Pandas dataframe get first row of each group

Question

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])

Excess

Answer 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

Graphemics

@Devin you are talking about DataFrame.first. The guy is using DataFrameGroupBy.first – Hickok

Answer 2

I'd suggest to use .nth(0) rather than .first() if you need to get the first row.

The difference between them is how they handle NaNs, so .nth(0) will return the first row of group no matter what are the values in this row, while .first() will eventually return the first not NaN value in each column.

E.g. if your dataset is :

df = pd.DataFrame({'id' : [1,1,1,2,2,3,3,3,3,4,4],
            'value'  : ["first","second","third", np.NaN,
                        "second","first","second","third",
                        "fourth","first","second"]})

>>> df.groupby('id').nth(0)
    value
id        
1    first
2    NaN
3    first
4    first

And

>>> df.groupby('id').first()
    value
id        
1    first
2    second
3    first
4    first

Calamint

Answer 3

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

Laskowski

But then it should be nth(0) not nth(1) – Modiste

Answer 4

If you only need the first row from each group we can do with drop_duplicates, Notice the function default method keep='first'.

df.drop_duplicates('id')
Out[1027]: 
    id   value
0    1   first
3    2   first
5    3   first
9    4  second
11   5   first
12   6   first
15   7  fourth

Damages

Answer 5

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

Vladikavkaz

Answer 6

I suppose "first" means you have already sorted your DataFrame as you want.

What I do is :

df.groupby('id').agg('first') I suppose "first" means you have already sorted your DataFrame as you want. What I do is :

df.groupby('id').agg('first')
     value
id        
1    first
2    first
3    first
4   second
5    first
6    first
7   fourth

the nice thing is that you can plug any function you want :

df.groupby('id').agg(['first','last','count']))
     value              
     first    last count
id                      
1    first  second     3
2    first  second     2
3    first   fifth     4
4   second   fifth     2
5    first   first     1
6    first   third     3
7   fourth   fifth     2

Output DataFrame has MultiIndex columns

MultiIndex([('value', 'first'),
            ('value',  'last'),
            ('value', 'count')],
           )

Frippery

I noted in the comment for another answer that .first() should not be used for this purpose. .nth() is the method that should be used instead. – Nica

Answer 7

You can use the method take that accepts a list of indices of elements to select:

df.groupby('id').take([0])

Grocer

Answer 8

Considering that the 'id' column is of numeric type, such as int32/int64, one might also use groupby.rank() as following

[In]: df[df.groupby('value')['id'].rank() == 1]
[Out]:
   id   value
0   1   first
6   3   third
7   3  fourth
8   3   fifth

If one wants to reset the index, just pass .reset_index() such as

[In]: df[df.groupby('value')['id'].rank() == 1].reset_index()
[Out]:
   index  id   value
0      0   1   first
1      6   3   third
2      7   3  fourth
3      8   3   fifth

If the index and id columns are not needed

[In]: df.drop(['index', 'id'], axis=1, inplace=True)
[Out]:
    value
0   first
1   third
2  fourth
3   fifth

Barling