如何根据 Python 中的非空列的字典创建 dataframe 列

Question

I have a data frame and a dictionary like this:

df:
ID   Science  Social 
1      12       24   
2      NaN      13   
3      26       NaN  
4      23       35   

count_dict = {Science:30, Social: 40}

For every course column in the data frame, I want to create 2 new columns such that:

Col-1(Course_Count): If the course column is not null, then the new column gets the value from the dictionary, else it will remain Null.

Col-2(Course_%): Course/Course_Count

The output looks like this:

df:
ID   Science Science_Count Science_% Social Social_Count Social_%
1      12         30          12/30    24        40        24/40    
2      NaN                             13        40        13/40
3      26         30          26/30    NaN               
4      23         30          23/30    35        40        35/40

Can anyone help me with this?

Answer 1

If not any column in your dataframe is a course column, you can specify only the course column names in the courses list. Now I am just skipping the first column there ('ID'):

courses = df.columns[1:]


order = ['ID'] + [col for course in courses for col in (course, course+'_Count', course+'_%')]

for course in courses:
    df[course + '_Count'] = count_dict[course]
    df.loc[df[course].isna(), course + '_Count'] = np.nan
    df[course + '_%'] = df[course] / df[course + '_Count']

df = df[order]  # reorder the columns

Result:

   ID  Science  Science_Count  Science_%  Social  Social_Count  Social_%
0   1     12.0           30.0   0.400000    24.0          40.0     0.600
1   2      NaN            NaN        NaN    13.0          40.0     0.325
2   3     26.0           30.0   0.866667     NaN           NaN       NaN
3   4     23.0           30.0   0.766667    35.0          40.0     0.875

Answer 2

try this:

column_name=list(df.columns)
for column in column_name:
  df[f"{column}_Count"]=df.apply(lambda x:count_dict[column] if x==None else None,axis=1)