Python Pandas : pivot table with aggfunc = count unique distinct

Question

df2 = pd.DataFrame({\'X\' : [\'X1\', \'X1\', \'X1\', \'X1\'], \'Y\' : [\'Y2\',\'Y1\',\'Y1\',\'Y1\'], \'Z\' : [\'Z3\',\'Z1\',\'Z1\',\'Z2\']})

    X   Y   Z
0  X1  Y2         


        

Answer 1

You can construct a pivot table for each distinct value of X. In this case,

for xval, xgroup in g:
    ptable = pd.pivot_table(xgroup, rows='Y', cols='Z', 
        margins=False, aggfunc=numpy.size)

will construct a pivot table for each value of X. You may want to index ptable using the xvalue. With this code, I get (for X1)

     X        
Z   Z1  Z2  Z3
Y             
Y1   2   1 NaN
Y2 NaN NaN   1

Answer 2

Since at least version 0.16 of pandas, it does not take the parameter "rows"

As of 0.23, the solution would be:

df2.pivot_table(values='X', index='Y', columns='Z', aggfunc=pd.Series.nunique)

which returns:

Z    Z1   Z2   Z3
Y                
Y1  1.0  1.0  NaN
Y2  NaN  NaN  1.0