python pandas, DF.groupby().agg(), column reference in agg()

Question

On a concrete problem, say I have a DataFrame DF

     word  tag count
0    a     S    30
1    the   S    20
2    a     T    60
3    an    T    5
4    the   T    10

I want to find, for every “word”, the “tag” that has the most “count”. So the return would be something like

     word  tag count
1    the   S    20
2    a     T    60
3    an    T    5

I don’t care about the count column or if the order/Index is original or messed up. Returning a dictionary {‘the’ : ‘S’, …} is just fine.

I hope I can do

DF.groupby(['word']).agg(lambda x: x['tag'][ x['count'].argmax() ] )

but it doesn’t work. I can’t access column information.

More abstractly, what does the function in agg(function) see as its argument?

btw, is .agg() the same as .aggregate() ?

Many thanks.

Asked By: jf328

||

Source

Answer 1

agg is the same as aggregate. It’s callable is passed the columns (Series objects) of the DataFrame, one at a time.

You could use idxmax to collect the index labels of the rows with the maximum
count:

idx = df.groupby('word')['count'].idxmax()
print(idx)

yields

word
a       2
an      3
the     1
Name: count

and then use loc to select those rows in the word and tag columns:

print(df.loc[idx, ['word', 'tag']])

yields

  word tag
2    a   T
3   an   T
1  the   S

Note that idxmax returns index labels. df.loc can be used to select rows
by label. But if the index is not unique — that is, if there are rows with duplicate index labels — then df.loc will select all rows with the labels listed in idx. So be careful that df.index.is_unique is True if you use idxmax with df.loc

Alternative, you could use apply. apply‘s callable is passed a sub-DataFrame which gives you access to all the columns:

import pandas as pd
df = pd.DataFrame({'word':'a the a an the'.split(),
                   'tag': list('SSTTT'),
                   'count': [30, 20, 60, 5, 10]})

print(df.groupby('word').apply(lambda subf: subf['tag'][subf['count'].idxmax()]))

yields

word
a       T
an      T
the     S

Using idxmax and loc is typically faster than apply, especially for large DataFrames. Using IPython’s %timeit:

N = 10000
df = pd.DataFrame({'word':'a the a an the'.split()*N,
                   'tag': list('SSTTT')*N,
                   'count': [30, 20, 60, 5, 10]*N})
def using_apply(df):
    return (df.groupby('word').apply(lambda subf: subf['tag'][subf['count'].idxmax()]))

def using_idxmax_loc(df):
    idx = df.groupby('word')['count'].idxmax()
    return df.loc[idx, ['word', 'tag']]

In [22]: %timeit using_apply(df)
100 loops, best of 3: 7.68 ms per loop

In [23]: %timeit using_idxmax_loc(df)
100 loops, best of 3: 5.43 ms per loop

If you want a dictionary mapping words to tags, then you could use set_index
and to_dict like this:

In [36]: df2 = df.loc[idx, ['word', 'tag']].set_index('word')

In [37]: df2
Out[37]: 
     tag
word    
a      T
an     T
the    S

In [38]: df2.to_dict()['tag']
Out[38]: {'a': 'T', 'an': 'T', 'the': 'S'}

Answered By: unutbu

Answer 2

Here’s a simple way to figure out what is being passed (the unutbu) solution then ‘applies’!

In [33]: def f(x):
....:     print type(x)
....:     print x
....:     

In [34]: df.groupby('word').apply(f)
<class 'pandas.core.frame.DataFrame'>
  word tag  count
0    a   S     30
2    a   T     60
<class 'pandas.core.frame.DataFrame'>
  word tag  count
0    a   S     30
2    a   T     60
<class 'pandas.core.frame.DataFrame'>
  word tag  count
3   an   T      5
<class 'pandas.core.frame.DataFrame'>
  word tag  count
1  the   S     20
4  the   T     10

your function just operates (in this case) on a sub-section of the frame with the grouped variable all having the same value (in this cas ‘word’), if you are passing a function, then you have to deal with the aggregation of potentially non-string columns; standard functions, like ‘sum’ do this for you

Automatically does NOT aggregate on the string columns

In [41]: df.groupby('word').sum()
Out[41]: 
      count
word       
a        90
an        5
the      30

You ARE aggregating on all columns

In [42]: df.groupby('word').apply(lambda x: x.sum())
Out[42]: 
        word tag count
word                  
a         aa  ST    90
an        an   T     5
the   thethe  ST    30

You can do pretty much anything within the function

In [43]: df.groupby('word').apply(lambda x: x['count'].sum())
Out[43]: 
word
a       90
an       5
the     30

Answered By: Jeff

python pandas, DF.groupby().agg(), column reference in agg()

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