如何计算大 pandas groupby中的所有正值和负值? [英] How to count all positive and negative values in a pandas groupby?

问题描述

假设我们有一张桌子:

df = pd.DataFrame({'A' : ['foo', 'bar', 'foo', 'bar', 'foo', 'bar', 'foo', 'foo'],
                   'B' : ['one', 'one', 'two', 'three', 'two', 'two', 'one', 'three'],
                   'C' : np.random.randn(8), 'D' : np.random.randn(8)})

输出:

      A       B          C           D
0    foo     one    -1.304026    0.237045
1    bar     one     0.030488   -0.672931
2    foo     two     0.530976   -0.669559
3    bar     three  -0.004624   -1.604039
4    foo     two    -0.247809   -1.571291
5    bar     two    -0.570580    1.454514
6    foo     one     1.441081    0.096880
7    foo     three   0.296377    1.575791

我想计算C列中的正数和负数分别属于A列中的每个组，并按比例计算. A中的组比foo和bar多得多，因此组名不应出现在代码中.

I want to count how many positive and negative numbers in column C belong to each group in column A and in what proportion. There are much more groups in A than foo and bar, so group names shouldn't be in the code.

我试图对A进行分组，然后进行过滤，但没有找到正确的方法.还尝试使用一些聪明的lambda进行汇总，但没有成功.

I was trying to groupby A and then filter, but didn't find the right way. Also tried to aggregate with some smart lambda, but didn't succeed.

推荐答案

您可以将其作为一行应用(第一列为负，第二列为正):

You could do this as a one line apply (the first column being negative, the second positive):

In [11]: df.groupby('A').C.apply(lambda x: pd.Series([(x < 0).sum(), (x >= 0).sum()])).unstack()
Out[111]: 
     0  1
A        
bar  2  1
foo  2  3

[2 rows x 2 columns]

但是，我认为更巧妙的方法是使用虚拟列并使用value_counts:

However, I think a neater way is to use a dummy column and use value_counts:

In [21]: df['C_sign'] = np.sign(df.C)

In [22]: df.groupby('A').C_sign.value_counts()
Out[22]: 
A      
bar  -1    2
      1    1
foo   1    3
     -1    2
dtype: int64

In [23]: df.groupby('A').C_sign.value_counts().unstack()
Out[23]: 
     -1   1
A          
bar   2   1
foo   2   3

[2 rows x 2 columns]

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