similar to: Histogram in Lattice with 3 factors

Hi all, I am getting the following error message: > mymodel = glm(response ~ . , family=binomial, data=C); Error: cannot allocate vector of size 734.2 Mb In addition: Warning messages: 1: In array(0, c(n, n), list(levs, levs)) : Reached total allocation of 1535Mb: see help(memory.size) 2: In array(0, c(n, n), list(levs, levs)) : Reached total allocation of 1535Mb: see help(memory.size) 3:

Dear List, creating factors in a given non-default orders is notoriously difficult to explain in a course. Students love the ifelse construct given below most, but I remember some comment from Martin M?chler (?) that ifelse should be banned from courses. Any better idea? Not necessarily short, easy to remember is important. Dieter data = c(1,7,10,50,70) levs =

I sent this to the list last week, and haven't seen it pop up. Either I deleted it when it did appear, or possibly it was destroyed as spam...? If it did appear and I somehow missed it, appologies. In a nutshell, can the function FUN supplied to by() deduce what level of factor by() was on when FUN was called? I've been digging through the functions, and can't see where the

Lattice experts: Can you think of a way to produce a levelplot as below and then add a histogram of the z variable to the top margin of the plot that would sit on top of the color key? x <- seq(pi/4, 5 * pi, length.out = 100) y <- seq(pi/4, 5 * pi, length.out = 100) r <- as.vector(sqrt(outer(x^2, y^2, "+"))) grid <- expand.grid(x=x, y=y) grid$z <- cos(r^2) *

I am new to using S4 methods and have run into this problem (on Windows XP using R 2.4.1): I am writing a package in which I use the graph package. I define my own classes of graphs as: setOldClass("graphsh") setOldClass("ugsh") setIs("ugsh", "graphsh") (I know that I "should have" used setClass instead - and I will eventually - but right now

Given the example: R> (levs <- quantile(volcano,c(0,0.1,0.5,0.9,0.99,1))) 0% 10% 50% 90% 99% 100% 94 100 124 170 189 195 R> levelplot(volcano,at=levs) How can I make the key categorical with the size of the divisions equally spaced in the key? E.g., five equal size rectangles with labels at levs c(100,124,170,189,195)? Apologies if this is obvious. -A R> version

For ordered factor the natural contrast coding would be to parametrize by the succsessive differences between levels, which does not assume equal spacing of factor levels as does the polynomial contrasts (implicitly at least). This requires the contr.cum, which could be: contr.cum <- function (n, contrasts = TRUE) { if (is.numeric(n) && length(n) == 1) levs <- 1:n

I am trying once again. By just changing f <- match(xlevs[f], nlevs) to f <- match(xlevs, nlevs)[f] , function 'factor' in R devel could be made more consistent and back-compatible. Why not picking it? -------------------------------------------- On Sat, 25/11/17, Suharto Anggono Suharto Anggono <suharto_anggono at yahoo.com> wrote: Subject: Re: [Rd] Function

In R devel, function 'factor' has been changed, allowing and merging duplicated 'labels'. Issue 1: Handling of specified 'labels' without duplicates is slower than before. Example: x <- rep(1:26, 40000) system.time(factor(x, levels=1:26, labels=letters)) Function 'factor' is already rather slow because of conversion to character. Please don't add slowdown.

Hi I think I have been struggling to use replace correctly, I usually work my way around this using a loop, but I think this is in fact inefficient. I have a dataset with runoff from three plots and associated rainfall. However either the datarecording was sloppy, or the rainfall very patchy. So I am trying to remove data from my dataset for which the runoff is larger than the rainfall on the

Hi, the following R lines work fine in R 2.4.0, but not in R 2.4.1 or any devel versions of R 2.5.0 (see below for details). library(drc) # to load the dataset 'PestSci' library(nlme) ## Setting starting values sv <- c(0.43355869, 2.49963220, 0.05861799, 1.73290589, 0.38153146, 0.24316978) ## No error m1 <- nlme(SLOPE ~ c + (d-c)/(1+exp(b*(log(DOSE)-log(e)))), fixed =

Hi Dimitris, thank you for your reply, > does the following work in your data: > > levs <- unique.default(Names) > factor(Names, levs[order(unique.default(Weights))]) your solution is really shorter, but has two issues to be meant here: 1. "unique.default" is applied twice, what might be a bit expensive for strings. 2. your solution brings an implicit

Dear CSAG R users, I will be glad if someone can point out what I am doing wrong or not doing at all in this. I am trying to write out netcdf file in R. I have 26 time step but only the first time step is written. For example: >library(ncdf) >path <- '/home/work/' >forecast <- open.ncdf(paste(path,'cam.1980.2005.nc',sep="")) > fore <-

Dear CSAG R users, I will be glad if someone can point out what I am doing wrong or not doing at all in this. I am trying to write out netcdf file in R. I have 26 time step but only the first time step is written. For example: >library(ncdf) >path <- '/home/work/' >forecast <- open.ncdf(paste(path,'cam.1980.2005.nc',sep="")) > fore <-

With 2.7.0 patched (not tested with 2.0.0), I get an error message in a program that ran correctly in R 2.6.2 when the grouping factor of a stripplot contains an Umlaut: I am aware that there are a few locale-changes in R 2.7.0, but I could not easily locate who's at fault Dieter library(lattice) dt = data.frame(x=rnorm(100),y=1:100,levs= as.factor(c("Gru","Gr?")))

Dear R-devel list members, I'd like to suggest a more flexible procedure for generating contrast names. I apologise for a relatively long message -- I want my proposal to be clear. I've never liked the current approach. For example, the names generated by contr.treatment paste factor to level names with no separation between the two; contr.sum simply numbers contrasts (I recall an

Dear all, I am struggling to understand this. What happens when you raise a negative value to a power and the result is a very large number? B [1] 47.73092 > -51^B [1] -3.190824e+81 # seems fine # now this: > x <- seq(-51,-49,length=100) > x^B [1] NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN <snip> > is.numeric(x^B) [1] TRUE > is.real(x^B) [1]

Dear All, I encountered the following problem: fun1 <- function(x){ 2*cos(x) } >library(debug) > mtrace(fun1) > fun1(0.1) Error in all.levs[[j]] : subscript out of bounds > sessionInfo() R version 2.5.0 (2007-04-23) i386-pc-mingw32 locale: LC_COLLATE=Polish_Poland.1250;LC_CTYPE=Polish_Poland.1250; LC_MONETARY=Polish_Poland.1250;LC_NUMERIC=C;LC_TIME=Polish_Poland.1250 attached

*Dear **Madam**/**Sir,* *What would happen if we irrigate the degraded land using the desalinated saline water in Minqin Oasis, Gansu Province, China? Can we prevent the disappearing of Minqin Qasis?* * * *Background:* *1. **The Minqin Oasis is surrounded by two deserts, Badain Jaran Desert and Tengger Desert. * *2. **Water is the root of ecological

Dear R users, It is a pleasure for me to announce the availability of the new package kwb.hantush (0.2.1)? on CRAN. Its objective is the calculation of groundwater mounding beneath an (stormwater) infiltration basin by solving the Hantush (1967) equation. For checking the correct implementation of the algorithm the R modelling results were cross-checked against alternative models assessed in