similar to: Xapply question

Dear list and Hadley, The new plyr package seems to provide a clean and consistent way to apply a function on several arguments. However, I don't understand why the following example does not work like the standard mapply, library(plyr) df <- data.frame(a=1:10 , b=1:10) foo1 <- function(a, b, cc=0, d=0){ a + b + cc + d } mdply(df, foo1, cc=1) # fine mdply(df, foo1, d=1) #

Hi All I have regression coefficients from an experiment and I want to plot them in lattice using panel curve but I have run into error messages. I want an 3 panel conditioned plot of 2 curves of Treatment 2 in each panel conditioned by Treatment1, the example curve expression is x+value*x^2 A rough toy example to give an idea of what I want is: Data: data = expand.grid(Treatment1 =

Hi, R users, Here is an example. k <- c(1,2,3,4,5) i <- c(0,1,3,2,1) if k=1, then i=0 if k=2, then i=0, 1 if k=3, then i=0, 1, 2, 3 if k=4, then i=0, 1, 2 if k=5, then i=0, 1 so i'd like to create a list like below. > list k i 1 1 0 2 2 0 3 2 1 4 3 0 5 3 1 6 3 2 7 3 3 8 4 0 9 4 1 10 4 2 11 5 0 12 5 1 I tried expand.grid, but I can't. Any suggestion will be

Dear R users, I have two datasets: id1 <- c(rep(1,10), rep(2,10), rep(3,10)) value1 <- sample(1:100, 30, replace=TRUE) dataset1 <- cbind(id1,value1) id2 <- c(1,2,3) subtract.value <- c(1,3,5) dataset2 <- cbind(id2, subtract.value) I want to subtract the number of rows in the subtract.value that corresponds to the id value in dataset1. So for the 1 in id1, I want to

Hi, I want to sample from a distribution (say a normal distribution, for example) using vectors of the different parameters (i.e. the mean and standard deviation). That is, I have a list/vector of say 100 means and another of the corresponding 100 SD's, and I want a matrix of 100 rows (one for each mean and SD pair) each having 1000 random samples. Something like: sample_matrix =

I think need to do something like this: dat<-data.frame(state=sample(id=rep(1:5,each=200),1:3, 1000, replace=T,prob=c(0.7,0.05,0.25)),V1=runif(1,10,1000),V2=rnorm(1000)) rle.dat<-rle(dat$state) temp<-1 out<-data.frame(id=1:length(rle.dat$length)) for(i in 1:length(rle.dat$length)){ temp2<-temp+rle.dat$length[[i]] out$V1[i]<-mean(dat$V1[temp:temp2])

Dear R Helpers, First, I apologize for asking for help on the first of my topics. I have been looking at the posts and pages for apply, tapply etc, and I know that the solution to this must be ridiculously easy, but I just can't seem to get my brain around it. If I want to produce a set of tables for all the variables in my data, how can I do that without having to type them into the table

I have a data frame called e, dim is 27,3, the first 5 lines look like this: V1 V2 V3 V4 1 1673 0.36 0.08 Smith 2 167 0.36 0.08 Allen 3 99 0.37 0.06 Allen 4 116 0.38 0.07 Allen 5 95 0.41 0.08 Allen I am trying to calculate the proportion/percentage of V1 which would have values >0.42 if V2 was the mean of a normal distribution with V1

Hello. This is a follow-up to a question I posted last week. With some previous suggestions from the R-help community, I have been able to plot survival (, hazard, and density) curves using published data for Siler hazard parameters from a number of ethnographic populations. Can the function below be modified, perhaps with a "for" statement, so that multiple curves (different line

Greetings all, I am curious to know if either of these two sets of code is more efficient? Example1: ## t-test ## colA <- temp [ , j ] colB <- temp [ , k ] ttr <- t.test ( colA, colB, var.equal=TRUE) tt_pvalue [ i ] <- ttr$p.value or Example2: tt_pvalue [ i ] <- t.test ( temp[ , j ], temp[ , k ], var.equal=TRUE) ------------- I have three loops, i, j, k. One to test the all of

