熊猫数据帧得到每组的第一行

我有一个熊猫数据框架如下:

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"]来分组，并获得每个组的第一行:

        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

我试着跟随，它只给出了DataFrame的第一行。任何关于这方面的帮助都是感激的。

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

当前回答

考虑到'id'列是数字类型，例如int32/int64，也可以使用groupby.rank()，如下所示

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

如果想要重置索引，只需传递.reset_index()，例如

[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

如果不需要索引列和id列

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

2021-05-03 07:44:43

其他回答

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

如果你需要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

要获得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

2013-11-19 09:25:52

你可以使用接受元素索引列表的方法take来选择:

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

2022-05-13 21:31:29

如果需要获取第一行，我建议使用.nth(0)而不是.first()。

它们之间的区别在于如何处理NaN，因此.nth(0)将返回组的第一行，无论这一行中的值是什么，而.first()最终将返回每列中的第一个非NaN值。

例如，如果你的数据集是:

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

And

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

2018-03-07 09:54:56

这将为您提供每个组的第二行(零索引，第n(0)与第一个()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

2016-03-18 00:03:33

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)

流行 12 county2 15 county3 65 county4 42 州县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

2016-10-28 18:39:39

熊猫数据帧得到每组的第一行

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