Multi-column calculation in pandas












1















I've got this long algebra formula that I need to apply to a dataframe:



def experience_mod(A, B, C, D, T, W):
E = (T-A)
F = (C-D)

xmod = (A + B + (E*W) + ((1-W)*F))/(D + B + (F*W) + ((1-W)*F))

return xmod

A = loss['actual_primary_losses']
B = loss['ballast']
C = loss['ExpectedLosses']
D = loss['ExpectedPrimaryLosses']
T = loss['ActualIncurred']
W = loss['weight']


How would I write this to calculate the experience_mod() for every row?



something like this?



loss['ExperienceRating'] = loss.apply(experience_mod(A,B,C,D,T,W) axis = 0)









share|improve this question


















  • 1





    Pandas support vectorized operations, so if you have a dataframe A and a dataframe B, A - B, A + B, etc.. are valid operations.

    – MedAli
    Nov 21 '18 at 17:01











  • @MedAli ahh, ty

    – James
    Nov 21 '18 at 17:07
















1















I've got this long algebra formula that I need to apply to a dataframe:



def experience_mod(A, B, C, D, T, W):
E = (T-A)
F = (C-D)

xmod = (A + B + (E*W) + ((1-W)*F))/(D + B + (F*W) + ((1-W)*F))

return xmod

A = loss['actual_primary_losses']
B = loss['ballast']
C = loss['ExpectedLosses']
D = loss['ExpectedPrimaryLosses']
T = loss['ActualIncurred']
W = loss['weight']


How would I write this to calculate the experience_mod() for every row?



something like this?



loss['ExperienceRating'] = loss.apply(experience_mod(A,B,C,D,T,W) axis = 0)









share|improve this question


















  • 1





    Pandas support vectorized operations, so if you have a dataframe A and a dataframe B, A - B, A + B, etc.. are valid operations.

    – MedAli
    Nov 21 '18 at 17:01











  • @MedAli ahh, ty

    – James
    Nov 21 '18 at 17:07














1












1








1


0






I've got this long algebra formula that I need to apply to a dataframe:



def experience_mod(A, B, C, D, T, W):
E = (T-A)
F = (C-D)

xmod = (A + B + (E*W) + ((1-W)*F))/(D + B + (F*W) + ((1-W)*F))

return xmod

A = loss['actual_primary_losses']
B = loss['ballast']
C = loss['ExpectedLosses']
D = loss['ExpectedPrimaryLosses']
T = loss['ActualIncurred']
W = loss['weight']


How would I write this to calculate the experience_mod() for every row?



something like this?



loss['ExperienceRating'] = loss.apply(experience_mod(A,B,C,D,T,W) axis = 0)









share|improve this question














I've got this long algebra formula that I need to apply to a dataframe:



def experience_mod(A, B, C, D, T, W):
E = (T-A)
F = (C-D)

xmod = (A + B + (E*W) + ((1-W)*F))/(D + B + (F*W) + ((1-W)*F))

return xmod

A = loss['actual_primary_losses']
B = loss['ballast']
C = loss['ExpectedLosses']
D = loss['ExpectedPrimaryLosses']
T = loss['ActualIncurred']
W = loss['weight']


How would I write this to calculate the experience_mod() for every row?



something like this?



loss['ExperienceRating'] = loss.apply(experience_mod(A,B,C,D,T,W) axis = 0)






pandas






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 21 '18 at 16:58









JamesJames

30127




30127








  • 1





    Pandas support vectorized operations, so if you have a dataframe A and a dataframe B, A - B, A + B, etc.. are valid operations.

    – MedAli
    Nov 21 '18 at 17:01











  • @MedAli ahh, ty

    – James
    Nov 21 '18 at 17:07














  • 1





    Pandas support vectorized operations, so if you have a dataframe A and a dataframe B, A - B, A + B, etc.. are valid operations.

    – MedAli
    Nov 21 '18 at 17:01











  • @MedAli ahh, ty

    – James
    Nov 21 '18 at 17:07








1




1





Pandas support vectorized operations, so if you have a dataframe A and a dataframe B, A - B, A + B, etc.. are valid operations.

– MedAli
Nov 21 '18 at 17:01





Pandas support vectorized operations, so if you have a dataframe A and a dataframe B, A - B, A + B, etc.. are valid operations.

