I think you can check groupby docs.

1. Indexing, iteration
2. Function application
3. Computations / Descriptive Stats

I think you got it in upper case. Just to get the mean, I tried this code and it seems to work. Correct me if I am misunderstanding something here.

A=[1,1,3,4,3]
B=[6,7,8,9,10]
C=[8,9,101,1,2]

df = pd.DataFrame(data=[A,B,C]).T
df.groupby(by=[0]).agg(['mean','min'])

First option: iterate over all groups.

for name, group in df.groupby(column):
    print(name)
    print(group)
    print('\n')

Second option: if you want to see the group for a specific value, use the get_group method.

df.groupby(column).get_group(name)

There is string 'pct', need variable pct - lambda function by removing '':

aggs = {'B':pct}
print(df.groupby('A').agg(aggs))

          B
A          
1  0.333333
4  0.333333
7  0.333333

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