similar to: rnorm using vector of means and SDs

I have the following xts objetct "temp" > str(temp) An ?xts? object from 2010-12-26 to 2011-03-05 containing: Data: num [1:70, 1] 2.95 0.852 -0.139 1.347 2.485 ... - attr(*, "dimnames")=List of 2 ..$ : NULL ..$ : chr "t_n" Indexed by objects of class: [POSIXct,POSIXt] TZ: GMT xts Attributes: NULL > temp t_n 2010-12-26

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, I have a very large data set stored as an xts object. xts is very nice about showing row labels as "human readable" dates and times. I want the actual epoch values that are stored internally. The only way I can find to access them is one-at-a-time using the internal function: xcoredata() Calling this in an entire column, the "R" way doesn't work. It will only

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 =

Dear list, I'm stuck with looking for a function of the *apply family, which I suppose exists already ? just I can't find it: What I'm looking for is somewhere between sweep and mapply that does a calculation vectorized over a matrix and a vector: It should work complementary to sweep: for each row of the matrix, a different value of the vector should be handed over. Close to

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

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,

Hi Folks, I am dealing with data which have been presented as at each x_i, mean m_i of the y-values at x_i, sd s_i of the y-values at x_i number n_i of the y-values at x_i and I want to linearly regress y on x. There does not seem to be an option to 'lm' which can deal with such data directly, though the regression problem could be algebraically

Dear R-community, I'm struggling with a paper that reports only fragmented results of a 2by2by3 experimental design. However, Means and SDs for all cells are given. Does anyone know a package/function that helps computing an ANOVA with only Means and SDs as input (resulting in some sort of effect size and significance test)? The reason why I'm interested is simple: I'm conducting a

Hi R users This looks a simple question Is there any difference between between rnorm(1000,0,1) and running rnorm(500,0,1) twice in terms of outcome ? TM

First: R core: (Thank You!)^HUGE_VAL Now, down to business: In a loop or in repeated command lines: system.time(lm(rnorm(1000)~rnorm(1000))) ( or lm(rnorm(1000)~rnorm(1000))$coef ) fails after several iterations with the Windows message 'This program has performed an illegal operation and will be shut down. If the problem persists, please contact the vendor'. clicking on DETAILS

I've been trying to generate a series of random normal distributions using the rnorm function. After I generated the first set: set1<-rnorm(150, 505.5, 15.2) I checked the mean and standard deviation of set1. I was surprised to find the mean at 502.37 (rounded) and the standard deviation at 13.3 (again, rounded). While I know that it is unrealistic to expect the distribution of the

Thanks to all who replied to my query about the "normality" of rnorm. I am using R 1.1.1 (the version released last week) under Windows 98. It has been unusually hot here and I noticed that the CPU fan seemed to be running at high speed for a long time. The CPU probe that I use to monitor the temperature emitted a warning shortly after I sent off the email. After shutting my