Nilani Algiriyage - 1 year ago 222

Python Question

I have a pandas

`DataFrame`

`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`

`In [25]: for index, row in df.iterrows():`

....: df2 = pd.DataFrame(df.groupby(['id','value']).reset_index().ix[0])

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Answer Source

```
>>> 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
```

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