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How to find the moving standard deviation in an R matrix?
To find the moving standard deviation in a matrix is done in the same way as in a data frame, we just need to use the matrix object name in place of data frame name. Hence, we can make use of rollapply function of zoo package for this purpose.
For example, if we have a matrix called M and we want to find the 2 moving standard deviations then we can use the below given command −
rollapply(M,width=2,FUN=sd,fill=0,align="r")
Example 1
Following snippet creates a matrix −
M1<-matrix(rnorm(40),ncol=2) M1
The following matrix is created −
[,1] [,2] [1,] 0.345787411 0.43349738 [2,] 0.430162396 0.27067794 [3,] 1.103564081 -1.04460809 [4,] -0.580082754 -0.25060257 [5,] 1.927945086 -0.89618197 [6,] -0.280682051 -1.80365388 [7,] -0.001555098 -0.39360239 [8,] 1.326034225 -1.07766271 [9,] -1.397108815 0.23207688 [10,] 1.011509167 0.34147468 [11,] 0.334467618 1.26899363 [12,] 0.108641768 0.11525723 [13,] -0.002091529 0.93016179 [14,] -0.714714754 1.31114310 [15,] -1.186013755 -0.63420972 [16,] -0.126330767 -0.07257558 [17,] -0.971670037 1.03083570 [18,] -1.216119558 -0.03943005 [19,] 1.016730169 1.06252932 [20,] -0.706784197 0.09479385
To load zoo package and find 2-moving standard deviation for matrix M1, add the following code to the above snippet −
library(zoo) rollapply(M1,width=2,FUN=sd,fill=0,align="r")
Output
If you execute all the above given snippets as a single program, it generates the following output −
[,1] [,2] [1,] 0.00000000 0.00000000 [2,] 0.05966212 0.11513073 [3,] 0.47616690 0.93004767 [4,] 1.19051809 0.56144668 [5,] 1.77344349 0.45649357 [6,] 1.56173523 0.64167954 [7,] 0.19737256 0.99705697 [8,] 0.93874741 0.48370369 [9,] 1.92555291 0.92612575 [10,] 1.70315011 0.07735593 [11,] 0.47874067 0.65585494 [12,] 0.15968299 0.81581483 [13,] 0.07830026 0.57622454 [14,] 0.50390071 0.26939447 [15,] 0.33325872 1.37557217 [16,] 0.74930903 0.39713531 [17,] 0.59774513 0.78022959 [18,] 0.17285191 0.75679217 [19,] 1.57886318 0.77920294 [20,] 1.21870870 0.68429231
Example 2
Following snippet creates a matrix −
M2<-matrix(rpois(60,10),ncol=3) M2
The following matrix is created −
[,1][,2][,3] [1,] 11 9 6 [2,] 3 10 8 [3,] 12 10 5 [4,] 8 10 4 [5,] 14 7 10 [6,] 3 10 10 [7,] 10 10 9 [8,] 7 9 8 [9,] 11 11 14 [10,] 6 9 9 [11,] 14 13 14 [12,] 9 11 9 [13,] 12 11 6 [14,] 10 14 7 [15,] 8 14 13 [16,] 21 10 8 [17,] 15 7 10 [18,] 9 15 9 [19,] 10 8 8 [20,] 17 10 7
To find 4-moving standard deviation for matrix M2, add the following code to the above snippet −
rollapply(M2,width=4,FUN=sd,fill=0,align="r")
Output
If you execute all the above given snippets as a single program, it generates the following output −
[,1] [,2] [,3] [1,] 0.000000 0.0000000 0.0000000 [2,] 0.000000 0.0000000 0.0000000 [3,] 0.000000 0.0000000 0.0000000 [4,] 4.041452 0.5000000 1.7078251 [5,] 4.856267 1.5000000 2.7537853 [6,] 4.856267 1.5000000 3.2015621 [7,] 4.573474 1.5000000 2.8722813 [8,] 4.654747 1.4142136 0.9574271 [9,] 3.593976 0.8164966 2.6299556 [10,] 2.380476 0.9574271 2.7080128 [11,] 3.696846 1.9148542 3.2015621 [12,] 3.366502 1.6329932 2.8867513 [13,] 3.500000 1.6329932 3.3166248 [14,] 2.217356 1.5000000 3.5590261 [15,] 1.707825 1.7320508 3.0956959 [16,] 5.737305 2.0615528 3.1091264 [17,] 5.802298 3.4034296 2.6457513 [18,] 6.020797 3.6968455 2.1602469 [19,] 5.500000 3.5590261 0.9574271 [20,] 3.862210 3.5590261 1.2909944
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