Pandas groupby 并获取 dataframe 中多列的唯一值

Question

I have a dataframe like as below

stu_id,Mat_grade,sci_grade,eng_grade
1,A,C,A
1,A,C,A
1,B,C,A
1,C,C,A
2,D,B,B
2,D,C,B
2,D,D,C
2,D,A,C

tf = pd.read_clipboard(sep=',')

My objective is to

a) Find out how many different unique grades that a student got under Mat_grade , sci_grade and eng_grade

So, I tried the below

tf['mat_cnt'] = tf.groupby(['stu_id'])['Mat_grade'].nunique()
tf['sci_cnt'] = tf.groupby(['stu_id'])['sci_grade'].nunique()
tf['eng_cnt'] = tf.groupby(['stu_id'])['eng_grade'].nunique() 

But this doesn't provide the expected output. Since, I have more than 100K unique ids, any efficient and elegant solution is really helpful

I expect my output to be like as below

Answer 1

You can specify columns names in list and for column cols call DataFrameGroupBy.nunique with rename :

cols = ['Mat_grade','sci_grade', 'eng_grade']
new = ['mat_cnt','sci_cnt','eng_cnt']
d = dict(zip(cols, new))
df = tf.groupby(['stu_id'], as_index=False)[cols].nunique().rename(columns=d)
print (df)
   stu_id  mat_cnt  sci_cnt  eng_cnt
0       1        3        1        1
1       2        1        4        2

Another idea is used named aggregation:

cols = ['Mat_grade','sci_grade', 'eng_grade']
new = ['mat_cnt','sci_cnt','eng_cnt']
d = {v: (k,'nunique') for k, v in zip(cols, new)}
print (d)
{'mat_cnt': ('Mat_grade', 'nunique'), 
 'sci_cnt': ('sci_grade', 'nunique'), 
 'eng_cnt': ('eng_grade', 'nunique')}

df = tf.groupby(['stu_id'], as_index=False).agg(**d)
print (df)
   stu_id  mat_cnt  sci_cnt  eng_cnt
0       1        3        1        1
1       2        1        4        2