问题:熊猫数据框获取每个组的第一行

我有DataFrame下面的熊猫。

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"]})

我想通过[“ id”,“ value”]对此分组,并获得每个分组的第一行。

        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

预期结果

    id   value
     1   first
     2   first
     3   first
     4  second
     5  first
     6  first
     7  fourth

我尝试了以下操作,仅给出的第一行DataFrame。任何有关此的帮助表示赞赏。

In [25]: for index, row in df.iterrows():
   ....:     df2 = pd.DataFrame(df.groupby(['id','value']).reset_index().ix[0])

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])

回答 0

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

如果需要id作为列:

>>> 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

要获取n条第一条记录,可以使用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
>>> 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

回答 1

这将为您提供每组的第二行(零索引,nth(0)与first()相同):

df.groupby('id').nth(1) 

文档:http : //pandas.pydata.org/pandas-docs/stable/groupby.html#taking-the-nth-row-of-each-group

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


回答 2

我建议使用.nth(0)而不是.first()如果您需要获得第一行。

它们之间的区别在于它们处理NaN的方式,因此.nth(0)无论该行中的值是什么,都将返回组的第一行,而.first()最终将返回每列中的第一个not NaN值。

例如,如果您的数据集是:

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

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

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

回答 3

也许这就是你想要的

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

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

回答 4

如果只需要我们可以处理的每个组的第一行,请注意函数default方法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

If you only need the first row from each group we can do with , 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

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