How to remove double loop?
On Thu, 20 Mar 2008, Alberto Monteiro wrote:
Jonas Malmros wrote:
But what if I had a matrix, where the last column was filled with values first (again, a for loop), and the rest was filled by using a double loop?
Killing one of those loops is quite simple. The other may be harder.
OVal <- matrix(0, n+1, n+1)
for(i in 0:n){
OVal[i+1, n+1] <- max(Val[i+1, n+1]-K, 0)
}
This is straightforward: notice that most R arithmetic
functions that operate in scalars also operate in vectors,
matrices and arrays. For example, sin(pi) is zero, but
sin(c(0,pi/2,pi,3*pi/2)) is c(0, 1, 0, -1) (with some
rounding errors).
The loop in _i_ is just to take the max of a column? So:
OVal[1:(n+1), n+1] <- # this is a vector
max(Val[1:(n+1), n+1] - # this is another vector
- K, 0) # vector - scalar = vector, max(vector) = vector
or, in one line:
OVal[1:(n+1), n+1] <- max(Val[1:(n+1), n+1] - K, 0)
You can drop the indices, as you are taking all of them:
OVal[,n+1] <- max(Val[,n+1] - K, 0)
for(i in seq(n,1, by=-1)){
for(j in 0:(i-1)){
OVal[j+1, i] <- a*((1-p)*OVal[j+1, i+1]+p*OVal[j+2, i+1])
}
}
The inner loop is simple, but somehow tricky.
You are computing each j-th term as the same j-th term combined
with the (j+1)-th term. So, you take a combination of js in
the 1:i range and combine with js in the 2:(i+1) range... So:
OVal[1:i, i] <- a*((1-p)*OVal[1:i, i+1] + p*OVal[2:(i+1), i+1])
The outer loop (in i) probably can't be optimized.
Alberto Monteiro
Unfortunately, neither of these loop removals gives the same answers as the
original code :-( [Hence the request for reproducible code]. Try:
TreeMy(K=0.5)
What does work though is the following:
TreeMy <-
function(S=50,K=50,sigma=0.4,r=0.1,Time=1,n=3){
dtime <- Time/n
u <- exp(sigma*sqrt(dtime))
d <- 1/u
p <- (exp(r*dtime)-d)/(u-d)
a <- exp(-r*dtime)
OVal <- matrix(0, n+1, n+1)
Val <- outer(0:n, 0:n, function(i, j) ifelse(i <= j, u^i*d^(j-i), 0))
OVal[, n+1] <- pmax(Val[, n+1]-K, 0)
for (i in n:1) {
OVal[1:i, i] <- colSums(matrix(a*c((1 - p), p)*
OVal[, i + 1][rep(1:2, i) + rep(0:(i - 1), each=2)], ncol=i))
}
list("Stock Price"=Val, "Option Price"=OVal)
}
Here the first loop is removed with a simple pmax() (the 'vector' form of
max()). The second loop uses the trick of expanding the column operations
into a vector using recycling then restructuring them back into a matrix.
This provides a significant speed improvement.
HTH
Ray Brownrigg