How to speed up cumulative concatenation with Pandas?

Question

I have this function which is very slow, essentially I need to create cumulative sequence of string.

I tried with NumPy but there was no time improvement.

def create_sequence_shifted(df) -> list:
    articles = []
    customers = df["CUSTOMER"].drop_duplicates()
    for customer in customers:
        tmp = df[df["CUSTOMER"]==customer].sort_values(by="DATE").reset_index(drop=True)
        i=0
        for x in tmp["DOCUMENT_CODE"].drop_duplicates():
            if i==0: 
                articles += [0]
                sequence = tmp[tmp["DOCUMENT_CODE"]==x]["PRODUCT_ID"].str.cat(sep =", ")
                i = 1
            else:
                articles += [sequence]
                sequence = articles[-1] +", " + tmp[tmp["DOCUMENT_CODE"]==x]["PRODUCT_ID"].str.cat(sep =", ")

    return articles

Starting data:

CUSTOMER	DATE	DOCUMENT_CODE	ARTICLE_CODE
1	01/01/22	a	x
1	01/01/22	a	y
1	02/01/22	b	z
1	03/01/22	d	w
2	01/01/22	c	x

To:

CUSTOMER	DATE	DOCUMENT_CODE	ARTICLE_CODE
1	01/01/22	a	0
1	02/01/22	b	x,y
1	03/01/22	d	x,y,z
2	01/01/22	c	0

Answer 1

IIUC you need a double groupby:

out = (
 df.groupby(['CUSTOMER', 'DATE', 'DOCUMENT_CODE'])
   ['ARTICLE_CODE'].agg(','.join)
   .groupby('CUSTOMER').shift(fill_value=0)
   .reset_index()
)

Output:

   CUSTOMER      DATE DOCUMENT_CODE ARTICLE_CODE
0         1  01/01/22             a            0
1         1  02/01/22             b          x,y
2         2  01/01/22             c            0

cumulated aggregation

sep = ','

out = (
 df.groupby(['CUSTOMER', 'DATE', 'DOCUMENT_CODE'])
   ['ARTICLE_CODE'].agg(sep.join)
   .groupby('CUSTOMER')
   .apply(lambda s: s.add(sep).cumsum().str[:-len(sep)].shift(fill_value=0))
   .reset_index()
)

Output:

   CUSTOMER      DATE DOCUMENT_CODE ARTICLE_CODE
0         1  01/01/22             a            0
1         1  02/01/22             b          x,y
2         1  03/01/22             d        x,y,z
3         2  01/01/22             c            0