如何计算n个连续列的平均值？

Question

I have a dataframe like this:

import pandas as pd

df = pd.DataFrame({'A_1': [1, 2], 'A_2': [3, 4], 'A_3': [5, 6], 'A_4': [7, 8],
                   'B_1': [0, 2], 'B_2': [4, 4], 'B_3': [9, 6], 'B_4': [5, 8]})

   A_1  A_2  A_3  A_4  B_1  B_2  B_3  B_4
0    1    3    5    7    0    4    9    5
1    2    4    6    8    2    4    6    8

which I would like to convert into a dataframe that looks like this:

   A_G1  A_G2  B_G1  B_G2
0     2     6     2     7
1     3     7     3     7

Thereby, A_G1 is the mean of the columns A_1 and A_2 , A_G2 is the mean of the columns A_3 and A_4 ; the same applies to B_G1 and B_G2 . So what I would like to do is to calculate the mean of two consecutive columns and add the result as a new column into a dataframe.

A straightforward implementation could look like this:

res_df = pd.DataFrame()
for i in range(0, len(df.columns), 2):
    temp_df = df[[i, i + 1]].mean(axis=1)
    res_df = pd.concat([res_df, temp_df], axis=1)

which gives me the desired output (except for the column names):

   0  0  0  0
0  2  6  2  7
1  3  7  3  7

Is there any better way of doing this ie a vectorized way?

Answer 1

This might work for you:

In [15]: df.rolling(window=2,axis=1).mean().iloc[:,1::2]
Out[15]:
   A_2  A_4  B_2  B_4
0  2.0  6.0  2.0  7.0
1  3.0  7.0  3.0  7.0

But I haven't tested it against your "straightforward" implementation.

Answer 2

Here's a NumPy based vectorized solution using reshaping -

pd.DataFrame(df.values.reshape(-1,df.shape[1]//n,n).mean(2))

Sample runs -

In [65]: df
Out[65]: 
   A_1  A_2  A_3  A_4  B_1  B_2  B_3  B_4
0    1    3    5    7    0    4    9    5
1    2    4    6    8    2    4    6    8

In [66]: n = 2

In [67]: pd.DataFrame(df.values.reshape(-1,df.shape[1]//n,n).mean(2))
Out[67]: 
     0    1    2    3
0  2.0  6.0  2.0  7.0
1  3.0  7.0  3.0  7.0

In [68]: n = 4

In [69]: pd.DataFrame(df.values.reshape(-1,df.shape[1]//n,n).mean(2))
Out[69]: 
     0    1
0  4.0  4.5
1  5.0  5.0

Runtime test -

In [71]: df = pd.DataFrame(np.random.randint(0,9,(200,800)))

In [72]: %timeit df.rolling(window=2,axis=1).mean().iloc[:,1::2]
100 loops, best of 3: 11 ms per loop # @juanpa.arrivillaga's soln

In [73]: n = 2

In [74]: %timeit pd.DataFrame(df.values.reshape(-1,df.shape[1]//n,n).mean(2))
100 loops, best of 3: 2.6 ms per loop