Question

我使用：

bins = pd.cut(data['R10rank'], list(np.arange(0.0, 1.1, 0.1)))
sum=data.groupby(bins)['Ret20d'].agg(['count', 'mean'])

创建如下统计信息：

    count   mean
R10rank     
(0.0, 0.1]  1044    4.782833
(0.1, 0.2]  809 5.527745
(0.2, 0.3]  746 5.181306
(0.3, 0.4]  706 4.034747
(0.4, 0.5]  627 3.119654
(0.5, 0.6]  585 1.977387
(0.6, 0.7]  609 -0.602742
(0.7, 0.8]  493 -2.745312
(0.8, 0.9]  412 -2.476791
(0.9, 1.0]  374 -6.364374

接下来，我想查看可以汇总不同值间隔内的统计信息的垃圾箱。

赞：

<0.1
<0.3
<0.5
>0.5
>0.7
etc

因此，第二行将包含R10rank中所有值为0-3的值的计数和均值。第四行将为R10rank中所有值> 0.5

的值创建计数和均值

我也可以使用pd.cut吗？如果没有，哪种方法更简单？

谢谢。

Answer 1

您可以使用expanding

进行检查

df['New']=df['count']*df['mean']
df.expanding(min_periods=1).sum().assign(mean=lambda x : x['New']/x['count'])
Out[105]: 
            count      mean           New
R10rank                                  
(0.0,0.1]  1044.0  4.782833   4993.277652
(0.1,0.2]  1853.0  5.108054   9465.223357
(0.2,0.3]  2599.0  5.129080  13330.477633
(0.3,0.4]  3305.0  4.895313  16179.009015
(0.4,0.5]  3932.0  4.612165  18135.032073
(0.5,0.6]  4517.0  4.270933  19291.803468
(0.6,0.7]  5126.0  3.691911  18924.733590
(0.7,0.8]  5619.0  3.127121  17571.294774
(0.8,0.9]  6031.0  2.744297  16550.856882
(0.9,1.0]  6405.0  2.212425  14170.581006

熊猫中的自定义范围仓（间隔始终从零开始）

1 个答案: