Reputation: 2545
I have a DataFrame of the form:
What I would like to achieve is a DataFrame that has unique groups and a count of non-zeros for each column 1-9. It would look like the below.
I'm pretty much at a loss on this. I saw the question here, but this doesn't address the grouping aspect. How can I achieve this using pandas?
Upvotes: 3
Views: 1097
Reputation: 210832
Assuming you have the following DF:
In [82]: df
Out[82]:
Group 1 2 3 4 5
0 Group1 0 1 4 0 1
1 Group1 3 0 4 1 5
2 Group2 0 1 4 3 6
3 Group2 5 1 4 0 7
4 Group3 0 0 4 7 8
5 Group3 7 1 4 7 9
Solution:
In [83]: df.set_index('Group').ne(0).groupby(level=0).sum().reset_index()
Out[83]:
Group 1 2 3 4 5
0 Group1 1.0 1.0 2.0 1.0 2.0
1 Group2 1.0 2.0 2.0 1.0 2.0
2 Group3 1.0 1.0 2.0 2.0 2.0
as integers:
In [84]: df.set_index('Group').ne(0).groupby(level=0).sum().astype(int).reset_index()
Out[84]:
Group 1 2 3 4 5
0 Group1 1 1 2 1 2
1 Group2 1 2 2 1 2
2 Group3 1 1 2 2 2
UPDATE:
In [92]: df
Out[92]:
Group 1 2 3 4 5
0 Group1 0 1 4 0 0
1 Group1 3 0 4 1 0
2 Group2 0 1 4 3 0
3 Group2 5 1 4 0 0
4 Group3 0 0 4 7 0
5 Group3 7 1 4 7 0
In [93]: df.set_index('Group').ne(0).groupby(level=0).sum().astype(int).reset_index()
Out[93]:
Group 1 2 3 4 5
0 Group1 1 1 2 1 0
1 Group2 1 2 2 1 0
2 Group3 1 1 2 2 0
Timing:
In [37]: %timeit df.set_index('Group').ne(0).groupby(level=0).sum().astype(int).reset_index()
100 loops, best of 3: 6.24 ms per loop
In [38]: %timeit df.pivot_table(index='Group', aggfunc=np.count_nonzero)
100 loops, best of 3: 19.2 ms per loop
let's test it against a bigger (60.000 rows) DF:
In [39]: df = pd.concat([df] * 10**4, ignore_index=True)
In [40]: df.shape
Out[40]: (60000, 6)
In [42]: %timeit df.set_index('Group').ne(0).groupby(level=0).sum().astype(int).reset_index()
10 loops, best of 3: 22.4 ms per loop
In [43]: %timeit df.pivot_table(index='Group', aggfunc=np.count_nonzero)
10 loops, best of 3: 43 ms per loop
Conclusion:
IMO high-level function pivot_table
is slower due to additional overhead, but t he difference is not that huge for bigger data sets...
Upvotes: 3
Reputation: 153460
Using df from @MaxU solution:
df.pivot_table(index='Group', aggfunc=np.count_nonzero)
1 2 3 4 5
Group
Group1 1 1 2 1 2
Group2 1 2 2 1 2
Group3 1 1 2 2 2
Upvotes: 2