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from what I've read from different answers on stackoverflow and other resources, when providing the .transform() with a UDF, each column is passed one by one for each Group

But when i tried it myself, i saw a Dataframe being passed into the UDF

df = pd.Dataframe({'State':['Texas', 'Texas', 'Florida', 'Florida'], 
                   'a':[4,5,1,3], 'b':[6,10,3,11]}
def inspect(x):
    print(type(x))

df.groupby('State').transform(inspect)

# Output 
# <class 'pandas.core.series.Series'>
# <class 'pandas.core.series.Series'>
# <class 'pandas.core.frame.DataFrame'>
# <class 'pandas.core.series.Series'>
# <class 'pandas.core.series.Series'>

the Dataframe passed to the inspect happens to be the Dataframe of the first group (State = Florida). But no one has mentioned and talked about a Dataframe being passed when working with UDFs while using .transform().

my question is :

  • Why is a Dataframe passed to the inspect function when everyone says a Series (each column) is passed to the UDF?
  • why was the Dataframe of the first groupby object passed to the inspect? why wasn't the second groupby passed to the inspect ?

1 Answers1

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According to the groupby.transform documentation (see the highlighted part):

The current implementation imposes three requirements on f:

  • f must return a value that either has the same shape as the input subframe or can be broadcast to the shape of the input subframe. For example, if f returns a scalar it will be broadcast to have the same shape as the input subframe.
  • if this is a DataFrame, f must support application column-by-column in the subframe. If f also supports application to the entire subframe, then a fast path is used starting from the second chunk.
  • f must not mutate groups. Mutation is not supported and may produce unexpected results. See Mutating with User Defined Function (UDF) methods for more details.

I thus believe that transform is performing this check. Indeed, if we identify the order of the groups transforms using a counter, we indeed have successive numbers, except after the first group:

from itertools import count

df = pd.DataFrame({'State':['Texas', 'Texas', 'Florida', 'Florida', 'Washington'], 
                   'a':[4,5,1,3,2], 'b':[6,10,3,11,12]})

c = count()
def inspect(x):
    x = next(c)
    return x

df.groupby('State').transform(inspect)

Output, notice that step 2 is missing, likely when the check for a DataFrame happens:

   a  b
0  3  4  # second group (3 and 4)
1  3  4
2  0  1  # first group (0 and 1)
3  0  1
4  5  6  # third group (5 and 6)
mozway
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  • i still can't understand why the DataFrame is being passed. to what i can understand from the doc, it's saying if `f` can be applied to the whole the DataFrame, then a faster path is used from the second group. so that means passing the the first group as a DataFrame and seeing if `f` can be applied to it, is how it's choosing to use the fast path? or have i misunderstood it completely? – AmirMohammad Shakeri May 19 '23 at 09:58