Pandas dataframe get first row of each group
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])
Solution 1:
>>> 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
Solution 2:
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