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