Want to update column Values for each occurrence of column AppName where at least one of the values in column Values is “Yes” s.t all values are ‘Yes’

I have a table

And I want to update this table so that for each occurrence of AppName where at least one of the values in the “Values” column is ‘Yes’ then we update all of the values for that specific AppName to “Yes”.

Updated Table:

I tried

        if df2['Values'] == 'Yes':
           df2['Values'] == 'Yes'  ```
But this throws a ```ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().``` for ```if df2['Values'] == 'Yes':```

Answers:

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Method 1

Approach using groupby.transform, placing results into a new column for easy comparison with original Values column:

df['app_has_any_yes'] = (df.groupby('AppName')['Values']
                           .transform(lambda x: (x == 'Yes').any())
                           .replace({True: 'Yes', False: 'No'})

Result:

Method 2

Conveniently, “Yes” is lexicographically sorted after “No”, so Yes>No and you can do:

df['Values'] = df.groupby('AppName')['Values'].transform('max')

NB. this is working for this particular example, but you can use the same logic with other words, for example in German “Ja” (Yes) is before “Nein” (No), so you would use transform('min') in this case

output:

  AppName Values
0     JBX    Yes
1     JBX    Yes
2     JBX    Yes
3     ABD     No
4     ABD     No
5     CBR    Yes
6     LBA    Yes
7     LBA    Yes

Method 3

You can try groupby then apply

out = df.groupby('AppName').apply(lambda g: g.assign(Values=['Yes']*len(g) if g['Values'].eq('Yes').any() else ['No']*len(g)))

print(out)

  AppName Values
0     JBX    Yes
1     JBX    Yes
2     JBX    Yes
3     ABD     No
4     ABD     No
5     CBR    Yes
6     LBA    Yes
7     LBA    Yes

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0