13

I have a simple dataframe as such:

df = [    {'col1' : 'A', 'col2': 'B', 'col3':   'C', 'col4':'0'},
          {'col1' : 'M', 'col2':   '0', 'col3': 'M', 'col4':'0'},
          {'col1' : 'B', 'col2':  'B', 'col3':  '0', 'col4':'B'},
          {'col1' : 'X', 'col2':  '0', 'col3':  'Y', 'col4':'0'}
          ]
df = pd.DataFrame(df)
df = df[['col1', 'col2', 'col3', 'col4']]
df  

Which looks like this:

| col1 | col2 | col3 | col4 |
|------|------|------|------|
| A    | B    | C    | 0    |
| M    | 0    | M    | 0    |
| B    | B    | 0    | B    |
| X    | 0    | Y    | 0    |

I just want to replace repeated characters with the character '0', across the rows. It boils down to keeping the first duplicate value we come across, as like this:

| col1 | col2 | col3 | col4 |
|------|------|------|------|
| A    | B    | C    | 0    |
| M    | 0    | 0    | 0    |
| B    | 0    | 0    | 0    |
| X    | 0    | Y    | 0    |

This seems so simple but I'm stuck. Any nudges in the right direction would be really appreciated.

0

1 Answer 1

18

You can use the duplicated method to return a boolean indexer of whether elements are duplicates or not:

In [214]: pd.Series(['M', '0', 'M', '0']).duplicated()
Out[214]:
0    False
1    False
2     True
3     True
dtype: bool

Then you could create a mask by mapping this across the rows of your dataframe, and using where to perform your substitution:

is_duplicate = df.apply(pd.Series.duplicated, axis=1)
df.where(~is_duplicate, 0)

  col1 col2 col3 col4
0    A    B    C    0
1    M    0    0    0
2    B    0    0    0
3    X    0    Y    0
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1 Comment

Nice -- I was thinking of stacking/groupby/pivoting, but this is much, much cleaner. You could avoid the lambda by using pd.Series.duplicated directly, but that's minor.

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