Creating names from URLs

Question

I am creating a NAME column in a DataFrame and set its value based on substrings contained in another column.

Is there a more efficient way to do this?

import pandas as pd
df = pd.DataFrame([['www.pandas.org','low'], ['www.python.org','high']],
     columns=['URL','speed'])
print(df.head())


df['Name'] = df['URL']  
print(df.head())

#set Name based on substring in URL
df.loc[df['Name'].str.contains("pandas", na=False), 'Name'] = "PANDAS"
df.loc[df['Name'].str.contains("python|pitone", na=False), 'Name'] = "PYTHON"
print(df.head())

G. Sliepen · Accepted Answer · 2020-06-25 16:05:23Z

2

Yes. Try to automate converting a URL to a name, instead of hardcoding the mapping. With only two URLs, your approach is doable, but as soon as you have to handle lots of different URLs, it will quickly become very painful.

Also, avoid first copying a column and then replacing every item in it. Rather, construct the column with the right contents directly. Here is an example:

df['Name'] = [url.split('.')[-2].upper() for url in df['URL']]

This will fail for proper URLs like https://www.pandas.org/index.html. If you want to handle that automatically, you'll have to properly parse them, for example using urllib.parse.

answered Jun 25, 2020 at 16:05

G. Sliepen

69.3k3 gold badges75 silver badges180 bronze badges

\$\begingroup\$ You should probably use df['URL'].str.split('.').str[-2].str.upper() instead of a list comprehension for some more speed. \$\endgroup\$

Graipher
– Graipher

2020-06-26 06:50:15 +00:00
Commented Jun 26, 2020 at 6:50

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