Dear All I have the following code set up: Code #1 a <-matrix(seq(0,8, by = sign(8-0)*0.25)) b <-matrix(seq(8,16, by = sign(16-8)*0.25)) c <-runif(1000,50,60) d <-exp(-c*a)+exp(-c*b) This will give me the obvious error message of lengths not matching. What I am trying to do here is to have 33 rows x 1000 columns d values calculated in total. As an eaxmple for visual, this is what it supposed to look like where d1,d2,d3 and so on up to d1000 are calculated using the equation and the respective a and b values: Code #2 d1 <-runif(33,10,20) d2 <-runif(33,15,20) d3 <-runif(33,18,20) e <-cbind(a,b,d1,d2,d3) Next, I would like to generate 33 x 1000 newd values again using newa and newb values and the same c values, eg: Code #3 newa <-matrix(seq(16,24, by = sign(24-16)*0.25)) newb <-matrix(seq(24,36, by = sign(36-24)*0.25)) c <-runif(1000,50,60) newd <-exp(-c*newa)+exp(-c*newb) Next I would like to sum d(ith) and newd(ith) according to their by sequence location in the matrices (or whatever type of objects we are working at this point) based on a and newa . As an example, Code #1 should generate a of 0,0.25,0.5, and so on with respective d for the a value 0, 0.25, 0.5, and so on..., while Code #3 should generate newa of 16,16.25,16.5, and so on with respective newd for the a value 0, 0.25, 0.5, and so on..... In the next object generated, we should again have 33 rows x 1000 columns with values, the first row in the first column would have the result of (d at 0+newd at 16), the second row would have the result of (d at 0.25+newd at 16.25), while the 3rd row would have the result of (d at 0.5+newd at 16.5), and so on... Last, after the sums established, I would need to find the value/s that are >x in each column of the latest object, and multiply by 0.25 to generate a list or row 34 for that matter for each 1000 columns, than in this row or list I would need to find all values that are >y. I apologize for the lengthy question, and all of your help would be greatly apreciated, sincerely, andras [[alternative HTML version deleted]]
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