> y <-matrix(1:50,ncol=5) > x <- rowSums(y > 13) # sum of logicals counts number of TRUEs > x # your first 'x': [1] 3 3 3 4 4 4 4 4 4 4 > sum(0.25 * x > 0.8) # sum of logicals counts number of TRUEs [1] 7
Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com > -----Original Message----- > From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On > Behalf > Of Andras Farkas > Sent: Wednesday, August 29, 2012 9:19 AM > To: PIKAL Petr; r-help@r-project.org > Subject: Re: [R] Help on not matching object lengths > > Petr, > > thank you very much for the help, I think this will work... It seems like > this will generate > 33 columns and 1000 rows with calculated values, which is fine. I also simply > used the > function x=d + dnext, which generated another 33 columns and 1000 rows object > with > the cell contents added up. In my example moving forward from here I will cut > down the > sample size as you suggested, so using the following code we will get a > matrix of 10 rows > and 5 columns: > > y <-matrix(1:50,ncol=5) > > In the next step i would need to find in each row the values that are greater > then a pre- > specified value (any random number can be picked for the purpose of > presenting the > example, lets pick the number 13). I do not actually need to know what the > value is, I > would just need to know the total number of cells in a row which contain > values of > 13 > out of the 5 of them per row, then I need to multiply that count by 0.25 for > each row and > place them in an object , lets call that x (this could also be a matrix > perhaps with 1 > column). > > 1. So in the 1st row of y in my example this would be 3, 2nd row is 3, third > is 3, 4th is 4, > 5th is 4, and so on... > 2. let's multiply them with 0.25 to create x : so 1st is 3*0.25, 2nd is > 3*0.25, 3rd is 3*0.25, > 4th is 4*0.25, 5th is 4*0.25, and so on.... > > finally, I would need to find the total number of cells in this newly created > x that are > greater then z, lets make that z = 0.8. based on the example data matrix of y > this final > number should be 7. > > Would you mind sharing your input on how to accomplish that? > > thanks, > > Andras > > > ________________________________ > From: PIKAL Petr <petr.pi...@precheza.cz> > > roject.org> > Sent: Wednesday, August 29, 2012 11:11 AM > Subject: RE: [R] Help on not matching object lengths > > Hi > > I am not sure if I understand it correctly but maybe > a <-seq(0,8, by = sign(8-0)*0.25) > b <-seq(8,16, by = sign(16-8)*0.25) > const <-runif(1000,50,60) > const <- rep(const, 33) > d <-exp(-const*a)+exp(-const*b) > dim(d) <- c(33,1000) > > gives you matrix 33x1000 > > Maybe you can simplyfy the example to 10 columns and 3 rows and show better > what do > you want to do with it. > > Regards > Petr > > > -----Original Message----- > > From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- > > project.org] On Behalf Of Andras Farkas > > Sent: Wednesday, August 29, 2012 12:25 PM > > To: r-help@r-project.org > > Subject: [R] Help on not matching object lengths > > > > 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]] > [[alternative HTML version deleted]] ______________________________________________ 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.
