I am checking on which words the SpaCy Spanish lemmatizer works on using the .has_vector method. In the two columns of the datafame I have the output of the function that indicates which words can be lemmatized and in the other one the corresponding phrase.
I would like to know how I can extract all the words that have False output to correct them so that I can lemmatize.
So I created the function:
def lemmatizer(text):
doc = nlp(text)
return ' '.join([str(word.has_vector) for word in doc])
And applied it to the column sentences in the DataFrame
df["Vectors"] = df.reviews.apply(lemmatizer)
And put in another data frame as:
df2= pd.DataFrame(df[['Vectors', 'reviews']])
The output is
index Vectors reviews
1 True True True False 'La pelicula es aburridora'
CodePudding user response:
Two ways to do this:
import pandas
import spacy
nlp = spacy.load('en_core_web_lg')
df = pandas.DataFrame({'reviews': ["aaabbbcccc some example words xxxxyyyz"]})
If you want to use has_vector:
def get_oov1(text):
return [word.text for word in nlp(text) if not word.has_vector]
Alternatively you can use the is_oov attribute:
def get_oov2(text):
return [word.text for word in nlp(text) if word.is_oov]
Then as you already did:
df["oov_words1"] = df.reviews.apply(get_oov1)
df["oov_words2"] = df.reviews.apply(get_oov2)
Which will return:
> reviews oov_words1 oov_words2
0 aaabbbcccc some example words xxxxyyyz [aaabbbcccc, xxxxyyyz] [aaabbbcccc, xxxxyyyz]
Note:
When working with both of these ways it is important to know that this is model dependent, and usually has no backbone in smaller models and will always return a default value!
That means when you run the exact same code but e.g. with en_core_web_sm
you get this:
> reviews oov_words1 oov_words2
0 aaabbbcccc some example words xxxxyyyz [] [aaabbbcccc, some, example, words, xxxxyyyz]
Which is because has_vector
has a default value of False
and is then not set by the model. is_oov
has a default value of True
and then is not by the model either. So with the has_vector
model it wrongly shows all words as unknown and with is_oov
it wrongly shows all as known.