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Hello, I have the following loop for two data sets: diveData_2008 and diveData_2009. It uses two other data: diveCond_all and fishTable. The problem is at the point to identify the dive_id for the given index (index is timestamp). It keeps on saying for the1st loop Error in fishReport$dive_id[i] <- dive_id : replacement has length zero for the 2nd loop Error in fishReport$dive_id[i + j] <-

I was wondering why I get <NA> instead of the timestamp in the following. Thanks. > dataDir <- file.path(wd) > localRaw <- read.csv(file.path(dataDir,"LOCAL.csv"), as.is=T,stringsAsFactors = FALSE) > localRaw[1:2,] Year Month Day hour minute second Temp1mab Temp7mab Temp14mab Salinity1mab 1 2009 10 5 0 0 0 11.288 13.675 13.743 33.513

If I use latex(summary(X)) where X is a data frame with four variables I get something like Rainfall Education Popden Nonwhite Min. :10.00 Min. : 9.00 Min. :1441 Min. : 0.80 1st Qu.:32.75 1st Qu.:10.40 1st Qu.:3104 1st Qu.: 4.95 Median :38.00 Median :11.05 Median :3567 Median :10.40 Mean :37.37 Mean :10.97 Mean :3866

#this is my stab at - I am sure that I am missing something. If this doesn't make sense then please ask for more details. #This may show my low level of programing knowledge hester. <- c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4) value <- rnorm(16) x <- data.frame(value, hester.) z <- (if(x[,"hester."]==1){ x[,"value"]*6.250} else if(x[,"hester."]==2){

What is equivalent to DROP or KEEP statements of SAS in R? -- This message was sent on behalf of sarjinder at yahoo.com at openSubscriber.com http://www.opensubscriber.com/messages/r-help at r-project.org/topic.html

I have a file, each column of which is a separate year, and each row of each column is mean precipitation for that month. Looks like this (except it goes back to 1964). month X2000 X2001 X2002 X2003 X2004 X2005 X2006 X2007 X2008 X2009 1 1.600 1.010 4.320 2.110 0.925 3.275 3.460 0.675 1.315 2.920 2 2.960 3.905 3.230 2.380 2.720 1.880 2.430 1.380

I have found Sweave() to be great for producing PDF documents. I have been experimenting with RweaveHTML (from the R2HTML) package and have had moderate success. My main issue has been that I simply want the R output to be shown verbatim in the HTML document but RweaveHTML tends to convert most output to a table, for example. So, is there a way to force the RweaveHTML driver to simply provide

Dear All: In the attached file, I have 3 group patients, and there are 5 in each group (the groups are decided by the prefix of the idno). I want draw a repeat measurement comparison figure. My goal is to list 5 patients from same group on one horizontal line. But xyplot sounds pick them randomly (or I was confused?). Could you please help me modify the following code to accomplish this?

Just wanted to start a discussion on whether R could ship with more appropriate GC parameters. Right now, loading the recommended package Matrix leads to: > library(Matrix) > gc() used (Mb) gc trigger (Mb) max used (Mb) Ncells 1076796 57.6 1368491 73.1 1198505 64.1 Vcells 1671329 12.8 2685683 20.5 1932418 14.8 Results may vary, but here R needed 64MB of N cells and 15MB

When I attach data frames I often want to be able to refer to the whole data frame rather then one of its components. For example: attach (my.data.frame) summary(my.data.frame) That's fine but often the frame has a very long name so I'd prefer some shorthand way of referring to it by its position on the search list. This applies especially to cases where I have a nested data frame

We would like to load data from Statistics Canada (http://www.statcan.gc.ca/) using R, for example, Employment and unemployment rates. It seems to me that the tables are displayed in HTML. I was wondering if you know how to load these tables. Thanks, -james

Dear R users, I am using lmer combined with AIC model selection and averaging (in the MuMIn package) to try and assess how isotope values (which indicate diet) vary within a population of animals. I have multiple measures from individuals (variable 'Tattoo') and multiple individuals within social groups within 4 locations (A, B, C ,D) crucially I am interested if there are

Martin Morgan discussed this a year or so ago and as I recall bumped up these values to the current defaults. I don't recall details about why we didn't go higher -- maybe Martin does. I suspect the main concern would be with small memory machines in student labs and less developed countries. If there was a way on all platforms to identify how much memory is available that might help to

Dear All, I'm not sure of the interpretation of interactions with contrasts. Can anyone help? I do an ANCOVA, dryweight is covariate, block and treatment are factors, c4 the response variable. model<-aov(log(c4+1)~dryweight+treatment+block+treatment:block) summary(model); Df Sum Sq Mean Sq F value Pr(>F) dryweight 1 3.947 3.947 6.6268 0.01076 *

Dear All, I was trying to follow some of the examples in Venables and Ripley "Modern applied ... with S-plus" I have downloaded a copy of the iris data set and loaded into R. : however I cannot use the apply command (from p47): > apply (iris, 2 ,mean) Error in sum(..., na.rm = na.rm) : invalid "mode" of argument > apply (iris, c(2) ,mean) Error in sum(..., na.rm =

I created a tex file following the example in the Sweave help which produced the following files in my working directory. Sweave-test-1-006.eps Sweave-test-1-006.pdf Sweave-test-1-007.eps Sweave-test-1-007.pdf Sweave-test-1.tex When I run latex on this, I get a latex error, log file below. I am running R 2.1.1 on Windows XP. I have installed "small MiKTeX" and I have added

I use by() to generate a summary statistics like so: Lbys <- by(dat[Nidx], dat$LipTest, summary) where Nidx is an index vector with names picking out the columns in the data frame dat. This returns a list of character arrays (see below for str() output) where the columns are named correctly but the rownames are empty strings and the values are strings prepended with the summary

I was wondering if somebody could explain why I get different results here: >treats[,2]<-as.factor(treats[,2]) >treats[,5]<-as.factor(treats[,5]) >treats[,4]<-as.factor(treats[,4]) #there are 'c' on more days than I have 'h2o2', where treats[,4] is the day. I only want 'c' that correspond to the same days that I have a 'h2o2' also.