Re: [R] Loop question?

2013-01-26 Thread Jeff Newmiller
The unlist-sapply-seq_len bit is unnecessarily convoluted, since the infcprodessa function can accept vector inputs. z <- infcprodessa( ab$a, TINF, ab$b, ab$b-TINF ) possibles <- ab[ z >= 15 & z <= 20, ] possibles[ which.min( possibles$a ), ] --

Re: [R] Loop question?

2013-01-26 Thread Berend Hasselman
On 26-01-2013, at 12:31, Andras Farkas wrote: > > Sorry Jeff, probably the new version of Yahoo mail doing the html, I switched > back to the older one hope that takes care of the problem. Let me clarify the > code below: > > TINF <-1 > a <-c(500,750,1000,1250,1500,1750,2000) > b <-c(8,12,18

Re: [R] Loop question?

2013-01-26 Thread Andras Farkas
Andras --- On Sat, 1/26/13, Jeff Newmiller wrote: > From: Jeff Newmiller > Subject: Re: [R] Loop question? > To: "Andras Farkas" , "r-help@r-project.org" > > Date: Saturday, January 26, 2013, 2:09 AM > Please read the Posting Guide > no html email

Re: [R] Loop question?

2013-01-26 Thread David Hugh-Jones
Here's a toy example which you can apply the logic of: dfr <- expand.grid(1:3,1:2) results <- apply(dfr, 1, sum) dfr[results==4,] On 25 January 2013 22:19, Andras Farkas wrote: > > Dear All > > I have the following data (somewhat simplyfied): > > TINF <-1 > a <-c(500,750,1000,1250,1500,1750,

Re: [R] Loop question?

2013-01-25 Thread Jeff Newmiller
Please read the Posting Guide no html email reproducible example please In general, you can use expand.grid to generate all combinations of inputs, compute results as a vector just as long as the expand.grid data frame has rows, and identify which results meet your criteria by a logical test, an

[R] Loop question?

2013-01-25 Thread Andras Farkas
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(-0.0

Re: [R] Loop question

2012-05-30 Thread R. Michael Weylandt
Note that in R >= 2.15 you can also use paste0 for this operation more efficiently. Michael On Thu, May 31, 2012 at 1:58 AM, Özgür Asar wrote: > Dear Sebastian, > > The following will create the names > > paste("sb",1:5,sep="") > paste("sw",1:5,sep="") > paste("Lw",1:5,sep="") > paste("Lb",1:5,s

Re: [R] Loop question

2012-05-30 Thread Özgür Asar
Dear Sebastian, The following will create the names paste("sb",1:5,sep="") paste("sw",1:5,sep="") paste("Lw",1:5,sep="") paste("Lb",1:5,sep="") Then you can easily combine and/or order them in R. Hope this helps. Ozgur - Ozgur ASAR Research Assistant M

[R] Loop question

2012-05-30 Thread Sebastián Daza
Hello, I have a dataframe (Lx) with 5 Lb, and 5 Lw variables. I want to create several variables according to the loop presented below (data attached). Lx <- read.csv("Lx.csv", header=T, sep=",") for (i in 1:20) { Lx$sb1[i] <- Lx$Lb1[i+1]/Lx$Lb1[i] Lx$sb2[i] <- Lx$Lb2[i+1]/Lx$Lb2[i] Lx$sb

[R] loop question

2011-04-14 Thread Thomas
Dear all, I am trying to implement the following in a loop: g <- cbind(c(1, 2, 3), c(1, 2, 3)) h <- cbind(c(1, 2, 3), c(1, 2, 3)) c <- cbind(g[,1]*h[,1], g[,2]*h[,2]) g<-rowSums(c) My attempt looks like this but does not produce the desired results as above. for (i in 1:2) {g<- rowSums(cbind(

Re: [R] loop question

2011-04-05 Thread Joshua Wiley
Dear Thomas, On Tue, Apr 5, 2011 at 8:33 AM, Thomas wrote: > Dear all, > > I am trying to set up a list with 1:c objects each meant to capture the > coefficients for one coefficient and 100 replications. I receive the > following error message: > > Error in betaboot[[p]] : subscript out of bounds

[R] loop question

2011-04-05 Thread Thomas
Dear all, I am trying to set up a list with 1:c objects each meant to capture the coefficients for one coefficient and 100 replications. I receive the following error message: Error in betaboot[[p]] : subscript out of bounds. My code is below. Where is my mistake? Many thanks, Thomas

Re: [R] Loop question

2009-04-18 Thread Stavros Macrakis
On Fri, Apr 17, 2009 at 10:12 PM, Brendan Morse wrote: > ...I would like to automatically generate a series of matrices and > give them successive names. Here is what I thought at first: > > t1<-matrix(0, nrow=250, ncol=1) > > for(i in 1:10){ >        t1[i]<-rnorm(250) > } > > What I intended was

Re: [R] Loop question

2009-04-18 Thread Jun Shen
Brendan, Matrix is atomic. Once you define t1 in matrix, t1[1]=0 rather than the whole column. I would just convert t1 to a data frame, which is a special list, by adding t1<- data.frame(t1). Now t1[1] represents the whole column. Then you can use your loop to add more columns. Jun On Fri, Apr

Re: [R] Loop question

2009-04-17 Thread Ben Bolker
Brendan Morse wrote: > > Hi everyone, I am trying to accomplish a small task that is giving me > quite a headache. I would like to automatically generate a series of > matrices and give them successive names. Here is what I thought at > first: > > t1<-matrix(0, nrow=250, ncol=1) > > for

Re: [R] Loop question

2009-04-17 Thread Jorge Ivan Velez
Dear Brendan, One way could be either bigt <- sapply(1:10,function(x) rnorm(250)) colnames(bigt) <- paste('t',1:10,sep="") bigt or bigt2 <- NULL for(i in 1:10) bigt2 <- cbind( bigt2, rnorm(250) ) colnames(bigt2) <- paste('t',1:10,sep="") bigt2 or bigt3 <- matrix(rnorm(250*10),ncol=10) colnames

[R] Loop question

2009-04-17 Thread Brendan Morse
Hi everyone, I am trying to accomplish a small task that is giving me quite a headache. I would like to automatically generate a series of matrices and give them successive names. Here is what I thought at first: t1<-matrix(0, nrow=250, ncol=1) for(i in 1:10){ t1[i]<-rnorm(250) }