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)
pandas
asked Nov 21 '18 at 16:58
JamesJames
30127
add a comment |
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)
pandas
asked Nov 21 '18 at 16:58
JamesJames
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
add a comment |
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)
pandas
asked Nov 21 '18 at 16:58
JamesJames
30127
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
pandas
asked Nov 21 '18 at 16:58
JamesJames
30127
asked Nov 21 '18 at 16:58
JamesJames
30127
asked Nov 21 '18 at 16:58
JamesJames
30127
asked Nov 21 '18 at 16:58
JamesJames
30127
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
add a comment |
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
add a comment |
1 Answer
1
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oldest
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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
answered Nov 21 '18 at 17:14
MedAliMedAli
7,17874182
add a comment |
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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
answered Nov 21 '18 at 17:14
MedAliMedAli
7,17874182
add a comment |
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
answered Nov 21 '18 at 17:14
MedAliMedAli
7,17874182
add a comment |
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
answered Nov 21 '18 at 17:14
MedAliMedAli
7,17874182
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
answered Nov 21 '18 at 17:14
MedAliMedAli
7,17874182
answered Nov 21 '18 at 17:14
MedAliMedAli
7,17874182
answered Nov 21 '18 at 17:14
MedAliMedAli
7,17874182
answered Nov 21 '18 at 17:14
MedAliMedAli
7,17874182
7,17874182
add a comment |
add a comment |
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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