为什么不先在一个团队中给我第一个和最后一个 [英] Why doesn't first and last in a groupby give me first and last
问题描述
考虑数据框<$ p>
c $ c $ d $
pre $ d $ df = list('xxxyyy'),
B = [np.nan,1,2,3,4,np.nan]
))
AB
0 x NaN
1 x 1.0
2 x 2.0
3 y 3.0
4 y 4.0
5 y NaN
我想得到列'A'
定义的每个组的第一行和最后一行。 p>
我试过了
df.groupby('A')。 agg(['first','last'])
第一个最后
A
x 1.0 2.0 $ b $由3.0 4.0
然而,这并没有给我预期的 np.NaN
s。 p>
我如何获得每组中的实际第一个和最后一个值? 一个选择n是使用 .nth
方法:
>> > gb = df.groupby('A')
>>> gb.nth(0)
B
A
x NaN
y 3.0
>>> gb.nth(-1)
B
A b b b x 2.0
y NaN
>>>
然而,我还没有找到一种方法来整合它们。当然,也可以使用 pd.DataFrame
构造函数:
>>> pd.DataFrame({'first':gb.B.nth(0),'last':gb.B.nth(-1)})
第一个最后
A
x NaN 2.0
y 3.0 NaN
注意:我明确地使用了 gb.B
属性,否则你必须使用 .squeeze
I'm posting this because the topic just got brought up in another question/answer and the behavior isn't very well documented.
Consider the dataframe df
df = pd.DataFrame(dict(
A=list('xxxyyy'),
B=[np.nan, 1, 2, 3, 4, np.nan]
))
A B
0 x NaN
1 x 1.0
2 x 2.0
3 y 3.0
4 y 4.0
5 y NaN
I wanted to get the first and last rows of each group defined by column 'A'
.
I tried
df.groupby('A').B.agg(['first', 'last'])
first last
A
x 1.0 2.0
y 3.0 4.0
However, This doesn't give me the np.NaN
s that I expected.
How do I get the actual first and last values in each group?
One option is to use the .nth
method:
>>> gb = df.groupby('A')
>>> gb.nth(0)
B
A
x NaN
y 3.0
>>> gb.nth(-1)
B
A
x 2.0
y NaN
>>>
However, I haven't found a way to aggregate them neatly. Of course, one can always use a pd.DataFrame
constructor:
>>> pd.DataFrame({'first':gb.B.nth(0), 'last':gb.B.nth(-1)})
first last
A
x NaN 2.0
y 3.0 NaN
Note: I explicitly used the gb.B
attribute, or else you have to use .squeeze
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