pandas groupby with a lambda parameter

Question

I can't understand the code:

pivot = pd.pivot_table(subset, values='count', rows=['date'], cols=['sample'], fill_value=0)

by = lambda x: lambda y: getattr(y, x)

grouped = pivot.groupby([by('year'),by('month')]).sum()

subset in the code is a DataFrame which have a column named "date"(e.g.2013-02-04 06:20:49.634244), and do not have a column named "year" and "month".

where I have trouble with

I can't figure out the "year" and "month" in:

grouped = pivot.groupby([by('year'),by('month')]).sum()

What the meaning of

grouped = pivot.groupby([by('year'),by('month')]).sum()

What I have done:

In the pandas pandas document says: the first parame of the pandas.DataFrame.groupby can be

by : mapping function / list of functions, dict, Series, or tuple /
by = lambda x: lambda y: getattr(y, x)

means by('bar') returns a function that returns the attribute 'bar' from an object

Can you post a sample data set for the `subset` DF? For example if you have `year` and `month` __values__ in the `subset['sample']` column, then you will have such columns after pivoting... — MaxU - stand with Ukraine, Aug 20 '16 at 16:26
[How to make good reproducible pandas examples](http://stackoverflow.com/questions/20109391/how-to-make-good-reproducible-pandas-examples) — MaxU - stand with Ukraine, Aug 20 '16 at 16:27

score 1 · Accepted Answer · answered Aug 20 '16 at 16:46

If a callable is passed to groupby, it is called on the DataFrame's index, so this code is is grouping by the year and month of a datetimelike index.

In [55]: df = pd.DataFrame({'a': 1.0}, 
                           index=pd.date_range('2014-01-01', periods=13, freq='M'))

In [56]: df.groupby([by('year'), by('month')]).sum()
Out[56]: 
           a
2014 1   1.0
     2   1.0
     3   1.0
     4   1.0
     5   1.0
     6   1.0
     7   1.0
     8   1.0
     9   1.0
     10  1.0
     11  1.0
     12  1.0
2015 1   1.0

More explicitly

In [57]: df.groupby([df.index.year, df.index.month]).sum()
Out[57]: 
           a
2014 1   1.0
     2   1.0
     3   1.0
     4   1.0
     5   1.0
     6   1.0
     7   1.0
     8   1.0
     9   1.0
     10  1.0
     11  1.0
     12  1.0
2015 1   1.0

pandas groupby with a lambda parameter

1 Answers1