如何正确循环熊猫数据框？

Question

I have a dataframe that looks like this:

for user in df_enlaces['CorreoElectronico'].values:

  to_mail=(df_enlaces.set_index('CorreoElectronico').loc[user].index[0])

  print(to_mail, df_enlaces.set_index('CorreoElectronico').loc[user]['Enlace'])

The output of the above code is:

example1@mail.com CorreoElectronico
example1@mail.com    link1
example1@mail.com    link2
Name: Enlace, dtype: object
example1@mail.com CorreoElectronico
example1@mail.com    link1
example1@mail.com    link2
Name: Enlace, dtype: object
example2@mail.com CorreoElectronico
example2@mail.com    link3
example2@mail.com    link4
example2@mail.com    link5
Name: Enlace, dtype: object
example2@mail.com CorreoElectronico
example2@mail.com    link3
example2@mail.com    link4
example2@mail.com    link5
Name: Enlace, dtype: object
example2@mail.com CorreoElectronico
example2@mail.com    link3
example2@mail.com    link4
example2@mail.com    link5
Name: Enlace, dtype: object

However, it is repeated. Something I am doing wrong when looping. I just want to match the user email with its link in the column ['Enlace']. The desired output would be:

example1@mail.com    link1
example1@mail.com    link2
example2@mail.com    link3
example2@mail.com    link4
example2@mail.com    link5

Answer 1

try this

for user in df_enlaces['CorreoElectronico'].unique():
    ....

You are looping through each value in the CorreoElectronico column and then when you perform .loc it pulls out every row which matches that user (and there are several)

Answer 2

It seems like this should do the trick:

users = ['user1', 'user1', 'user2', 'user2', 'user3']
links = ['link1', 'link2', 'link3', 'link4', 'link5']
df = pd.DataFrame({'user':users,'link':links})

for index, row in df.iterrows():
    print(row['user'], row['link'])

At least it does when you just want to print the user and link from each row in the dataframe? This outputs:

user1 link1
user1 link2
user2 link3
user2 link4
user3 link5

Answer 3

As I suppose, you want to:

• iterate over rows of df_enlaces ,
• from each row print the e-mail address ( CorreoElectronico ) and the link ( Enlace ).

To do it, use the following loop:

for _, row in df_enlaces.iterrows():
    print(f'{row.CorreoElectronico}    {row.Enlace}')