pandas.core.groupby.DataFrameGroupBy.nunique

DataFrameGroupBy.nunique(dropna=True)[source]

Return DataFrame with counts of unique elements in each position.

Parameters

dropna:bool, default True

Don’t include NaN in the counts.

Returns

nunique: DataFrame

Examples

>>> df = pd.DataFrame({'id': ['spam', 'egg', 'egg', 'spam',
...                           'ham', 'ham'],
...                    'value1': [1, 5, 5, 2, 5, 5],
...                    'value2': list('abbaxy')})
>>> df
     id  value1 value2
0  spam       1      a
1   egg       5      b
2   egg       5      b
3  spam       2      a
4   ham       5      x
5   ham       5      y

>>> df.groupby('id').nunique()
      value1  value2
id
egg        1       1
ham        1       2
spam       2       1

Check for rows with the same id but conflicting values:

>>> df.groupby('id').filter(lambda g: (g.nunique() > 1).any())
     id  value1 value2
0  spam       1      a
3  spam       2      a
4   ham       5      x
5   ham       5      y