# python - pandas groupby first - Pandas dataframe get first row of each group

2
Answers

This will give you the second row of each group (zero indexed, nth(0) is the same as first()):

```
df.groupby('id').nth(1)
```

Documentation: http://pandas.pydata.org/pandas-docs/stable/groupby.html#taking-the-nth-row-of-each-group

pandas groupby first()

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])
```

63 votes

maybe this is what you want

```
import pandas as pd
idx = pd.MultiIndex.from_product([['state1','state2'], ['county1','county2','county3','county4']])
df = pd.DataFrame({'pop': [12,15,65,42,78,67,55,31]}, index=idx)
```

`pop state1 county1 12 county2 15 county3 65 county4 42 state2 county1 78 county2 67 county3 55 county4 31`

```
df.groupby(level=0, group_keys=False).apply(lambda x: x.sort_values('pop', ascending=False)).groupby(level=0).head(3)
> Out[29]:
pop
state1 county3 65
county4 42
county2 15
state2 county1 78
county2 67
county3 55
```