Python Forum

I have a dataframe column for emal addresses and I am trying to find the count of the top domains.
I am sppliting the email using the '@" as the separator and trying to grab the domain part as follows:

df['Email'].str.split('@')[1]

# I get the following list instead of the second part of the list:
[output]

['anthony41', 'reed.com']

[/output]

how can I just get the domain part? in this case 'reed.com'?

import pandas as pd

df = pd.DataFrame({"Email": ["[email protected]", "[email protected]", "[email protected]"]})
df["domain"] = df["Email"].apply(lambda a: a.split("@")[1])
df["name"] = df["Email"].map(lambda a: a.split("@")[0])
print(df)

Output:
           Email       domain name
0    [email protected]    gmail.com    A
1  [email protected]  outlook.com    B
2      [email protected]      aol.com    C

Or

import pandas as pd

df = pd.DataFrame({"Email": ["[email protected]", "[email protected]", "[email protected]"]})
df[["Name", "Domain"]] = df["Email"].str.split("@", expand=True)
print(df)

Output:
           Email Name       Domain
0    [email protected]    A    gmail.com
1  [email protected]    B  outlook.com
2      [email protected]    C      aol.com

(Jan-27-2023, 09:25 PM)deanhystad Wrote: [ -> ]

import pandas as pd

df = pd.DataFrame({"Email": ["[email protected]", "[email protected]", "[email protected]"]})
df["domain"] = df["Email"].apply(lambda a: a.split("@")[1])
df["name"] = df["Email"].map(lambda a: a.split("@")[0])
print(df)

Output:
           Email       domain name
0    [email protected]    gmail.com    A
1  [email protected]  outlook.com    B
2      [email protected]      aol.com    C

Or

import pandas as pd

df = pd.DataFrame({"Email": ["[email protected]", "[email protected]", "[email protected]"]})
df[["Name", "Domain"]] = df["Email"].str.split("@", expand=True)
print(df)

Output:
           Email Name       Domain
0    [email protected]    A    gmail.com
1  [email protected]    B  outlook.com
2      [email protected]    C      aol.com

deanhystad - Thank kindly for the help. - Both versions worked very well.