Getting groups by group index - pandas

Question

I want to access the group by group index. My dataframe is as given below
import pandas as pd
from io import StringIO
import numpy as np
data = """
id,name
100,A
100,B
100,C
100,D
100,pp;
212,E
212,F
212,ds
212,G
212, dsds
212, sas
300,Endüstrisi`
"""
df = pd.read_csv(StringIO(data))
I want to groupby 'id' and access the groups by its group index.
dfg=df.groupby('id',sort=False,as_index=False)
dfg.get_group(0)
I was expecting this to return the first group which is the group for id =1 (which is the first group)

Answer 1

You need pass value of id:
dfg=df.groupby('id',sort=False)
a = dfg.get_group(100)
print (a)
id name
0 100 A
1 100 B
2 100 C
3 100 D
4 100 pp;
dfg=df.groupby('id',sort=False)
a = dfg.get_group(df.loc[0, 'id'])
print (a)
id name
0 100 A
1 100 B
2 100 C
3 100 D
4 100 pp;
If need enumerate groups is possible use GroupBy.ngroup:
dfg=df.groupby('id',sort=False)
a = df[dfg.ngroup() == 0]
print (a)
id name
0 100 A
1 100 B
2 100 C
3 100 D
4 100 pp;
Detail:
print (dfg.ngroup())
0 0
1 0
2 0
3 0
4 0
5 1
6 1
7 1
8 1
9 1
10 1
11 2
dtype: int64
EDIT: Another idea is if need select groups by positions (all id are consecutive groups) with compare by unique values of id selected by positions:
ids = df['id'].unique()
print (ids)
[100 212 300]
print (df[df['id'].eq(ids[0])])
id name
0 100 A
1 100 B
2 100 C
3 100 D
4 100 pp;
print (df[df['id'].eq(ids[1])])
id name
5 212 E
6 212 F
7 212 ds
8 212 G
9 212 dsds
10 212 sas