Rolling Standard Deviation in a Matrix in R
Asked Answered
M

2

6

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

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] ...

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 ?

zoo code bellow :

require(zoo)

apply(ret_matriz, 2, function(x) rollapply(x, width = 5, FUN = sd, fill = 0, align = 'r')) 
Mollymollycoddle answered 5/6, 2014 at 17:3 Comment(3)
No need to use apply, I think rollapply(df, width=5, FUN=sd, fill=0, align="r") is enough if by.column = TRUE (default value)Groat
How big is your data?Carnarvon
4062 observations of 143 variablesMollymollycoddle
C
9

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

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

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))
Choreodrama answered 5/6, 2014 at 17:24 Comment(0)
T
8

You can use TTR::runSD instead.

library(quantmod)
getSymbols("SPY")
spy <- apply(ROC(SPY), 2, runSD, n=5)
# head(spy)
#         SPY.Open    SPY.High     SPY.Low   SPY.Close SPY.Volume SPY.Adjusted
# [1,]          NA          NA          NA          NA         NA           NA
# [2,]          NA          NA          NA          NA         NA           NA
# [3,]          NA          NA          NA          NA         NA           NA
# [4,]          NA          NA          NA          NA         NA           NA
# [5,]          NA          NA          NA          NA         NA           NA
# [6,] 0.004369094 0.003112967 0.001064232 0.005035266  0.1577499  0.005063025
Toole answered 5/6, 2014 at 17:7 Comment(0)

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