【发布时间】:2014-06-05 17:03:34
【问题描述】:
下面是股票每日收益矩阵示例 (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
例如,考虑一个 5 天的移动窗口,因此我想要一个新的矩阵,如下所述:
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 包我能够达到结果,但它有点慢,关于如何提高速度以达到相同结果的任何想法?
动物园代码如下:
require(zoo)
apply(ret_matriz, 2, function(x) rollapply(x, width = 5, FUN = sd, fill = 0, align = 'r'))
【问题讨论】:
-
不用
apply,我觉得rollapply(df, width=5, FUN=sd, fill=0, align="r")如果by.column = TRUE就够用了(默认值) -
您的数据有多大?
-
143 个变量的 4062 次观察
标签: r matrix zoo standard-deviation rollapply