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 ______________________________________________ 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.