How to implement SQL Groupby in RAPIDS












1















I'm seeking to translate an SQL query to use RAPIDS. Consider the simplified query below:



(SELECT min(a), max(b), c

FROM T

GROUP BY c) AS result


I have validated the code below, but is this the optimal solution? Is sorting on the group key necessary? Is there a cleaner / more idiomatic way to write it?



from pygdf import DataFrame as gdf



T = gdf(...)

df = gdf({'a':T.a, 'c':T.c}).groupby('c').min().sort_values(by='c')

df['max_b'] = gdf({'b':T.b, 'c':T.c}).groupby('c').max().sort_values(by='c').max_b

result = gdf({'a': df.min_a, 'b': df.max_b, 'c':df.c})





python rapids







share|improve this question





























    1















    I'm seeking to translate an SQL query to use RAPIDS. Consider the simplified query below:



    (SELECT min(a), max(b), c
    
FROM T
    
GROUP BY c) AS result


    I have validated the code below, but is this the optimal solution? Is sorting on the group key necessary? Is there a cleaner / more idiomatic way to write it?



    from pygdf import DataFrame as gdf
    

    
T = gdf(...)
    
df = gdf({'a':T.a, 'c':T.c}).groupby('c').min().sort_values(by='c')
    
df['max_b'] = gdf({'b':T.b, 'c':T.c}).groupby('c').max().sort_values(by='c').max_b
    
result = gdf({'a': df.min_a, 'b': df.max_b, 'c':df.c})





    python rapids







    share|improve this question



























      1












      1








      1








      I'm seeking to translate an SQL query to use RAPIDS. Consider the simplified query below:



      (SELECT min(a), max(b), c
      
FROM T
      
GROUP BY c) AS result


      I have validated the code below, but is this the optimal solution? Is sorting on the group key necessary? Is there a cleaner / more idiomatic way to write it?



      from pygdf import DataFrame as gdf
      

      
T = gdf(...)
      
df = gdf({'a':T.a, 'c':T.c}).groupby('c').min().sort_values(by='c')
      
df['max_b'] = gdf({'b':T.b, 'c':T.c}).groupby('c').max().sort_values(by='c').max_b
      
result = gdf({'a': df.min_a, 'b': df.max_b, 'c':df.c})





      python rapids







      share|improve this question
















      I'm seeking to translate an SQL query to use RAPIDS. Consider the simplified query below:



      (SELECT min(a), max(b), c
      
FROM T
      
GROUP BY c) AS result


      I have validated the code below, but is this the optimal solution? Is sorting on the group key necessary? Is there a cleaner / more idiomatic way to write it?



      from pygdf import DataFrame as gdf
      

      
T = gdf(...)
      
df = gdf({'a':T.a, 'c':T.c}).groupby('c').min().sort_values(by='c')
      
df['max_b'] = gdf({'b':T.b, 'c':T.c}).groupby('c').max().sort_values(by='c').max_b
      
result = gdf({'a': df.min_a, 'b': df.max_b, 'c':df.c})





      python rapids




      python rapids






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 20 '18 at 16:59







      Thomas Ryan Stovall

















      asked Nov 20 '18 at 16:49









      Thomas Ryan StovallThomas Ryan Stovall

      405




      405
























          1 Answer
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          oldest

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          2














          You can rewrite your aggregation using the .agg function to make it more straightforward:



          from pygdf import DataFrame as gdf
          

          
T = gdf(...)
          
df = gdf({'a':T.a, 'b': T.b, 'c':T.c}).groupby('c').agg({'a': 'min', 'b': 'max'})
          
result = gdf({'a': df.min_a, 'b': df.max_b, 'c':df.c})





          share|improve this answer























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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2














            You can rewrite your aggregation using the .agg function to make it more straightforward:



            from pygdf import DataFrame as gdf
            

            
T = gdf(...)
            
df = gdf({'a':T.a, 'b': T.b, 'c':T.c}).groupby('c').agg({'a': 'min', 'b': 'max'})
            
result = gdf({'a': df.min_a, 'b': df.max_b, 'c':df.c})





            share|improve this answer




























              2














              You can rewrite your aggregation using the .agg function to make it more straightforward:



              from pygdf import DataFrame as gdf
              

              
T = gdf(...)
              
df = gdf({'a':T.a, 'b': T.b, 'c':T.c}).groupby('c').agg({'a': 'min', 'b': 'max'})
              
result = gdf({'a': df.min_a, 'b': df.max_b, 'c':df.c})





              share|improve this answer


























                2












                2








                2







                You can rewrite your aggregation using the .agg function to make it more straightforward:



                from pygdf import DataFrame as gdf
                

                
T = gdf(...)
                
df = gdf({'a':T.a, 'b': T.b, 'c':T.c}).groupby('c').agg({'a': 'min', 'b': 'max'})
                
result = gdf({'a': df.min_a, 'b': df.max_b, 'c':df.c})





                share|improve this answer













                You can rewrite your aggregation using the .agg function to make it more straightforward:



                from pygdf import DataFrame as gdf
                

                
T = gdf(...)
                
df = gdf({'a':T.a, 'b': T.b, 'c':T.c}).groupby('c').agg({'a': 'min', 'b': 'max'})
                
result = gdf({'a': df.min_a, 'b': df.max_b, 'c':df.c})






                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Nov 23 '18 at 20:35









                Keith KrausKeith Kraus

                361




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