– MedAli
Nov 21 '18 at 17:01













@MedAli ahh, ty

– James
Nov 21 '18 at 17:07





@MedAli ahh, ty

– James
Nov 21 '18 at 17:07












1 Answer
1






active

oldest

votes


















2














Pandas and the underlying library, numpy, it's using, support vectorized operations, so given two dataframes A and B, operations like A + B, A - B etc are valid.



Your code works fine, you need to apply the function to the columns directly and assign the results back to the new column ExperienceRating,





Here's a working example:



In [1]: import pandas as pd 

In [2]: import numpy as np

In [3]: df = pd.DataFrame(np.random.randn(6,6), columns=list('ABCDTW'))

In [4]: df
Out[4]:
A B C D T W
0 0.049617 0.082861 2.289549 -0.783082 -0.691990 -0.071152
1 0.722605 0.209683 -0.347372 0.254951 0.468615 -0.132794
2 -0.301469 -1.849026 -0.334381 -0.365116 -0.238384 -1.999025
3 -0.554925 -0.859044 -0.637079 -1.040336 0.627027 -0.955889
4 -2.024621 -0.539384 0.006734 0.117628 -0.215070 -0.661466
5 1.942926 -0.433067 -1.034814 -0.292179 0.744039 0.233953

In [5]: def experience_mod(A, B, C, D, T, W):
...: E = (T-A)
...: F = (C-D)
...:
...: xmod = (A + B + (E*W) + ((1-W)*F))/(D + B + (F*W) + ((1-W)*F))
...:
...: return xmod
...:

In [6]: experience_mod(df["A"], df["B"], df["C"], df["D"], df["T"], df["W"])
Out[6]:
0 1.465387
1 -2.060483
2 1.000469
3 1.173070
4 7.406756
5 -0.449957
dtype: float64

In [7]: df['ExperienceRating'] = experience_mod(df["A"], df["B"], df["C"], df["D"], df["T"], df["W"])

In [8]: df
Out[8]:
A B C D T W ExperienceRating
0 0.049617 0.082861 2.289549 -0.783082 -0.691990 -0.071152 1.465387
1 0.722605 0.209683 -0.347372 0.254951 0.468615 -0.132794 -2.060483
2 -0.301469 -1.849026 -0.334381 -0.365116 -0.238384 -1.999025 1.000469
3 -0.554925 -0.859044 -0.637079 -1.040336 0.627027 -0.955889 1.173070
4 -2.024621 -0.539384 0.006734 0.117628 -0.215070 -0.661466 7.406756
5 1.942926 -0.433067 -1.034814 -0.292179 0.744039 0.233953 -0.449957





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    1 Answer
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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    Pandas and the underlying library, numpy, it's using, support vectorized operations, so given two dataframes A and B, operations like A + B, A - B etc are valid.



    Your code works fine, you need to apply the function to the columns directly and assign the results back to the new column ExperienceRating,





    Here's a working example:



    In [1]: import pandas as pd 

    In [2]: import numpy as np

    In [3]: df = pd.DataFrame(np.random.randn(6,6), columns=list('ABCDTW'))

    In [4]: df
    Out[4]:
    A B C D T W
    0 0.049617 0.082861 2.289549 -0.783082 -0.691990 -0.071152
    1 0.722605 0.209683 -0.347372 0.254951 0.468615 -0.132794
    2 -0.301469 -1.849026 -0.334381 -0.365116 -0.238384 -1.999025
    3 -0.554925 -0.859044 -0.637079 -1.040336 0.627027 -0.955889
    4 -2.024621 -0.539384 0.006734 0.117628 -0.215070 -0.661466
    5 1.942926 -0.433067 -1.034814 -0.292179 0.744039 0.233953

    In [5]: def experience_mod(A, B, C, D, T, W):
    ...: E = (T-A)
    ...: F = (C-D)
    ...:
    ...: xmod = (A + B + (E*W) + ((1-W)*F))/(D + B + (F*W) + ((1-W)*F))
    ...:
    ...: return xmod
    ...:

    In [6]: experience_mod(df["A"], df["B"], df["C"], df["D"], df["T"], df["W"])
    Out[6]:
    0 1.465387
    1 -2.060483
    2 1.000469
    3 1.173070
    4 7.406756
    5 -0.449957
    dtype: float64

