Chaining grouping, filtration and aggregation

Question

DataFrameGroupby.filter method filters the groups, and returns the DataFrame that contains the rows that passed the filter.

But what can I do to obtain a new DataFrameGroupBy object instead of a DataFrame after filtration?

For example, let's say I have a DataFrame df with two columns A and B. I want to obtain average value of column B for each value of column A, as long as there's at least 5 rows in that group:

# pandas 0.18.0
# doesn't work because `filter` returns a DF not a GroupBy object
df.groupby('A').filter(lambda x: len(x)>=5).mean()
# works but slower and awkward to write because needs to groupby('A') twice
df.groupby('A').filter(lambda x: len(x)>=5).reset_index().groupby('A').mean()
# works but more verbose than chaining
groups = df.groupby('A')
groups.mean()[groups.size() >= 5]

Answer 1

You can do it this way:

In [310]: df
Out[310]:
    a  b
0   1  4
1   7  3
2   6  9
3   4  4
4   0  2
5   8  4
6   7  7
7   0  5
8   8  5
9   8  7
10  6  1
11  3  8
12  7  4
13  8  0
14  5  3
15  5  3
16  8  1
17  7  2
18  9  9
19  3  2
20  9  1
21  1  2
22  0  3
23  8  9
24  7  7
25  8  1
26  5  8
27  9  6
28  2  8
29  9  0

In [314]: r = df.groupby('a').apply(lambda x: x.b.mean() if len(x)>=5 else -1)

In [315]: r
Out[315]:
a
0   -1.000000
1   -1.000000
2   -1.000000
3   -1.000000
4   -1.000000
5   -1.000000
6   -1.000000
7    4.600000
8    3.857143
9   -1.000000
dtype: float64

In [316]: r[r>0]
Out[316]:
a
7    4.600000
8    3.857143
dtype: float64

One-liner, which returns data frame instead of series:

df.groupby('a') \
  .apply(lambda x: x.b.mean() if len(x)>=5 else -1) \
  .to_frame() \
  .rename(columns={0:'mean'}) \
  .query('mean > 0')

Timeit comparison against a DF with 100.000 rows:

def maxu():
    r = df.groupby('a').apply(lambda x: x.b.mean() if len(x)>=5 else -1)
    return r[r>0]

def maxu2():
    return df.groupby('a') \
             .apply(lambda x: x.b.mean() if len(x)>=5 else -1) \
             .to_frame() \
             .rename(columns={0:'mean'}) \
             .query('mean > 0')

def alexander():
    return df.groupby('a', as_index=False).filter(lambda group: group.a.count() >= 5).groupby('a').mean()

def alexander2():
    vc = df.a.value_counts()
    return df.loc[df.a.isin(vc[vc >= 5].index)].groupby('a').mean()

Results:

In [419]: %timeit maxu()
1 loop, best of 3: 1.12 s per loop

In [420]: %timeit maxu2()
1 loop, best of 3: 1.12 s per loop

In [421]: %timeit alexander()
1 loop, best of 3: 34.9 s per loop

In [422]: %timeit alexander2()
10 loops, best of 3: 66.6 ms per loop

Check:

In [423]: alexander2().sum()
Out[423]:
b   19220943.162
dtype: float64

In [424]: maxu2().sum()
Out[424]:
mean   19220943.162
dtype: float64

Conclusion:

clear winner is alexander2() function

@Alexander, congratulations!

Answer 2

Here is some reproduceable data:

np.random.seed(0)

df = pd.DataFrame(np.random.randint(0, 10, (10, 2)), columns=list('AB'))

>>> df
   A  B
0  5  0
1  3  3
2  7  9
3  3  5
4  2  4
5  7  6
6  8  8
7  1  6
8  7  7
9  8  1

A sample filter application demonstrating that it works on the data.

gb = df.groupby('A')
>>> gb.filter(lambda group: group.A.count() >= 3)
   A  B
2  7  9
5  7  6
8  7  7

Here are some of your options:

1) You can also first filter based on the value counts, and then group.

vc = df.A.value_counts()

>>> df.loc[df.A.isin(vc[vc >= 2].index)].groupby('A').mean()
          B
A          
3  4.000000
7  7.333333
8  4.500000

2) Perform groupby twice, before and after the filter:

>>> (df.groupby('A', as_index=False)
       .filter(lambda group: group.A.count() >= 2)
       .groupby('A')
       .mean())
          B
A          
3  4.000000
7  7.333333
8  4.500000

3) Given that your first groupby returns the groups, you can also filter on those:

d = {k: v 
     for k, v in df.groupby('A').groups.items() 
     if len(v) >= 2}  # gb.groups.iteritems() for Python 2

>>> d
{3: [1, 3], 7: [2, 5, 8], 8: [6, 9]}

This is a bit of a hack, but should be relatively efficient as you don't need to regroup.

>>> pd.DataFrame({col: [df.ix[d[col], 'B'].mean()] for col in d}).T.rename(columns={0: 'B'})
          B
3  4.000000
7  7.333333
8  4.500000

Timings with 100k rows

np.random.seed(0)
df = pd.DataFrame(np.random.randint(0, 10, (100000, 2)), columns=list('AB'))

%timeit df.groupby('A', as_index=False).filter(lambda group: group['A'].count() >= 5).groupby('A').mean()
100 loops, best of 3: 18 ms per loop

%%timeit
vc = df.A.value_counts()
df.loc[df.A.isin(vc[vc >= 2].index)].groupby('A').mean()
100 loops, best of 3: 15.7 ms per loop