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 ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.