Question

我有一个像这样的熊猫df：

%diff          

-0.087704164    
0.003908466 
-0.032150706    
-0.035684163    
0.001682029 
-0.072205803    
0.031636864 
-0.069263158    
-0.214883511    
-0.109286469    
0.274932615 
-0.016913319    
-0.075268817    
0.191906977 
0.043861703 
-0.048598131    
0.01280943  
0.014509621 
0.075564054 
-0.024034701    
0.009107468     
0.023465704

我想通过'％diff列中最后20个值的标准偏差来计算252倍的平方根。

    %diff          std      

    -0.087704164    
    0.003908466 
    -0.032150706    
    -0.035684163    
    0.001682029 
    -0.072205803    
    0.031636864 
    -0.069263158    
    -0.214883511    
    -0.109286469    
    0.274932615 
    -0.016913319    
    -0.075268817    
    0.191906977 
    0.043861703 
    -0.048598131    
    0.01280943  
    0.014509621 
    0.075564054 
   -0.024034701     165.9%
    0.009107468     163.2%
    0.023465704     163.4%

我尝试的代码是：

df1['std'] = 252**(1.0/2) * df1['%diff'].std().split(20)

但是我收到了不受支持的操作数错误

Answer 1

您需要rolling的20窗口和std像

df1['std'] = 252**(1.0/2) * df1.rolling(20)['%diff'].std()
print (df1)
       %diff       std
0  -0.087704       NaN
1   0.003908       NaN
2  -0.032151       NaN
3  -0.035684       NaN
4   0.001682       NaN
5  -0.072206       NaN
6   0.031637       NaN
7  -0.069263       NaN
8  -0.214884       NaN
9  -0.109286       NaN
10  0.274933       NaN
11 -0.016913       NaN
12 -0.075269       NaN
13  0.191907       NaN
14  0.043862       NaN
15 -0.048598       NaN
16  0.012809       NaN
17  0.014510       NaN
18  0.075564       NaN
19 -0.024035  1.659144
20  0.009107  1.631865
21  0.023466  1.634266

基于熊猫行的计算和迭代

1 个答案: