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])
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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
 
						