I want to convert this lambda function to function so I write this:

from Sastrawi.Stemmer.StemmerFactory import StemmerFactory

def stemmer_remover(stemming):
  factory = StemmerFactory()
  stemmer = factory.create_stemmer()
  stemming = [stemmer.stem(stemming) for stemming in df['tweets_stopwords']]

  return stemming

df['tweets_stemming'] = df['tweets_stopwords'].apply(stemmer_remover)

If using lambda function my code runs fine. But, if using function (not lambda) giving an error TypeError: descriptor 'lower' requires a 'str' object but received a 'list'. So, I think because my pandas dataframe from df['tweets_stopwords'] (see the image)

How to fix this?

In your own words, when you use .apply, how many times do you think the passed-in function gets called? What do you think gets passed to it? In your own words, in your version of the code using the function, how do you think the function is using the passed-in parameter? — Karl Knechtel
– Karl Knechtel, Commented Oct 1, 2021 at 23:11
A lambda expression, lambda <args>: <lambda body> can be simply replaced by def some_name(<args>): return <lambda body> — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Oct 1, 2021 at 23:15

Samwise · Accepted Answer · 2021-10-01 23:13:01Z

1

Your named function doesn't work the same way as your lambda.

lambda x: [stemmer.stem(stem) for stem in x]

is the same as:

def stemmer_remover(x):
    return [stemmer.stem(stem) for stem in x]

If you replace your lambda expression with the above stemmer_remover you should find that it behaves the same way.

answered Oct 1, 2021 at 23:13

Samwise

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