pandas groupby without turning grouped by column into index
Another way to do this would be:
df.groupby(['col2', 'col3']).sum().reset_index()
The following, somewhat detailed answer, is added to help those who are still confused on which variant of the answers to use.
First, the suggested two solutions to this problem are:
- Solution 1:
df.groupby(['A', 'B'], as_index=False).sum()
- Solution 2:
df.groupby(['A', 'B']).sum().reset_index()
Both give the expected result.
Solution 1:
As explained in the documentation, as_index
will ask for SQL style grouped output, which will effectively ask pandas to preserve these grouped by columns in the output as it is prepared.
as_index: bool, default True
For aggregated output, return object with group labels as the index. Only relevant for DataFrame input. as_index=False is effectively “SQL-style” grouped output.
Example:
Given the following Dataframe:
A B C D
0 A 1 0.502130 0.959404
1 A 3 0.335416 0.087215
2 B 2 0.067308 0.084595
3 B 4 0.454158 0.723124
4 B 4 0.323326 0.895858
5 C 2 0.672375 0.356736
6 C 5 0.929655 0.371913
7 D 5 0.212634 0.540736
8 D 5 0.471418 0.268270
9 E 1 0.061270 0.739610
Applying the first solution gives:
>>> df.groupby(["A", "B"], as_index=False).sum()
A B C D
0 A 1 0.502130 0.959404
1 A 3 0.335416 0.087215
2 B 2 0.067308 0.084595
3 B 4 0.777483 1.618982
4 C 2 0.672375 0.356736
5 C 5 0.929655 0.371913
6 D 5 0.684052 0.809006
7 E 1 0.061270 0.739610
Where the groupby columns are preserved correctly.
Solution 2:
To understand the second solution, let's look at the output of the previous command with as_index = True
which is the default behavior of pandas.DataFrame.groupby
(check documentation):
>>> df.groupby(["A", "B"], as_index=True).sum()
C D
A B
A 1 0.502130 0.959404
3 0.335416 0.087215
B 2 0.067308 0.084595
4 0.777483 1.618982
C 2 0.672375 0.356736
5 0.929655 0.371913
D 5 0.684052 0.809006
E 1 0.061270 0.739610
As you can see, the groupby keys become the index of the dataframe. Using, pandas.DataFrame.reset_index
(check documentation) we can put back the indices of the dataframe as columns and use a default index. Which also leads us to the same results as in the previous step:
>>> df.groupby(['A', 'B']).sum().reset_index()
A B C D
0 A 1 0.502130 0.959404
1 A 3 0.335416 0.087215
2 B 2 0.067308 0.084595
3 B 4 0.777483 1.618982
4 C 2 0.672375 0.356736
5 C 5 0.929655 0.371913
6 D 5 0.684052 0.809006
7 E 1 0.061270 0.739610
Benchmark
Notice that since the first solution achieves the requirement in 1 step versus 2 steps in the second solution, the former is slightly faster:
%timeit df.groupby(["A", "B"], as_index=False).sum()
3.38 ms ± 21.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit df.groupby(["A", "B"]).sum().reset_index()
3.9 ms ± 365 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
df.groupby(['col2','col3'], as_index=False).sum()