熊猫：分组多列，连接一列，同时添加另一列

Question

If I had the following df:

      amount   name   role    desc
0        1.0    a      x       f
1        2.0    a      y       g
2        3.0    b      y       h
3        4.0    b      y       j
4        5.0    c      x       k
5        6.0    c      x       l
6        6.0    c      y       p

I want to group by the name and role columns, add up the amount , and also do a concatenation of the desc with a , :

      amount   name   role    desc
0        1.0    a      x       f
1        2.0    a      y       g
2        7.0    b      y       h,j
4        11.0   c      x       k,l
6        6.0    c      y       p

What would be the correct way of approaching this?

Side question: say if the df was being read from a .csv and it had other unrelated columns, how do I do this calculation and then write to a new .csv along with the other columns (same schema as the one read)?

Answer 1

May be not exact dupe but there are a lot of questions related to groupby agg

df.groupby(['name', 'role'], as_index=False)\
.agg({'amount':'sum', 'desc':lambda x: ','.join(x)})


    name    role    amount  desc
0   a       x       1.0     f
1   a       y       2.0     g
2   b       y       7.0     h,j
3   c       x       11.0    k,l
4   c       y       6.0     p

Edit: If there are other columns in the dataframe, you can aggregate them using 'first' or 'last' or if their values are identical, include them in grouping.

Option1:

df.groupby(['name', 'role'], as_index=False).agg({'amount':'sum', 'desc':lambda x: ','.join(x), 'other1':'first', 'other2':'first'})

Option 2:

df.groupby(['name', 'role', 'other1', 'other2'], as_index=False).agg({'amount':'sum', 'desc':lambda x: ','.join(x)})

Answer 2

Extending @Vaishali's answer. To handle the remaining columns without having to specify each one you could create a dictionary and have that as the argument for the agg(regate) function.

dict = {}
for col in df:
    if (col == 'column_you_wish_to_merge'):
        dict[col] = ' '.join
    else:
        dict[col] = 'first' # or any other group aggregation operation

df.groupby(['key1', 'key2'], as_index=False).agg(dict)