在满足条件的SeriesGroupBy对象上使用Apply

Question

I have a DataFrame df1 :

 df1.head() = 

           id      ret     eff
    1469  2300 -0.010879  4480.0
    328   2300 -0.000692 -4074.0
    1376  2300 -0.009551  4350.0
    2110  2300 -0.014013  5335.0
    849   2300 -0.286490 -9460.0

I would like to create a new column that contains the normalized values of the column df1['eff'] .
In other words, I would like to group df1['eff'] by df1['id'] , look for the max value ( mx = df1['eff'].max() ) and the min value ( mn = df2['eff'].min() ), and divide in a pairwise fashion each value of the column df1['eff'] by mn or mx depending if df1['eff'] > 0 or df1['eff']< 0 .

The code that I have written is the following:

df1['normd'] = df1.groupby('id')['eff'].apply(lambda x: x/x.max() if x > 0 else x/x.min())

However python throws the following error:

*** ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(),
 a.item(), a.any() or a.all().

Since df1.groupby('id')['eff'] is a SeriesGroupBy Object , i decided to use map() . But again python throws the following error:

 *** AttributeError: Cannot access callable attribute 'map' of 'SeriesGroupBy' ob
 jects, try using the 'apply' method

Many thanks in advance.

Answer 1

You can use custom function f , where is possible easy add print . So x is Series and you need compare each group by numpy.where . Output is numpy array and you need convert it to Series :

def f(x):
    #print (x)
    #print (x/x.max())
    #print (x/x.min())
    return pd.Series(np.where(x>0, x/x.max(), x/x.min()), index=x.index)


df1['normd'] = df1.groupby('id')['eff'].apply(f)
print (df1)
        id       ret     eff     normd
1469  2300 -0.010879  4480.0  0.839738
328   2300 -0.000692 -4074.0  0.430655
1376  2300 -0.009551  4350.0  0.815370
2110  2300 -0.014013  5335.0  1.000000
849   2300 -0.286490 -9460.0  1.000000

What is same as:

df1['normd'] = df1.groupby('id')['eff']
                  .apply(lambda x: pd.Series(np.where(x>0, 
                                                      x/x.max(), 
                                                      x/x.min()), index=x.index))
print (df1)
        id       ret     eff     normd
1469  2300 -0.010879  4480.0  0.839738
328   2300 -0.000692 -4074.0  0.430655
1376  2300 -0.009551  4350.0  0.815370
2110  2300 -0.014013  5335.0  1.000000
849   2300 -0.286490 -9460.0  1.000000