saving dataframe groupby rows to exactly two lines - pandas

Question

I got a dataframe and I want to groupby the rows based on a specific column. Number of rows in each group will be at least 4 and at most 50. I want to save one column from the group into two lines. If the groupsize is even, let us say 2n, then n rows in one line and the remaining n in the second line. If it is odd, n+1 and n or n and n+1 will do.
For example,
import pandas as pd
from io import StringIO
data = """
id,name
1,A
1,B
1,C
1,D
2,E
2,F
2,ds
2,G
2, dsds
"""
df = pd.read_csv(StringIO(data))
I want to groupby id
df.groupby('id',sort=False)
and then get a dataframe like
id name
0 1 A B
1 1 C D
2 2 E F ds
3 2 G dsds

Answer 1

Probably not the most efficient solution, but it works:
import numpy as np
df = df.sort_values('id')
# next 3 lines: for each group find the separation
df['range_idx'] = range(0, df.shape[0])
df['mean_rank_group'] = df.groupby(['id'])['range_idx'].transform(np.mean)
df['separate_column'] = df['range_idx'] < df['mean_rank_group']
# groupby itself with the help of additional column
df.groupby(['id', 'separate_column'], as_index=False)['name'].agg(','.join).drop(
columns='separate_column')

Answer 2

This is a bit convoluted approach but it does the work;
def func(s: pd.Series):
mid = max(s.shape[0]//2 ,1)
l1 = ' '.join(list(s[:mid]))
l2 = ' '.join(list(s[mid:]))
return [l1, l2]
df_new = df.groupby('id').agg(func)
df_new["name1"]= df_new["name"].apply(lambda x: x[0])
df_new["name2"]= df_new["name"].apply(lambda x: x[1])
df = df_new.drop(labels="name", axis=1).stack().reset_index().drop(labels = ["level_1"], axis=1).rename(columns={0:"name"}).set_index("id")