CODEWITHSUNDEEP
rloopsmatrix
glarys
glarys

Reputation: 131

apply function to each value of a matrix

I have a matrix pmatrix

sigma = 0.03
alpha = 0.01
sims = 6
N = 10
pmatrix = matrix(NA, ncol=sims, nrow = N)
for (i in 1:N){
x = rnorm(sims)
pmatrix[i,] <- x
}

And I need to use this matrix in order to get the xt values from the following expression:

xt = 0
for (i in 1:10){     
xt[i+1] = xt[i] * exp(-alpha*1) + sqrt(((sigma^2)/2*alpha)*(1-exp(-2*alpha*1)))*pmatrix[i]
  }

However the following loop only returns a xt vector. Ideally I would like to obtain a matrix which consists of 10 rows (N- number of years) and 6 columns (sims - the number of simulated scenarios). I believe it is doable via a second loop or an apply function.

Upvotes: 1

Views: 168

Answers (2)

coffeinjunky
coffeinjunky

Reputation: 11514

Just as an extended comment: the issue you were facing came from the fact that you were subsetting xt wrongly. If you want to do something on a matrix rowwise, use the entire row, which you can retrieve using xt[1,] as opposed to xt[1]. See:

sigma = 0.03; alpha = 0.01; sims = 6; N = 10
pmatrix = matrix(rnorm(sims*N), ncol=sims, nrow = N)
xt <- matrix(0, ncol=sims, nrow = N)

xt[1] # just one element
[1] 0
xt[1,] # entire row
[1] 0 0 0 0 0 0

Then it works also with your approach:

for(i in 2:N){
  xt[i,] <- xt[i-1,] * exp(-alpha) + sqrt(((sigma^2)/2*alpha)*(1-exp(-2*alpha)))*pmatrix[i-1,]
}

head(xt)
            [,1]          [,2]         [,3]          [,4]         [,5]         [,6]
[1,] 0.0000000000  0.000000e+00 0.0000000000  0.000000e+00 0.000000e+00 0.0000000000
[2,] 0.0005102757  9.680876e-05 0.0006264992  2.147689e-05 9.432744e-05 0.0004370840
[3,] 0.0005035357 -4.704802e-04 0.0003954507 -5.083994e-04 4.237845e-04 0.0004007277
[4,] 0.0009345963 -2.699634e-04 0.0003559880 -3.877696e-04 5.337181e-04 0.0005230792
[5,] 0.0016501564 -1.947569e-04 0.0005003010 -6.680216e-05 6.098233e-04 0.0007106022
[6,] 0.0015317653  4.316999e-04 0.0011361772 -2.209149e-04 6.881100e-04 0.0005893373

Upvotes: 1

Martin Schmelzer
Martin Schmelzer

Reputation: 23899

Hope this suits you. Sometimes its easier to use an apply function on column or row indices instead of a data object itself.

sigma = 0.03
alpha = 0.01
sims = 6
N = 10
pmatrix <- matrix(rnorm(N * sims), N)

xt <- matrix(nrow=N, ncol=6)
xt[1,] <- 0

sapply(2:N, FUN = function(x) {
    xt[x,] <<- xt[x-1,] * exp(-alpha*1) + sqrt(((sigma^2)/2*alpha)*(1-exp(-2*alpha*1)))*pmatrix[x-1,]
})


> xt
               [,1]          [,2]          [,3]         [,4]          [,5]          [,6]
 [1,]  0.0000000000  0.000000e+00  0.000000e+00 0.0000000000  0.0000000000  0.000000e+00
 [2,] -0.0006488202  4.730257e-04  4.707051e-04 0.0002174562  0.0001655868 -2.063875e-04
 [3,] -0.0007110547  3.792143e-04  3.922429e-04 0.0009465164 -0.0001667539 -1.165253e-05
 [4,] -0.0003679911 -3.596447e-05  5.490986e-06 0.0013176437 -0.0006049390  2.276431e-04
 [5,] -0.0007176342  1.754809e-05  3.647631e-04 0.0015136978 -0.0010303508  4.773186e-04
 [6,] -0.0007918234 -3.065909e-05 -3.703564e-05 0.0015006314 -0.0005650229  7.792698e-05
 [7,] -0.0008442265 -2.808698e-04  7.808261e-05 0.0015505998 -0.0005407453 -3.106797e-04
 [8,] -0.0010265038  8.540579e-05  2.547632e-04 0.0017364697 -0.0007112818 -3.972706e-04
 [9,] -0.0004011710  7.346707e-05  7.471667e-04 0.0014031268 -0.0008266330 -4.296555e-04
[10,] -0.0001490369 -3.189111e-04  1.133248e-03 0.0013038771 -0.0011771068 -2.719285e-04

Upvotes: 2

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