I have a pandas data frame. I want to group it by using one combination of columns and count distinct values of another combination of columns.
For example I have the following data frame:
a b c d e
0 1 10 100 1000 10000
1 1 10 100 1000 20000
2 1 20 100 1000 20000
3 1 20 100 2000 20000
I can group it by columns a
and b
and count distinct values in the column d
:
df.groupby(['a','b'])['d'].nunique().reset_index()
As a result I get:
a b d
0 1 10 1
1 1 20 2
However, I would like to count distinct values in a combination of columns. For example if I use c
and d
, then in the first group I have only one unique combination ((100, 1000)
) while in the second group I have two distinct combinations: (100, 1000)
and (100, 2000)
.
The following naive "generalization" does not work:
df.groupby(['a','b'])[['c','d']].nunique().reset_index()
because nunique()
is not applicable to data frames.
You can create combination of values converting to string
to new column e
and then use SeriesGroupBy.nunique
:
df['e'] = df.c.astype(str) + df.d.astype(str)
df = df.groupby(['a','b'])['e'].nunique().reset_index()
print (df)a b e
0 1 10 1
1 1 20 2
You can also use Series
without creating new column:
df =(df.c.astype(str)+df.d.astype(str)).groupby([df.a, df.b]).nunique().reset_index(name='f')
print (df)a b f
0 1 10 1
1 1 20 2
Another posible solution is create tuples:
df=(df[['c','d']].apply(tuple, axis=1)).groupby([df.a, df.b]).nunique().reset_index(name='f')
print (df)a b f
0 1 10 1
1 1 20 2
Another numpy solution by this answer:
def f(x):a = x.valuesc = len(np.unique(np.ascontiguousarray(a).view(np.dtype((np.void, a.dtype.itemsize * a.shape[1]))), return_counts=True)[1])return cprint (df.groupby(['a','b'])[['c','d']].apply(f))
Timings:
#[1000000 rows x 5 columns]
np.random.seed(123)
N = 1000000
df = pd.DataFrame(np.random.randint(30, size=(N,5)))
df.columns = list('abcde')
print (df) In [354]: %timeit (df.groupby(['a','b'])[['c','d']].apply(lambda g: len(g) - g.duplicated().sum()))
1 loop, best of 3: 663 ms per loopIn [355]: %timeit (df.groupby(['a','b'])[['c','d']].apply(f))
1 loop, best of 3: 387 ms per loopIn [356]: %timeit (df.groupby(['a', 'b', 'c', 'd']).size().groupby(level=['a', 'b']).size())
1 loop, best of 3: 441 ms per loopIn [357]: %timeit ((df.c.astype(str)+df.d.astype(str)).groupby([df.a, df.b]).nunique())
1 loop, best of 3: 4.95 s per loopIn [358]: %timeit ((df[['c','d']].apply(tuple, axis=1)).groupby([df.a, df.b]).nunique())
1 loop, best of 3: 17.6 s per loop