大 pandas 数据帧groupby并获得第n行 [英] pandas dataframe groupby and get nth row
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问题描述
df = pd.DataFrame([[1.1,1.1,1.1,2.6 ,2.5,3.4,2.6,2.6,3.4,3.4,2.6,1.1,1.1,3.3],(AAABBBBABCBDDD),[1.1,1.7,2.5,2.6,3.3,3.8,4.0,4.2,4.3,4.5, (x / y / z',x / y',x / y / z / n','x / u','x','x / u / v ','x / y / z','x','x / u / v / b',' - ','x / y','x / y / z' / v / w'],['1','3','3','2','4','2','5','3','6','3' ','1','1','1'],['200','400','404','200','200','404','200' ','200','500','200','200','400']])T
df.columns = ['col1','col2','col3' ,'col4','ID','col5']
并获得每组的第二排。后来我也需要得到第3和第4。只需解释一下,如何只得到每个组的第二行。
我尝试了以下第一和第二个。
$ b $ ($)
相反,我只需要获得第二行。由于ID 4和6没有第二行需要忽略它们。
col1 col2 col3 col4 ID col5
ID
1 0 1.1 A 1.1 x / y / z 1 200
11 1.1 D 4.7 x / y / z 1 200
2 3 2.6 B 2.6 x / u 2 200
5 3.4 B 3.8 x / u / v 2 404
3 1 1.1 A 1.7 x / y 3 400
2 1.1 A 2.5 x / y / z / n 3 404
4 4 2.5 B 3.3 x 4 200
5 6 2.6 B 4 x / y / z 5 200
10 2.6 B 4.6 x / y 5 500
6 8 3.4 B 4.3 x / u / v / b 6 500
解决方案
我认为第n种方法应该是这样做的:
在[11]中:g.nth(1).dropna()
输出[11]:
col1 col2 col3 col4 col5
ID
1 1.1 D 4.7 x / y / z 200
2 3.4 B 3.8 x / u / v 404
3 1.1 A 2.5 x / y / z / n 404
5 2.6 B 4.6 x / y 500
在0.13的另一种方法是使用cumcount:
df [g.cumcount()== n - 1]
...这是显着。
在[21]中:%timeit g.nth 1).dropna()
100循环,最好3:11.3 ms每循环
在[22]中:%timeit df [g.cumcount()== 1]
1000循环,最佳3:286μs每循环
I have a pandas DataFrame like following.
df = pd.DataFrame([[1.1, 1.1, 1.1, 2.6, 2.5, 3.4,2.6,2.6,3.4,3.4,2.6,1.1,1.1,3.3], list('AAABBBBABCBDDD'), [1.1, 1.7, 2.5, 2.6, 3.3, 3.8,4.0,4.2,4.3,4.5,4.6,4.7,4.7,4.8], ['x/y/z','x/y','x/y/z/n','x/u','x','x/u/v','x/y/z','x','x/u/v/b','-','x/y','x/y/z','x','x/u/v/w'],['1','3','3','2','4','2','5','3','6','3','5','1','1','1'],['200','400','404','200','200','404','200','404','500','200','500','200','200','400']]).T
df.columns = ['col1','col2','col3','col4','ID','col5']
I want group this by "ID" and get the 2nd row of each group. Later I will need to get 3rd and 4th also. Just explain me how to get only the 2nd row of each group.
I tried following which gives both first and second.
df.groupby('ID').head(2)
Instead I need to get only the second row. Since ID 4 and 6 has no second rows need to ignore them.
col1 col2 col3 col4 ID col5
ID
1 0 1.1 A 1.1 x/y/z 1 200
11 1.1 D 4.7 x/y/z 1 200
2 3 2.6 B 2.6 x/u 2 200
5 3.4 B 3.8 x/u/v 2 404
3 1 1.1 A 1.7 x/y 3 400
2 1.1 A 2.5 x/y/z/n 3 404
4 4 2.5 B 3.3 x 4 200
5 6 2.6 B 4 x/y/z 5 200
10 2.6 B 4.6 x/y 5 500
6 8 3.4 B 4.3 x/u/v/b 6 500
解决方案
I think the nth method is supposed to do just that:
In [11]: g.nth(1).dropna()
Out[11]:
col1 col2 col3 col4 col5
ID
1 1.1 D 4.7 x/y/z 200
2 3.4 B 3.8 x/u/v 404
3 1.1 A 2.5 x/y/z/n 404
5 2.6 B 4.6 x/y 500
In 0.13 another way to do this is to use cumcount:
df[g.cumcount() == n - 1]
...which is significantly faster.
In [21]: %timeit g.nth(1).dropna()
100 loops, best of 3: 11.3 ms per loop
In [22]: %timeit df[g.cumcount() == 1]
1000 loops, best of 3: 286 µs per loop
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