Multiple if conditions in pandas dataframe

I'm following the answer from this question

I have a df like this:

score_1   score_2  
1.11        NaN      
2.22        3.33      
NaN         3.33      
NaN         NaN
........

The rule for calculating final_score is that we require at least one of the scores to be non-null, if one of the scores in NULL, then final_score will equal to another score (it has all the weights) This is the code to replicate:

import numpy as np
import pandas as pd

df = pd.DataFrame({
            'score_1': [1.11, 2.22, np.nan],
            'score_2': [np.nan, 3.33, 3.33]
        })

def final_score(df):
    if (df['score_1'] != np.nan) and (df['score_2'] != np.nan):
        print('I am condition one')
        return df['score_1'] * 0.2   df['score_2'] * 0.8

    elif (df['score_1'] == np.nan) and (df['score_2'] != np.nan):
        print('I am the condition two')
        return df['score_2']

    elif (df['score_1'] != np.nan) and (df['score_2'] == np.nan):
        print('I am the condition three')
        return df['score_1']

    elif (df['score_1'] == np.nan) and (df['score_2'] == np.nan):
        print('I am the condition four')
        return np.nan

df['final_score'] = df.apply(final_score, axis=1)
print(df)

This gave me output:

score_1   score_2  final_score
1.11        NaN       NaN
2.22        3.33      3.108
NaN         3.33      NaN
NaN         NaN       NaN
........

But my expected output is below:

score_1   score_2  final_score
1.11        NaN       1.11
2.22        3.33      3.108
NaN         3.33      3.33
NaN         NaN       NaN
........

The first and third row are not the result I'm expecting, can someone help me, what's wrong with my code? Thanks a lot.

CodePudding user response：

Lets appy your conditions using np.where

df['final_score'] =np.where(df.notna().all(1),df['score_1'] * 0.2   df['score_2'] * 0.8,df.mean(1))



   score_1  score_2  final_score
0     1.11      NaN        1.110
1     2.22     3.33        3.108
2      NaN     3.33        3.330
3      NaN      NaN          NaN

CodePudding user response：

using np.isnan() for comparison should solve the problem