多个 if/then 评估数据框中列的值并有条件地修改另一列值的最佳方法是什么？

Question

I have a dataframe (1580 rows x 48 columns) where each column contains answers to questions, but not every row contains an answer to every question (leaving it NaN ). Groups of questions are related, and I'd like to tabulate the answers to the group of questions into new columns ( c_answers and i_answers ). I have generated lists of the correct answers for each group of questions. Here is an example of the data:

ex_df = pd.DataFrame([["a", "b", "d"],[np.nan, "a", "b"], ["c", "e", np.nan]], columns=["q1", "q2", "q3"])
correct_answers = ["a", "b", "c"]
ex_df

which generates the following dataframe:

    q1   q2   q3
0   a    b    d
1  NaN   a    b
2   e    c   NaN

What I would like to do, ideally, is to create a function that would score each column, and for each correct answer on a row (appears in the correct_answers list) it would increment a c_answers column by 1, for each answer that is not in correct_answers , it would increment a i_answers column by 1 instead, but if the provided answer is NaN , it would do neither (not counted as correct or incorrect). This function could then be applied to each group of questions, calculating the number of correct and incorrect answers for each row, for that group.

What I have been able to make a bit of progress with instead is something like this:

ex_df['q1score'] = np.where(ex_df['q1'].isna(), np.nan, 
                          np.where(ex_df['q1'].isin(correct_answers), 1, 100))

which updates the dataframe like so:

    q1   q2   q3   q1score
0   a    b    d    1.0
1  NaN   a    b    NaN
2   e    c   NaN   100.0

I could then re-use this code to score out q2 and q3 into their own new columns, which I could then sum up into a new column, and then from that column, I could generate two more columns which could calculate the number of correct and incorrect scores from that sum. Finally, I could go back and drop the other 4 columns that I created and keep only the two that I wanted in the first place.

Looking around and trying different methods for the last two hours, I'm finding a lot of answers that deal with one or another of the different issues I'm trying to deal with, but nothing that I could finagle to actually work for my situation. Maybe the solution I've kludged together is the best one, but I'm still relatively new to programming (<18 months) and it didn't seem like the most efficient or most Pythonic method to solve this problem. Hoping someone else has a better answer out there. Thank you!

Edit for more information regarding output: Regarding what I'd like the final output to look like, I'd like something that looks like this:

    q1   q2   q3   c_answers  i_answers
0   a    b    d    2          1
1  NaN   a    b    2          0
2   e    c   NaN   1          1

Like I said, I can kind of finagle that using the nested np.where() to create numeric columns that I can then sum up and reverse engineer to get a raw count from. While this is a solution, its cumbersome and seems like its probably not the optimal one, especially with the amount of repetition involved (I'll have to do this process for 9 different groups of columns, each being a cluster of questions).

Answer 1

Use sum for count True s values for correct and incorrect values per rows:

m1 = ex_df.isin(correct_answers)
m2 = ex_df.notna() & ~m1

df = ex_df.assign(c_answers=m1.sum(axis=1), i_answers=m2.sum(axis=1))
print (df)
    q1 q2   q3  c_answers  i_answers
0    a  b    d          2          1
1  NaN  a    b          2          0
2    c  e  NaN          1          1

Possible solution for multiple groups:

groups = {'g1':['q1','q2'], 'g2':['q2','q3'], 'g3':['q1','q2','q3']}

for k, v in groups.items():
    m1 = ex_df[v].isin(correct_answers)
    m2 = ex_df[v].notna() & ~m1
    
    ex_df = ex_df.assign(**{f'c_answers_{k}':m1.sum(axis=1), 
                            f'i_answers_{k}':m2.sum(axis=1)})
print (ex_df)
    q1 q2   q3  c_answers_g1  i_answers_g1  c_answers_g2  i_answers_g2  \
0    a  b    d             2             0             1             1   
1  NaN  a    b             1             0             2             0   
2    c  e  NaN             1             1             0             1   

   c_answers_g3  i_answers_g3  
0             2             1  
1             2             0  
2             1             1