How to Extract Domain from URL in Pandas   

In this short guide, I'll show you how to extract domain from a URL column in Pandas DataFrame. You can also find how to extract netloc, schema, path, params. So at the end you will get:

['https://www.datascientyst.com/cheatsheet','https://www.softhints.com/python']

to:

0    (https, www.datascientyst.com, /cheatsheet, , , )
1            (https, www.softhints.com, /python, , , )
Name: urls, dtype: object

or extracting only domains:

0    www.datascientyst.com
1        www.softhints.com
Name: urls, dtype: object

Setup

Let's have DataFrame with URL column from which we will extract list of domains:

import pandas as pd

data = {'urls': ['https://www.datascientyst.com/cheatsheet','https://www.softhints.com/python']}

df = pd.DataFrame(data)

DataFrame looks like:

Step 1: Extract domain from URL - urlparse

First way to extract domain from URL in Python is library - urlparse:

from urllib.parse import urlparse

df['urls'].apply(urlparse)

This will extract all information as Series of tuples:

0    (https, www.datascientyst.com, /cheatsheet, , , )
1            (https, www.softhints.com, /python, , , )
Name: urls, dtype: object

To extract only the netloc or the domain we can use:

df['urls'].apply(lambda x: urlparse(x)[1])

extracted netloc-s from the URL column:

0    www.datascientyst.com
1        www.softhints.com
Name: urls, dtype: object

We can extract the full ParseResult from the urlparse library by:

df['paths'] = df['urls'].apply(lambda x: urlparse(x))
df['paths'].to_dict()

result is full URL information:

{0: ParseResult(scheme='https', netloc='www.datascientyst.com', path='/cheatsheet', params='', query='', fragment=''),
 1: ParseResult(scheme='https', netloc='www.softhints.com', path='/python', params='', query='', fragment='')}

Step 2: Extract domain from URL - regex

We can use regular expression in order to extract patterns from the URL columns. Pandas offers .str.extract method:

df['urls'].str.extract(r'https://(.*)/')

would extract the domains plus the subdomains if any:

Or we can match up to symbol without extracting the symbol. In this case we will search for anything until we match char t:

df['urls'].str.extract(r'(https.*)(?:t)').head()

result:

Conclusion

We saw two different ways how to parse URL information with Python and Pandas.

We can extract from Pandas DataFrame information like:

• scheme='https'
• netloc='www.datascientyst.com'
• path='/cheatsheet'
• params=''
• query=''
• fragment=''