similar to: finding value for a parameter in an equation

Dear All,   I have a matrix with 33 columns and 5000 rows. I would like to find 2 specific columns in the set: the one that holds the highest values and the one that holds the lowest values. In this case the column's mean would be apropriate to use to try to find those specific columns because each columns mean is different and they all change together based on the same "change of rate

Dear All,   I am working with a Sweave document to be converted into PDF using Rstudio. It seems to me that my R code will also show up after conversion, which I would like not to happen. Is there a way to specify a command that I could place on the beggining and at the end of the R code that would allow R to recognize the area where my R code is and would ignor it when the data is converted?  

Dear All,   as a follow up to my previous e-mail (I think I am getting closer...):   I am trying to apply the trapezoidal functions to a matric column by column. I have the following code:   a <-matrix(c(1:100),ncol=10) b <-matrix(c(2,4,6,8,10,12,14,16,18,20))   apply(a,2,function(b,a) sum(diff(b)*(a[-1]+a[-length(a)]))/2)   for some reason i get an error message: Error in FUN(newX[[, i],

Dear All,   I have the following code set up:   x <-2000 y <-8 z <-3   I would need to use these numbers to show up in my plot title "mixed" with text. The x,y,z numbers would need to change, the text would not. So my title should look like this   "x txt1 y txt2 z txt3"   so if: txt1=hours txt2=minutes txt3=seconds   then my title of the plot should read:   2000 hours

Dear All,   sorry, got stuck again on the following: let us say we have:   a <-c(1:5) b <-c(6:10) d <-cbind(a,b)     from d I would like to remove total number of rows based on the length of f. So if:   f <-c(1)   my result is working great with the following solution:   d[-length(f),]   so I get: a b [1,] 2 7 [2,] 3 8 [3,] 4 9 [4,] 5 10 but if I do: f <-c(1,2) then I get: 

Dear All,   this is probably an easy one but I can not get a handle on it:   x <-c(1,2,3,4,5) y <-c(6,7,8,9,10) z <-15 w <-ifelse(z>14,x,y)   this will give me a value of 1 for w. What I would like to get is the whole string of x, so that w would become a numeric object of 5 characters exactly the same as x.   Apreciate the help,   Sincerely,   Andras [[alternative HTML version

Dear All   I have the following data (somewhat simplyfied):   TINF <-1 a <-c(500,750,1000,1250,1500,1750,2000) b <-c(8,12,18,24,36,48,60,72,96)   following function:   infcprodessa <-function (D, tin, tau, ts)   (D * (1 - exp(-0.048 * tin))/(tin * (0.048*79) * (1 - exp(-0.048 * tau)))) * exp(-0.048 * (ts - tin)) z <-sapply(1:1, function(n) infcprodessa(1000,TINF,12,12-TINF))   is

Dear all,   could you please give me some pointers on how I could make R screen for a value if it falls within a certain range? I looked at the subset function, but is not doing it, perhaps because I only have 1 value to screen?   aptreciate the input   ex: a <-16.5 I would like to screen to see if a is within the range of 15 to 20, (which it is:-)), and I would like the code to return a value

Dear All,   I often have to work with certain models in which I try to "reproduce" a distribution the best I can with very little known information avaible. Is there a package or function in R that could best reproduce a probability distribution using only the mean, median and SD values availble without knowing the actual distribution type to begin with and/or the covariance matrix (for

Dear All, would you please provide your thoughts on the following: let us say I have: a <-c(1,5,8,15,32,69) b <-c(8.5,33) and I would like to extract from "a" the two values that are closest to the values in "b", where the length of this vectors may change but b will allways be shorter than "a". So at the end based on this example I should have the result

Hi, I am using package deSolve to run some ordinary differential equations (ODE) as part of a mathematical modeling project. I have solved for the following equilibrium states: Seq1<-a*(1-Neq1)/(f*Veq1+m+d) Ceq1<-(f*Seq1*Veq1+g*Ieq1+r*(1-Neq1)-b1*Veq1*Ieq1)/(b2+m+d+g) Ieq1<-(-b2*Ceq1)-r*(1-Neq1)/(b1*Veq1-g-u) Veq1<-o*(Ceq1+Ieq1)/e I want to take the differential of both sides of

Dear All,   wonder if you have a thought on the following: I am using the round(x,digits=3) command, but some of my values come out as: 0.07099999999999999 AND 0.06900000000000001. Any thoughts on why this maty be happening or how to eliminate the problem?   apreciate the help,   Andras [[alternative HTML version deleted]]

Dear All,   any thoughts on how I can do the following:   let us say we have:   a <-c(2,4,16,28,48) b <-c(10,4,2,0.4,0.03) d <-cbind(a,b)   what I would like to do is to extract values of column b in the matrix based on the corresponding values of column a. For example: I would like to extract all b values that have a corresponding a value that is less than 24 into a numeric vector, so

Dear All,   I could use a bit of help here, this function is hard to figure out (for me at least) I have the following so far:   PKindex<-data.frame(Subject=c(1),time=c(1,2,3,4,6,10,12),conc=c(32,28,25,22,18,14,11)) Dose<-200 Tinf <-0.5   defun<- function(time, y, parms) {  dCpdt <- -parms["kel"] * y[1]  list(dCpdt)  } modfun <- function(time,kel, Vd) {   out <-

Dear All,   is there a way to set low and high limits to a simulation with rlnorm()?   as an example:  a <-rlnorm(500,0.7,1)     I get the summary of   Min. 1st Qu. Median Mean 3rd Qu. Max. 0.1175 1.0590 2.1270 3.4870 4.0260 45.3800 I would like to set limits so that the simulated values minimum would be greater then 0.5 and maximum of less than 30. If during simulation a

We've recently "upgraded" a server to solaris 8, and in reinstalling R I've encountered a problem in getting it to recognize the readline library. I have installed readline in what I believe to be the "usual" place: /usr/local/lib with include files in /usr/local/include/readline. But ./configure produces: ragnar.econ.uiuc.edu# grep readline hout checking for

Dear All   I would like to convert matrix rows to columns. I am thinking the t() function should help, but am having a hard time converting the matrix into the dimensions I would like them to. Example:   a <-matrix(c(1:30),ncol=3) gives me:[,1] [,2] [,3] [1,] 1 11 21 [2,] 2 12 22 [3,] 3 13 23 [4,] 4 14 24 [5,] 5 15 25 [6,] 6 16 26 [7,] 7 17

Dear R News, I am inquiring whether the articles that are published in the 'R News' have undergone a peer-review/refereeing process, for the purpose of reporting the academics publications here at the University of Sydney to the Australian Government. Thank you for your assistance. Regards Glenn Glenn Howell| Faculty Research Assistant FACULTY OF ECONOMICS AND BUSINESS

At the APL conference in Berlin on July 24 I am giving a talk about about a molecule-like notation for arrays for which some prototype code is available in R. It is a graphical notation for arrays of higher rank which makes the structure of arrays and their various concatenations intuitively apparent, and which, in my judgment, would make an excellent interface for array programming languages. I

Dear All,   another quick question, this one is on skipping part of my code, so let us say:   a <-5 b <-2 e <-0   d <-a+b f <-a-b   what I would like to do is to have R NOT to calculate the value for d in case the value of e equals to zero (essentially skip that "chunk"), but instead move on to calculate te value for f. In the code I am working with the value of e changes,