汇总熊猫中数据分组的分组列中的值

Question

Here is my pandas.DataFrame:

    a  b
0   1  5
1   1  7
2   2  3
3   1  3
4   2  5
5   2  6
6   1  4
7   1  3
8   2  7
9   2  4
10   2  5

I want to create a new DataFrame that will contain the data grouped by 'a' and contains the sum of the largest 3 values for each group.

Here is the output I expect. The largest 3 values of 'b' for group 1 are 7,5 and 4, and for group 2 are 7, 6 and 5.

a
1  16
2  18

df.groupby('a')['b'].nlargest(3)

gives me this output,

 a    
 1  1     7
    0     5
    6     4
 2  8     7
    5     6
    10    5

and

  df.groupby('a')['b'].nlargest(3).sum()

gives me the total sum 34 (16+18).

How can I get the expected output with pandas.DataFrame?

Thank you!

Answer 1

Using apply is one way to do it.

In [41]: df.groupby('a')['b'].apply(lambda x: x.nlargest(3).sum())
Out[41]:
a
1    16
2    18
Name: b, dtype: int64

Timings

In [42]: dff = pd.concat([df]*1000).reset_index(drop=True)

In [43]: dff.shape
Out[43]: (11000, 2)

In [44]: %timeit dff.groupby('a')['b'].apply(lambda x: x.nlargest(3).sum())
100 loops, best of 3: 2.44 ms per loop

In [45]: %timeit dff.groupby('a')['b'].nlargest(3).groupby(level='a').sum()
100 loops, best of 3: 3.44 ms per loop

Answer 2

Use double groupby - second by level a of MultiIndex :

s = df.groupby('a')['b'].nlargest(3).groupby(level='a').sum()
print (s)
a
1    16
2    18
Name: b, dtype: int64

But for me is nicer:

df.groupby('a')['b'].nlargest(3).sum(level=0)

thank you Nickil Maveli .

EDIT: If need top 3 again, use Series.nlargest :

df = pd.DataFrame({'a': [1, 1, 2, 3, 2, 2, 1, 3, 4, 3, 4],
                   'b': [5, 7, 3, 3, 5, 6, 4, 3, 7, 4, 5]})

print (df)
    a  b
0   1  5
1   1  7
2   2  3
3   3  3
4   2  5
5   2  6
6   1  4
7   3  3
8   4  7
9   3  4
10  4  5


df = df.groupby('a')['b'].nlargest(3).sum(level=0).nlargest(3)
print (df)
a
1    16
2    14
4    12
Name: b, dtype: int64

Timings :

np.random.seed(123)
N = 1000000

L2 = np.arange(100)

df = pd.DataFrame({'b':np.random.randint(20, size=N), 
                   'a': np.random.choice(L2, N)})

print (df)

In [22]: %timeit df.groupby('a')['b'].apply(lambda x: x.nlargest(3).sum())
10 loops, best of 3: 125 ms per loop

In [23]: %timeit df.groupby('a')['b'].nlargest(3).groupby(level='a').sum()
10 loops, best of 3: 121 ms per loop

In [29]: %timeit df.groupby('a')['b'].nlargest(3).sum(level=0)
10 loops, best of 3: 121 ms per loop

---

np.random.seed(123)
N = 1000000

L2 = list('abcdefghijklmno')

df = pd.DataFrame({'b':np.random.randint(20, size=N), 
                   'a': np.random.choice(L2, N)})

print (df)

In [19]: %timeit df.groupby('a')['b'].apply(lambda x: x.nlargest(3).sum())
10 loops, best of 3: 97.9 ms per loop

In [20]: %timeit df.groupby('a')['b'].nlargest(3).groupby(level='a').sum()
10 loops, best of 3: 96.5 ms per loop

In [31]: %timeit df.groupby('a')['b'].nlargest(3).sum(level=0)
10 loops, best of 3: 97.9 ms per loop

---

np.random.seed(123)
N = 1000000

L2 = list('abcde')

df = pd.DataFrame({'b':np.random.randint(20, size=N), 
                   'a': np.random.choice(L2, N)})

print (df)


In [25]: %timeit df.groupby('a')['b'].apply(lambda x: x.nlargest(3).sum())
10 loops, best of 3: 82 ms per loop

In [26]: %timeit df.groupby('a')['b'].nlargest(3).groupby(level='a').sum()
10 loops, best of 3: 81.9 ms per loop

In [33]: %timeit df.groupby('a')['b'].nlargest(3).sum(level=0)
10 loops, best of 3: 82.5 ms per loop