Access index in pandas.Series.apply

Make it a frame, return scalars if you want (so the result is a series)

Setup

In [11]: s = Series([1,2,3],dtype='float64',index=['a','b','c'])

In [12]: s
Out[12]: 
a    1
b    2
c    3
dtype: float64

Printing function

In [13]: def f(x):
    print type(x), x
    return x
   ....: 

In [14]: pd.DataFrame(s).apply(f)
<class 'pandas.core.series.Series'> a    1
b    2
c    3
Name: 0, dtype: float64
<class 'pandas.core.series.Series'> a    1
b    2
c    3
Name: 0, dtype: float64
Out[14]: 
   0
a  1
b  2
c  3

Since you can return anything here, just return the scalars (access the index via the name attribute)

In [15]: pd.DataFrame(s).apply(lambda x: 5 if x.name == 'a' else x[0] ,1)
Out[15]: 
a    5
b    2
c    3
dtype: float64

I don't believe apply has access to the index; it treats each row as a numpy object, not a Series, as you can see:

In [27]: s.apply(lambda x: type(x))
Out[27]: 
a  b
1  2    <type 'numpy.float64'>
3  6    <type 'numpy.float64'>
4  4    <type 'numpy.float64'>

To get around this limitation, promote the indexes to columns, apply your function, and recreate a Series with the original index.

Series(s.reset_index().apply(f, axis=1).values, index=s.index)

Other approaches might use s.get_level_values, which often gets a little ugly in my opinion, or s.iterrows(), which is likely to be slower -- perhaps depending on exactly what f does.