How to combine ';'.join and lambda x: x.tolist() inside an groupby.agg() function?

问题

Update down below!

I am trying to merge and sort a list of IDs and their connected unique Name_ID, separated by semicolons. For example:

Name_ID Adress_ID            Name_ID Adress_ID
Name1   5875383              Name1   5875383; 5901847
Name1   5901847              Name2   5285200
Name2   5285200      to      Name3   2342345; 6463736
Name3   2342345
Name3   6463736

This is my current code:

origin_file_path = Path("Folder/table.xlsx")
dest_file_path = Path("Folder/table_sorted.xlsx")

table = pd.read_excel(origin_file_path)
df1 = pd.DataFrame(table)

df1 = df1.groupby('Name_ID').agg(lambda x: x.tolist())

df1.to_excel(dest_file_path, sheet_name="Adress_IDs")

But it exports it like this to the excel file:

Name_ID Adress_ID
Name1   [5875383, 5901847]

Can someone tell me what the best way would be to get rid of the list format and separate by semicolons instead of commas?

Update:

The user Jezrael linked me this thread. But I can't seem to be able to combine ';'.join with lambda x: x.tolist().

df1 = df1.groupby('Kartenname').agg(';'.join, lambda x: x.tolist())

Produces TypeError: join() takes exactly one argument (2 given)

df1 = df1.groupby('Kartenname').agg(lambda x: x.tolist(), ';'.join)

Produces TypeError: () takes 1 positional argument but 2 were given.

I also tried other Combinations but none seem to even execute properly. Getting rid of the lambda function isn't an option because then it just pastes Name_ID Adress_ID a thousand times instead of the correct Name and correct IDs.

回答1:

You can pass to agg function tuples with new column names with aggregate functions:

df['Adress_ID'] = df['Adress_ID'].astype(str)
df1 = df.groupby('Name_ID')['Adress_ID'].agg([('a', ';'.join),
                                              ('b',  lambda x: x.tolist())])

print (df1)
                       a                   b
Name_ID                                     
Name1    5875383;5901847  [5875383, 5901847]
Name2            5285200           [5285200]
Name3    2342345;6463736  [2342345, 6463736]

If pass only aggregate functions in list (no tuples) get default columns names:

df2 = df.groupby('Name_ID')['Adress_ID'].agg([ ';'.join,lambda x: x.tolist()])

print (df2)
                    join          <lambda_0>
Name_ID                                     
Name1    5875383;5901847  [5875383, 5901847]
Name2            5285200           [5285200]
Name3    2342345;6463736  [2342345, 6463736]

回答2:

first you need to make sure the Address_ID is string

then you can apply this func:

df.groupby('Name_ID').agg(lambda x: ':'.join(list(x.values)))

more about 'str'.join method

回答3:

• The main issue
  • Can't join an int

Name_ID  Adress_ID
  Name1    5875383
  Name1    5901847
  Name2    5285200
  Name3    2342345
  Name3    6463736

def fix_my_stuff(x):
    x = x.tolist()
    x = '; '.join([str(y) for y in x])
    return(x)

df_updated = df.groupby('Name_ID').agg(lambda x: fix_my_stuff(x)).reset_index()
print(df_updated)

Name_ID         Adress_ID
  Name1  5875383; 5901847
  Name2           5285200
  Name3  2342345; 6463736

来源：https://stackoverflow.com/questions/58179593/how-to-combine-join-and-lambda-x-x-tolist-inside-an-groupby-agg-functio