How to apply "first" and "last" functions to columns while using group by in pandas?

Question

I have a data frame and I would like to group it by a particular column (or, in other words, by values from a particular column). I can do it in the following way: grouped = df.groupby(['ColumnName']).

I imagine the result of this operation as a table in which some cells can contain sets of values instead of single values. To get a usual table (i.e. a table in which every cell contains only one a single value) I need to indicate what function I want to use to transform the sets of values in the cells into single values.

For example I can replace sets of values by their sum, or by their minimal or maximal value. I can do it in the following way: grouped.sum() or grouped.min() and so on.

Now I want to use different functions for different columns. I figured out that I can do it in the following way: grouped.agg({'ColumnName1':sum, 'ColumnName2':min}).

However, because of some reasons I cannot use first. In more details, grouped.first() works, but grouped.agg({'ColumnName1':first, 'ColumnName2':first}) does not work. As a result I get a NameError: NameError: name 'first' is not defined. So, my question is: Why does it happen and how to resolve this problem.

ADDED

Here I found the following example:

grouped['D'].agg({'result1' : np.sum, 'result2' : np.mean}) 

May be I also need to use np? But in my case python does not recognize "np". Should I import it?

Answer 1

I think the issue is that there are two different first methods which share a name but act differently, one is for groupby objects and another for a Series/DataFrame (to do with timeseries).

To replicate the behaviour of the groupby first method over a DataFrame using agg you could use iloc[0] (which gets the first row in each group (DataFrame/Series) by index):

grouped.agg(lambda x: x.iloc[0]) 

For example:

In [1]: df = pd.DataFrame([[1, 2], [3, 4]])  In [2]: g = df.groupby(0)  In [3]: g.first() Out[3]:     1 0    1  2 3  4  In [4]: g.agg(lambda x: x.iloc[0]) Out[4]:     1 0    1  2 3  4 

Analogously you can replicate last using iloc[-1].

Note: This will works column-wise, et al:

g.agg({1: lambda x: x.iloc[0]}) 

In older version of pandas you could would use the irow method (e.g. x.irow(0), see previous edits.

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A couple of updated notes:

This is better done using the nth groupby method, which is much faster >=0.13:

g.nth(0)  # first g.nth(-1)  # last 

You have to take care a little, as the default behaviour for first and last ignores NaN rows... and IIRC for DataFrame groupbys it was broken pre-0.13... there's a dropna option for nth.

You can use the strings rather than built-ins (though IIRC pandas spots it's the sum builtin and applies np.sum):

grouped['D'].agg({'result1' : "sum", 'result2' : "mean"})