CODEWITHSUNDEEP
pythonregexpandasparsingdataframe
Omar14
Omar14

Reputation: 2055

Regex: Extract specific value from URL

I am having some trouble to exact the string from URL using re library.

here's an example:

http://www.example.it/[email protected]&direction=vente.aspx%3pid%xx123%63abcd"

I have a dataframe and i want to add a column using a value from another column, in this example df['URL_REG'] contains: '123'?

df['URL_REG'] = df['URL'].map(lambda x : re.findall(r'[REGEX]+', x)[0])

the structure of URL can change but the part that i want comes always between 'direction=vente.aspx%3pid%' and '%'.

Upvotes: 0

Views: 624

Answers (2)

MaxU - stand with Ukraine
MaxU - stand with Ukraine

Reputation: 210982

Use vectorized Series.str.extract() method:

In [50]: df['URL_REG'] = df.URL.str.extract(r'direction=vente.aspx\%3pid\%([^\%]+)\%*',
                                            expand=False)

In [51]: df
Out[51]:
                                                 URL URL_REG
0  http://www.example.it/remoteconnexion.aspx?u=x...   xx123

UPDATE:

i want only '123' part instead of 'xx123', where 'xx' is a hexademical number

In [53]: df['URL_REG'] = df.URL.str.extract(r'direction=vente.aspx\%3pid\%\w{2}(\d+)\%*', 
                                            expand=False)

In [54]: df
Out[54]:
                                                 URL URL_REG
0  http://www.example.it/remoteconnexion.aspx?u=x...     123

Upvotes: 2

Chiheb Nexus
Chiheb Nexus

Reputation: 9267

You can use this pattern:

import re

url='http://www.example.it/[email protected]&direction=vente.aspx%3pid%xx123%63abcd'
output = re.findall('3pid%(.*?)%', url)

print(output)

Output:

['xx123']

Then apply the same pattern to your DataFrame.

For example:

import pandas as pd
import re

df = pd.DataFrame(['http://www.example.it/[email protected]&direction=vente.aspx%3pid%xx123%63abcd'], columns = ['URL'])

output = df['URL'].apply(lambda x : re.findall('3pid%(.*?)%', x))

print(output)

# Or, maybe if you want to return the url and the data captured:
# output = df['URL'].apply(lambda x : (x, re.findall('3pid%(.*?)%', x)))
# output[0]
# >>> ('http://www.example.it/[email protected]&direction=vente.aspx%3pid%xx123%63abcd', 
#   ['xx123'])

Output:

0    [xx123]
Name: URL, dtype: object

Upvotes: 0

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