On Friday, 30 August 2019 00:49:32 UTC+10, Piet van Oostrum wrote: > Piet van Oostrum writes: > > > So the correct way to do this is to make df1 a copy rather than a view. > > > > df1 = df.loc[:, ('UID','Name','New Leader','Current Team', 'New Team')] > > Or maybe even make an explicit copy: > > df1 = df[['UID','Name','New Leader','Current Team', 'New Team']].copy() > -- > Piet van Oostrum > WWW: http://piet.vanoostrum.org/ > PGP key: [8DAE142BE17999C4]
I have tried both df1 = df.loc[:, ('UID','Name','New Leader','Current Team', 'New Team')] df1['Difference'] = df1.loc['Current Team'].str.lower().str.strip() == df1.loc['New Team'].str.lower().str.strip() and df1 = df[['UID','Name','New Leader','Current Team', 'New Team']].copy() df1['Difference'] = df1.loc['Current Team'].str.lower().str.strip() == df1.loc['New Team'].str.lower().str.strip() But on both occasions I receive this error. # KeyError: 'the label [Current Team] is not in the [index]' if I test df1 before trying to create the new column it works just fine. Sayth -- https://mail.python.org/mailman/listinfo/python-list