    In [7]: df['ExperienceRating'] = experience_mod(df["A"], df["B"], df["C"], df["D"], df["T"], df["W"])

    In [8]: df
    Out[8]:
    A B C D T W ExperienceRating
    0 0.049617 0.082861 2.289549 -0.783082 -0.691990 -0.071152 1.465387
    1 0.722605 0.209683 -0.347372 0.254951 0.468615 -0.132794 -2.060483
    2 -0.301469 -1.849026 -0.334381 -0.365116 -0.238384 -1.999025 1.000469
    3 -0.554925 -0.859044 -0.637079 -1.040336 0.627027 -0.955889 1.173070
    4 -2.024621 -0.539384 0.006734 0.117628 -0.215070 -0.661466 7.406756
    5 1.942926 -0.433067 -1.034814 -0.292179 0.744039 0.233953 -0.449957





    share|improve this answer




























      2














      Pandas and the underlying library, numpy, it's using, support vectorized operations, so given two dataframes A and B, operations like A + B, A - B etc are valid.



      Your code works fine, you need to apply the function to the columns directly and assign the results back to the new column ExperienceRating,





      Here's a working example:



      In [1]: import pandas as pd 

      In [2]: import numpy as np

      In [3]: df = pd.DataFrame(np.random.randn(6,6), columns=list('ABCDTW'))

      In [4]: df
      Out[4]:
      A B C D T W
      0 0.049617 0.082861 2.289549 -0.783082 -0.691990 -0.071152
      1 0.722605 0.209683 -0.347372 0.254951 0.468615 -0.132794
      2 -0.301469 -1.849026 -0.334381 -0.365116 -0.238384 -1.999025
      3 -0.554925 -0.859044 -0.637079 -1.040336 0.627027 -0.955889
      4 -2.024621 -0.539384 0.006734 0.117628 -0.215070 -0.661466
      5 1.942926 -0.433067 -1.034814 -0.292179 0.744039 0.233953

      In [5]: def experience_mod(A, B, C, D, T, W):
      ...: E = (T-A)
      ...: F = (C-D)
      ...:
      ...: xmod = (A + B + (E*W) + ((1-W)*F))/(D + B + (F*W) + ((1-W)*F))
      ...:
      ...: return xmod
      ...:

      In [6]: experience_mod(df["A"], df["B"], df["C"], df["D"], df["T"], df["W"])
      Out[6]:
      0 1.465387
      1 -2.060483
      2 1.000469
      3 1.173070
      4 7.406756
      5 -0.449957
      dtype: float64

      In [7]: df['ExperienceRating'] = experience_mod(df["A"], df["B"], df["C"], df["D"], df["T"], df["W"])

      In [8]: df
      Out[8]:
      A B C D T W ExperienceRating
      0 0.049617 0.082861 2.289549 -0.783082 -0.691990 -0.071152 1.465387
      1 0.722605 0.209683 -0.347372 0.254951 0.468615 -0.132794 -2.060483
      2 -0.301469 -1.849026 -0.334381 -0.365116 -0.238384 -1.999025 1.000469
      3 -0.554925 -0.859044 -0.637079 -1.040336 0.627027 -0.955889 1.173070
      4 -2.024621 -0.539384 0.006734 0.117628 -0.215070 -0.661466 7.406756
      5 1.942926 -0.433067 -1.034814 -0.292179 0.744039 0.233953 -0.449957





      share|improve this answer


























        2












        2








        2







        Pandas and the underlying library, numpy, it's using, support vectorized operations, so given two dataframes A and B, operations like A + B, A - B etc are valid.



        Your code works fine, you need to apply the function to the columns directly and assign the results back to the new column ExperienceRating,





        Here's a working example:



        In [1]: import pandas as pd 

        In [2]: import numpy as np

        In [3]: df = pd.DataFrame(np.random.randn(6,6), columns=list('ABCDTW'))

        In [4]: df
        Out[4]:
        A B C D T W
        0 0.049617 0.082861 2.289549 -0.783082 -0.691990 -0.071152
        1 0.722605 0.209683 -0.347372 0.254951 0.468615 -0.132794
        2 -0.301469 -1.849026 -0.334381 -0.365116 -0.238384 -1.999025
        3 -0.554925 -0.859044 -0.637079 -1.040336 0.627027 -0.955889
        4 -2.024621 -0.539384 0.006734 0.117628 -0.215070 -0.661466
        5 1.942926 -0.433067 -1.034814 -0.292179 0.744039 0.233953