Hello! I know that probably my question is rather simple but I' m a very beginner R-user. I have to numerically integrate the product of two function A(x) and B(x). The integretion limits are [X*; +inf] Function A(x) is a pdf function while B(x)=e*x is a linear function whose value is equal to 0 when the x < X* Moreover I have to iterate this process for different value of X* and for

Dear R helpers   (I have already written the required R code which is giving me correct results for a given single set of data. I just wish to wish to use it for multiple data.)   I have defined a function (as given below) which helps me calculate Macaulay Duration and Modified Duration and its working fine with given set of data.   My Code -   ## ONS - PPA   duration = function(par_value,

`outer` (and related functions like kronecker) require that their functional argument operate elementwise on arrays. This means for example that outer( 1:2, 3:4, list) or outer(1:2,3:4,function(a,b){1}) gives an error. Is there a version of `outer`/`kronecker`/etc. that takes arbitrary functions and does its own elementwise mapping? In the first example above, I'd expect the

Dear list, In a little numbers game, I've hit a performance snag and I'm not sure how to code this in C. The game is the following: how many 8-digit numbers have the sum of their digits equal to 17? The brute-force answer could be: maxi <- 9 # digits from 0 to 9 N <- 5 # 8 is too large test <- 17 # for example's sake sum(rowSums(do.call(expand.grid, c(list(1:maxi),

sorry, I forgot my sessionInfo: please see below. -------- Original Message -------- Subject: puzzled with plotmath Date: Thu, 20 Jan 2011 12:48:18 +0100 From: Claudia Beleites <cbeleites at units.it> To: R Help <r-help at r-project.org> Dear all, I'm puzzled with matrix indices in plotmath. I'm plotting matrix elements: Z [i, i], and I'd like to put that as label.

Hola a todos: Se me ha planteado un problema que no está ligado a ningún problema concreto. Es más teórico. Supongamos que tenemos tres variables: V1 <- c (47, 71, 41, 23, 83, 152, 82, 8, 160, 18) V2a <- c (NA, 36, 15, 5, 56, 18, NA, 5, NA, 5) V2b <- c (37, NA, 15, NA, NA, NA, 90, NA, 161, NA) Supongamos que tengo la expresión (que no puedo asignarlo a

Dear R users, Here is my problem: I have an array with at least four dimensions: > dim(myArray) [1] 20 17 3 6 I'd like to apply a function to each occurrence of the matrix (3x6) defined by the last two dimensions. This interpolation function always return a matrix of the same dimensions as its argument: > interpSecteurs.f(myArray[1, 1, , ]) secteur rotation 1 2 3

Dear all, I work with (vibrational) spectra: some kind of intensity (I) over frequency (nu), wavelength or the like. I want to do fourier transform for interpolation, smoothing, etc. My problem is that the spectra are often irregularly spaced in nu: the difference between 2 neighbouring nu varies across the spectrum, and data points may be missing. Searching for discrete fourier transform

---------- Weitergeleitete Nachricht ---------- Betreff: Re: [R] The end of Matlab Datum: Freitag 12 Dezember 2008 Von: Claudia Beleites <cbeleites at units.it> An: r-help at r-project.org Am Freitag 12 Dezember 2008 13:10:20 schrieb Patrick Burns: > How about: > > x[, -seq(to=ncol(x), length=n)] Doing it is not my problem. I just agree with Mike in that I would like if I could

A ver... con que xfunc() esté preparada para tomar un parámetro de tipo "carácter" y evaluarlo, claro que se puede hacer... Si el problema lo tienes en evaluar la expresión, la función "eval()" te lo hace. Si no te he entendido bien, explícate más ? Saludos Isidro -----Mensaje original----- De: R-help-es <r-help-es-bounces en r-project.org> En nombre de Griera Enviado