I want to take the surface-weighted average of the columns in my dataframe. I have two surface-columns and two U-value-columns. I want to create an extra column 'U_av' (surface-weighted-average U-value) and U_av = (A1*U1 + A2*U2) / (A1+A2). If NaN occurs in one of the columns, NaN should be returned.
Initial df:
ID A1 A2 U1 U2
0 14 2 1.0 10.0 11
1 16 2 2.0 12.0 12
2 18 2 3.0 24.0 13
3 20 2 NaN 8.0 14
4 22 4 5.0 84.0 15
5 24 4 6.0 84.0 16
Desired Output:
ID A1 A2 U1 U2 U_av
0 14 2 1.0 10.0 11 10.33
1 16 2 2.0 12.0 12 12
2 18 2 3.0 24.0 13 17.4
3 20 2 NaN 8.0 14 NaN
4 22 4 5.0 84.0 15 45.66
5 24 4 6.0 84.0 16 43.2
Code:
import numpy as np
import pandas as pd
df = pd.DataFrame({"ID": [14,16,18,20,22,24],
"A1": [2,2,2,2,4,4],
"U1": [10,12,24,8,84,84],
"A2": [1,2,3,np.nan,5,6],
"U2": [11,12,13,14,15,16]})
print(df)
#the mean of two columns U1 and U2 and dropping NaN is easy (U1+U2/2 in this case)
#but what to do for the surface-weighted mean (U_av = (A1*U1 + A2*U2) / (A1+A2))?
df.loc[:,'Umean'] = df[['U1','U2']].dropna().mean(axis=1)
EDIT:
adding to the solutions below:
df["U_av"] = (df.A1.mul(df.U1) + df.A2.mul(df.U2)).div(df[['A1','A2']].sum(axis=1))
Hope I got you correct:
df['U_av'] = (df['A1']*df['U1'] + df['A2']*df['U2']) / (df['A1']+df['A2'])
df
ID A1 U1 A2 U2 U_av
0 14 2 10 1.0 11 10.333333
1 16 2 12 2.0 12 12.000000
2 18 2 24 3.0 13 17.400000
3 20 2 8 NaN 14 NaN
4 22 4 84 5.0 15 45.666667
5 24 4 84 6.0 16 43.200000