在R中的矩阵中滚动标准偏差 [英] Rolling Standard Deviation in a Matrix in R

问题描述

Bellow是股票每日收益矩阵示例( ret_matriz )

Bellow is a stock daily returns matrix example (ret_matriz)

      IBOV        PETR4        VALE5        ITUB4        BBDC4        PETR3    
[1,] -0.040630825 -0.027795652 -0.052643733 -0.053488685 -0.048455772 -0.061668282
[2,] -0.030463489 -0.031010237 -0.047439725 -0.040229625 -0.030552275 -0.010409016
[3,] -0.022668170 -0.027012078 -0.022668170 -0.050372843 -0.080732363  0.005218051
[4,] -0.057468428 -0.074922051 -0.068414670 -0.044130126 -0.069032911 -0.057468428
[5,]  0.011897277 -0.004705891  0.035489885 -0.005934736 -0.006024115 -0.055017693
[6,]  0.020190656  0.038339130  0.009715552  0.014771317  0.023881732  0.011714308
[7,] -0.007047191  0.004529286  0.004135085  0.017442303 -0.005917177 -0.007047191
[8,] -0.022650593 -0.029481336 -0.019445057 -0.017442303 -0.011940440 -0.046076458
[9,]  0.033137223  0.035274722  0.038519205  0.060452104  0.017857617  0.046076458

出于示例目的，考虑一个5天的移动窗口，因此我想要一个如下所述的新矩阵:

For example purposes consider a 5 day moving window, i want as a result a new matrix as described bellow :

     IBOV        PETR4    ...       
[1,] 0           0        ...
[2,] 0           0        ... 
[3,] 0           0        ...
[4,] 0           0        ...
[5,] sd[1:5,1]  sd[1:5,2] ...
[6,] sd[2:6,1]  sd[2:6,2] ...
[7,] sd[3:7,1]  sd[3:7,2] ...
[8,] sd[4:8,1]  sd[4:8,2] ... 
[9,] sd[5:9,1]  sd[5:9,2] ...

使用zoo包我可以达到结果，但是有点慢，是否有任何关于如何提高达到相同结果速度的想法?

Using the zoo package i was able to reach the result but it is a little bit slow, any ideias on how to improve the speed to reach the same result ?

以下代码:

require(zoo)

apply(ret_matriz, 2, function(x) rollapply(x, width = 5, FUN = sd, fill = 0, align = 'r')) 

推荐答案

1)可以删除apply部分.为了简洁起见，我们还使用rollapplyr:

1) The apply part can be eliminated. We also use rollapplyr for brevity:

rollapplyr(ret_matriz, 5, sd, fill = 0)

2)另外，rollmean比rollapply快，因此我们可以使用公式sd = sqrt(n/(n-1) * (mean(x^2) - mean(x)^2))从其构建:

2) Also rollmean is faster than rollapply so we could construct it from that using the formula sd = sqrt(n/(n-1) * (mean(x^2) - mean(x)^2)):

sqrt((5/4) * (rollmeanr(ret_matriz^2, 5, fill = 0) - 
              rollmeanr(ret_matriz, 5, fill = 0)^2))

这篇关于在R中的矩阵中滚动标准偏差的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！