## Question

Asked By – Nilani Algiriyage

I have a pandas `DataFrame`

like following:

```
df = pd.DataFrame({'id' : [1,1,1,2,2,3,3,3,3,4,4,5,6,6,6,7,7],
'value' : ["first","second","second","first",
"second","first","third","fourth",
"fifth","second","fifth","first",
"first","second","third","fourth","fifth"]})
```

I want to group this by `["id","value"]`

and get the first row of each group:

```
id value
0 1 first
1 1 second
2 1 second
3 2 first
4 2 second
5 3 first
6 3 third
7 3 fourth
8 3 fifth
9 4 second
10 4 fifth
11 5 first
12 6 first
13 6 second
14 6 third
15 7 fourth
16 7 fifth
```

Expected outcome:

```
id value
1 first
2 first
3 first
4 second
5 first
6 first
7 fourth
```

I tried following, which only gives the first row of the `DataFrame`

. Any help regarding this is appreciated.

```
In [25]: for index, row in df.iterrows():
....: df2 = pd.DataFrame(df.groupby(['id','value']).reset_index().ix[0])
```

**Now we will see solution for issue: Pandas dataframe get first row of each group **

## Answer

```
>>> df.groupby('id').first()
value
id
1 first
2 first
3 first
4 second
5 first
6 first
7 fourth
```

If you need `id`

as column:

```
>>> df.groupby('id').first().reset_index()
id value
0 1 first
1 2 first
2 3 first
3 4 second
4 5 first
5 6 first
6 7 fourth
```

To get n first records, you can use head():

```
>>> df.groupby('id').head(2).reset_index(drop=True)
id value
0 1 first
1 1 second
2 2 first
3 2 second
4 3 first
5 3 third
6 4 second
7 4 fifth
8 5 first
9 6 first
10 6 second
11 7 fourth
12 7 fifth
```

This question is answered By – Roman Pekar

**This answer is collected from stackoverflow and reviewed by FixPython community admins, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0 **