Reputation: 33
Here is a simple dataset:
import pandas as pd
product = ['knife', 'box set', 'beautiful jewellery set on sale', 'green']
df = pd.DataFrame(product, columns = ['product_name'])
df
The output as follows:
product_name | |
---|---|
0 | knife |
1 | box set |
2 | beautiful jewellery set on sale |
3 | green |
I want to categorise these products by extracting two consecutive nouns if needed. So far I have the following, but the categories are presented by only one noun for all cases:
!pip install -q --upgrade spacy
import spacy
nlp = spacy.load('en_core_web_sm')
category=[]
for i in df['product_name'].tolist():
doc = nlp(i)
for t in doc:
if t.pos_ in ['NOUN']:
category.append(f'{t}')
break
if t.pos_ not in ['NOUN']:
category.append('NaN')
df1 = pd.DataFrame(category, columns =['product_category'])
df1
The output I have:
product_category | |
---|---|
0 | knife |
1 | set |
2 | jewellery |
3 | NaN |
The expected output:
product_category | |
---|---|
0 | knife |
1 | box set |
2 | jewellery set |
3 | NaN |
Is it possible to introduce some additional conditions to the code to extract two nouns if they follow one after another?
Upvotes: 3
Views: 368
Reputation: 626748
You can use
import spacy
import pandas as pd
import numpy as np
product = ['knife', 'box set', 'beautiful jewellery set on sale', 'green']
df = pd.DataFrame(product, columns = ['product_name'])
nlp = spacy.load('en_core_web_sm')
matcher = spacy.matcher.Matcher(nlp.vocab)
pattern = [{'POS': 'NOUN'},{'POS': 'NOUN','OP':'?'}]
matcher.add('NOUN_PATTERN', [pattern])
def get_two_nouns(x):
doc = nlp(x)
results = []
for match_id, start, end in matcher(doc):
span = doc[start:end]
results.append(span.text)
return max(results, key = lambda x: len(x.split()), default=np.nan)
df['product_name'].apply(get_two_nouns)
Output:
0 knife
1 box set
2 jewellery set
3 NaN
Name: product_name, dtype: object
The pattern = [{'POS': 'NOUN'},{'POS': 'NOUN','OP':'?'}]
pattern matches (combinations of) tokens that are both NOUN
s. The second one is optional due to the OP
operator set to ?
.
The return max(results, key = lambda x: len(x.split()), default=np.nan)
part returns the item with the longest length (length measured in whitespace separated token count here).
Upvotes: 2