similar to: naming things in functions

Dear all, I have written a function for calculating the volume of a tree (=trad) or snag (=h?gst). The included volume regreesion model includes ten parameter values, which are tree species specific. bj?rk.formh?jd.pars is an object which includes the parameter values (parameter set) for birch (=bj?rk). There is one row per tree in the data object. > relev.kols[1:5,

Dear all, I know there have been various questions posted over the years about loops but I'm afraid that I'm still stuck. I am using Windows XP and R 2.9.2. I am generating some data using the multivariate normal distribution (within the 'mnormt' package). [The numerical values of sanad and covmat are not important.] > datamat <-

Inthe code below I was trying to to obtain the GMM estimates for CAPM (REGRESSION) for 36 stocks each have 180 observations,however it only gives me one output rather than 36. In SAS i would just put in a *By statement*. I have a variable TICKER that categorize them into 36 groups. *How can I obtain all 36 output instead of just one.* **

Dear R-helpers, I am using a vrtest on time series data. My commands are as follows; read.table("B.txt",sep="\t",fill=TRUE, na.strings = "NA") require(vrtest) rm(list=ls(all=TRUE)) datamat <- read.table("B.txt",sep="\t",fill=TRUE, na.strings = "NA") column <- 1 nob <- nrow(datamat) y <-

Hi, I got a computational problem with lme (nlme library R 1.9.1) using corSymm(). Here is the data: [,1] [,2] [,3] [,4] [,5] [,6] [1,] 0.19639793 0.09127954 0.11733288 0.07598273 0.06545106 0.06211532 [2,] 0.22773467 0.10981912 0.16052847 0.38101187 0.18353474 0.24072918 [3,] 0.46743388 0.45733836 0.32191178 0.43356107 0.39159746 0.53984221 [4,]

A multinomial logit model can be specified as a conditional logit model after restructuring the data. Doing so gives flexibility in imposing restrictions on the dependent variable. One application is to specify a loglinear model for square tables, e.g. quasi-symmetry or quasi-independence, as a multinomial logit model with covariates. Further details on this technique and examples with several

Hello, I am trying to simulate 10 relicates of 100-tables. Each table is a 2 x 3 and 80% pf the tables are true nulls and 20% are non-nulls. The nulls follow the Hardy Weinberg distribution (ratio) 1:2:1. I have the code below but the p-values are not what I am expecting. I want to use the Cochran Armitage trend test to get the p-values. num.reps=10 num.vars=1000 pi0 = 80 num.subjects = 100

Dear all, I am trying to make a barplot with clustered pairs of bars, using class=numeric data and the following command: barplot(c(bline_precip[10,9], bline_runoff[10,9], cccma_precip[10,9], cccma_runoff[10,9], csiro_precip[10,9], csiro_runoff[10,9], ipsl_precip[10,9], ipsl_runoff[10,9], mpi_precip[10,9], mpi_runoff[10,9], ncar_precip[10,9], ncar_runoff[10,9], ukmo_precip[10,9],

Hello, I want to draw a histogram of the mean of sample observations drawn from multivariate t distribution. I am getting the following error corresponding to the code I used. Though I am getting the graph, but I am curious to know the warning message. Warning messages: 1: In if (freq) x$counts else { : the condition has length > 1 and only the first element will be used 2: In if (!freq)

Folks, I have been using the VAR {vars} program to find a fit for the following bi-variate time series (subset): bivariateTS<-structure(c(0.950415958293559, 0.96077848972081, 0.964348957109053, 0.967852884998915, 0.967773510751625, 0.970342843688257, 0.97613937178359, 0.980118627997436, 0.987059493773907, 0.99536830931504, 1.00622672085718, 1.01198013845981, 1.01866618122606,

Dear All, I've successfully import my synteny data to R by using scan command. Below show my results. My major problem with my data is how am i going to combine the column names with the data( splt) where i have tried on cbind but a warning message occur. I have realized that the splt data only have 5 column instead of 6. Please help me with this!! I want my data to be a numerical

I am using read.csv to read a CSV file (produced by saving an Excel file as a CSV file). The columns containing dates are being read as factors. Because of this, I can not compute follow-up time, i.e. Followup<-postDate-preDate. I would appreciate any suggestion that would help me read the dates as dates and thus allow me to calculate follow-up time. Thanks John John Sorkin M.D., Ph.D. Chief,

The menu directory won't compile with gcc 3.4.0 because gcc complains that "ebp" cannot be used as a constraint. If preserving ebp is necessary then it will have to be done in the assembler code. Gcc 3.3.3 just seems to ignore the constraint. It didn't do anything special to preserve the register. I also noticed that getnumrows returns the contents of 0x484 which is the

Hi, I have a matrix of data values like the example bellow. I would like to extract a subset of the matrix for the values where the first column is negative. I am using the subset function. However, I am getting an error message that the conditional variable doe snot exist. For some reason, the subset operation only works if I transform the matrix to a data set using as.data.set(). The help

Hi, On R 2.10.1 for Windows, when I do the following to duplicate the structure of a large numeric matrix called matrix1: matrix2 <- matrix(0,nrow=nrow(matrix1),ncol=ncol(matrix1)) and then rownames(matrix2) <- rownames(matrix1) I get a "cannot allocate vector of size xxMb" error but if I instead do: rnames <- list() rnames <- rownames(matrix1) matrix2 <-

I just started using nice package "sets" and I wonder if there are utilities to convert (some) sets to data frame (as in the example below) > library(sets) > a <- gset(elements = list(e('A', 0.1), e('B', 0.8))) > lst <- as.list(a) > nr <- length(lst) > rnames <- character() > for (i in 1:nr) rnames[i] <- lst[[i]] > df <-

Dear all, I have used 'read.table' to create a data frame of 720 columns and 360 rows (and assigned this to 'Jan'). The row and column names are numeric: > columnnames <- sprintf("%.2f", seq(from = -179.75, to = 179.75, length = 720)). > rnames <- sprintf("%.2f", seq(from = -89.75, to = 89.75, length = 360)) > colnames(Jan) <- columnnames

Hi, any good trick to get the column names for title() aside from running lapply on the column indexes? Thanks Nick. apply(X[,numCols],2,function(x){ nunqs <- length(unique(x)) nnans <- sum(is.na(x)) info <- paste("uniques:",nunqs,"(",nunqs/n,")","NAs:",nnans,"(",nnans/n,")") hist(x,xlab=info) #

Hello, Here a small improvement for R. When you use the function write.table, if the disk is full for example, the function doesn't return an error and the file is written but truncated. It can be a source of mistakes because you can then copy the output file and think everything is ok. How to reproduce ------------------------- >> write.csv(1:10000000, 'path') You must have

Looks like I?m bumping a lot into unexpected behaviour lately, but I think I found a bug again, but don?t have access to Bugzilla: Write.table (from core-package utils) doesn?t handle nested data.frames well, the quote arguments only marks top-level character (or-factor columns) for quoting, so this fails: df <- data.frame(a='One;Two;Three',