        In [5]: def experience_mod(A, B, C, D, T, W):
        ...: E = (T-A)
        ...: F = (C-D)
        ...:
        ...: xmod = (A + B + (E*W) + ((1-W)*F))/(D + B + (F*W) + ((1-W)*F))
        ...:
        ...: return xmod
        ...:

        In [6]: experience_mod(df["A"], df["B"], df["C"], df["D"], df["T"], df["W"])
        Out[6]:
        0 1.465387
        1 -2.060483
        2 1.000469
        3 1.173070
        4 7.406756
        5 -0.449957
        dtype: float64

        In [7]: df['ExperienceRating'] = experience_mod(df["A"], df["B"], df["C"], df["D"], df["T"], df["W"])

        In [8]: df
        Out[8]:
        A B C D T W ExperienceRating
        0 0.049617 0.082861 2.289549 -0.783082 -0.691990 -0.071152 1.465387
        1 0.722605 0.209683 -0.347372 0.254951 0.468615 -0.132794 -2.060483
        2 -0.301469 -1.849026 -0.334381 -0.365116 -0.238384 -1.999025 1.000469
        3 -0.554925 -0.859044 -0.637079 -1.040336 0.627027 -0.955889 1.173070
        4 -2.024621 -0.539384 0.006734 0.117628 -0.215070 -0.661466 7.406756
        5 1.942926 -0.433067 -1.034814 -0.292179 0.744039 0.233953 -0.449957





        share|improve this answer













        Pandas and the underlying library, numpy, it's using, support vectorized operations, so given two dataframes A and B, operations like A + B, A - B etc are valid.



        Your code works fine, you need to apply the function to the columns directly and assign the results back to the new column ExperienceRating,





        Here's a working example:



        In [1]: import pandas as pd 

        In [2]: import numpy as np

        In [3]: df = pd.DataFrame(np.random.randn(6,6), columns=list('ABCDTW'))

        In [4]: df
        Out[4]:
        A B C D T W
        0 0.049617 0.082861 2.289549 -0.783082 -0.691990 -0.071152
        1 0.722605 0.209683 -0.347372 0.254951 0.468615 -0.132794
        2 -0.301469 -1.849026 -0.334381 -0.365116 -0.238384 -1.999025
        3 -0.554925 -0.859044 -0.637079 -1.040336 0.627027 -0.955889
        4 -2.024621 -0.539384 0.006734 0.117628 -0.215070 -0.661466
        5 1.942926 -0.433067 -1.034814 -0.292179 0.744039 0.233953

        In [5]: def experience_mod(A, B, C, D, T, W):
        ...: E = (T-A)
        ...: F = (C-D)
        ...:
        ...: xmod = (A + B + (E*W) + ((1-W)*F))/(D + B + (F*W) + ((1-W)*F))
        ...:
        ...: return xmod
        ...:

        In [6]: experience_mod(df["A"], df["B"], df["C"], df["D"], df["T"], df["W"])
        Out[6]:
        0 1.465387
        1 -2.060483
        2 1.000469
        3 1.173070
        4 7.406756
        5 -0.449957
        dtype: float64

        In [7]: df['ExperienceRating'] = experience_mod(df["A"], df["B"], df["C"], df["D"], df["T"], df["W"])

        In [8]: df
        Out[8]:
        A B C D T W ExperienceRating
        0 0.049617 0.082861 2.289549 -0.783082 -0.691990 -0.071152 1.465387
        1 0.722605 0.209683 -0.347372 0.254951 0.468615 -0.132794 -2.060483
        2 -0.301469 -1.849026 -0.334381 -0.365116 -0.238384 -1.999025 1.000469
        3 -0.554925 -0.859044 -0.637079 -1.040336 0.627027 -0.955889 1.173070
        4 -2.024621 -0.539384 0.006734 0.117628 -0.215070 -0.661466 7.406756
        5 1.942926 -0.433067 -1.034814 -0.292179 0.744039 0.233953 -0.449957






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 21 '18 at 17:14









        MedAliMedAli

        7,17874182




        7,17874182
































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