Distinct of column along with aggregations on other columns

To get the distinct number of values for any column (CLIENTCODE in your case), we can use nunique. We can pass the input as a dictionary in agg function, along with aggregations on other columns:

grp_df = df.groupby('YEARMONTH').agg({'CLIENTCODE': ['nunique'],
                                      'other_col_1': ['sum', 'count']})

# to flatten the multi-level columns
grp_df.columns = ["_".join(col).strip() for col in grp_df.columns.values]

# if you wish to reset the index
grp_df.reset_index(inplace=True)

Here an approach to have count distinct over multiple columns. Let's have some data:

data = {'CLIENT_CODE':[1,1,2,1,2,2,3],
        'YEAR_MONTH':[201301,201301,201301,201302,201302,201302,201302],
        'PRODUCT_CODE': [100,150,220,400,50,80,100]
       }
table = pd.DataFrame(data)
table

CLIENT_CODE YEAR_MONTH  PRODUCT_CODE
0   1       201301      100
1   1       201301      150
2   2       201301      220
3   1       201302      400
4   2       201302      50
5   2       201302      80
6   3       201302      100

Now, list the columns of interest and use groupby in a slightly modified syntax:

columns = ['YEAR_MONTH', 'PRODUCT_CODE']
table[columns].groupby(table['CLIENT_CODE']).nunique()

We obtain:

YEAR_MONTH  PRODUCT_CODE CLIENT_CODE        
1           2            3
2           2            3
3           1            1

With new pandas version, it is easy to get as dataframe

unique_count = pd.groupby(['YEARMONTH'], as_index=False).agg(uniq_CLIENTCODE =('CLIENTCODE',pd.Series.count))

I am also using nunique but it will be very helpful if you have to use an aggregate function like 'min', 'max', 'count' or 'mean' etc.

df.groupby('YEARMONTH')['CLIENTCODE'].transform('nunique') #count(distinct)
df.groupby('YEARMONTH')['CLIENTCODE'].transform('min')     #min
df.groupby('YEARMONTH')['CLIENTCODE'].transform('max')     #max
df.groupby('YEARMONTH')['CLIENTCODE'].transform('mean')    #average
df.groupby('YEARMONTH')['CLIENTCODE'].transform('count')   #count

Using crosstab, this will return more information than groupby nunique

pd.crosstab(df.YEARMONTH,df.CLIENTCODE)
Out[196]: 
CLIENTCODE  1  2  3
YEARMONTH          
201301      2  1  0
201302      1  2  1

After a little bit modify ,yield the result

pd.crosstab(df.YEARMONTH,df.CLIENTCODE).ne(0).sum(1)
Out[197]: 
YEARMONTH
201301    2
201302    3
dtype: int64

Here is another method, much simple, lets say your dataframe name is daat and column name is YEARMONTH

daat.YEARMONTH.value_counts()

Interestingly enough, very often len(unique()) is a few times (3x-15x) faster than nunique().

I believe this is what you want:

table.groupby('YEARMONTH').CLIENTCODE.nunique()

Example:

In [2]: table
Out[2]: 
   CLIENTCODE  YEARMONTH
0           1     201301
1           1     201301
2           2     201301
3           1     201302
4           2     201302
5           2     201302
6           3     201302

In [3]: table.groupby('YEARMONTH').CLIENTCODE.nunique()
Out[3]: 
YEARMONTH
201301       2
201302       3