similar to: program avail. for simulating spatial patterns?

Hi I have the following code which I would like to simplify. Id does linear regressions and returns the r-squares, and the coefficients. It runs slow, as it is doing the regressions for each - is it possible to get the values in a dataframe which looks as follow: expert | xx | seeds | r.squared | slope | intercept Thanks in advance, Rainer library(reshape) rsqs <- as.data.frame(

If you tell me how to update R itself automatically, I will go for your advice. I am not aware of any method to do it... Bye Rick "ONKELINX, Thierry"

Hello My colleagues and I have recently established a large database (40 tables each with greater than 15 variables) in Microsoft's SQL Server 2000. Currently we are accessing this database via SQL client running an Windows XP. Our objectives are many fold including running SQL applications, outputting results to ARC/INFO IMS, production of summarizing tables - graphs and web interfaces for

Dear Happy R-users & experts, I am in need of advice, While working with spatial data (x & y coordinates of seed locations) I have come accross the problem that I need to convert my point data into a matrix or grid system. I then need to count how often a point falls into a certain position in the matrix or grid. I have searched all day online, asked collegeas but nothing works. Sadly

Hello, I use newly R! I'd like to plot several data set together in one output window! How can I do that? Best regards Hadi --------------------------------- Never miss an email again! [[alternative HTML version deleted]]

I am confused about why the following occurs: > version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 4.0

I want to make a plot, but instead of showing _numerical_ values, I would like to show _symbolic_ values. For example, I want to plot a function y = a x + b, where x varies between Xmin and Xmax. I would like the plot to show, in the x-axis, the strings Xmin and Xmax, instead of their numeric values. Is it possible? Alberto Monteiro

Hi R-users I'm a beginner with R and statistics, so I need some help to start my data analysis. I've been reading some docs and tutorials on R and cluster analysis. I've got a large dataset (102000 points) with values of longitude, latitude and temperature and want to see if I can find groups (clusters). Following some tutorials I can look for principal components but get an error

Hello all-- I've been using 1.13 (v1915) on my Mac pretty intensly for about a month, but today when I tried to generate a plot I recieved this message: Error in quartz(width=9) : too many open devices I was caught by surprise, because I only had the R console open at the time. I checked to be sure, by clicking "Window" at the top bar, and R console was the only window open. I

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I am trying to use apply and a list to supply names to a set of tables I want to generate. Below is an example that I hope mimics the larger original problem. EXAMPLE aa <- c( 2,2,1,1,2) bb <- c(5,6,6,7,4) aan <- c("yes", "no") bbn <- c("a", "b", "c", "d") mynames <- c("abby", "billy") mylist <-

Hi Sorry if this is a wrong post in the forum. Please suggest if this is a correct forum to discuss R related basic problem. I wanted to perform the following task by using R: e.g. input data.frame x y ============ a 10 b 20 a 10 a 10 b 15 b 15 b 20 In o/p i need x

Hi, I'm interested in using apply() with t.test() on a data.frame. Specifically, I'd like to use apply() to do the following: t.test(raw.sample[1,alive],raw.sample[1,dead]) t.test(raw.sample[2,alive],raw.sample[2,dead]) t.test(raw.sample[3,alive],raw.sample[3,dead]) etc. I tried the following, apply(raw.sample,1,function(x) t.test(raw.sample[,alive],raw.sample[,dead])) but it gives

Why will the following command not work sapply(objects(),dim) What does it say about the objects list? What does it say about the dim command? Likewise, the following also does not work all<-ls() for (f in all) print(dim(f)) -- Farrel Buchinsky [[alternative HTML version deleted]]

I have these two dataframes in which 'id' is the key field > tabella id nome 1 1 PIEMONTE 2 2 VALLED'AOSTA 3 3 LOMBARDIA 4 4 TRENTINO 5 5 VENETO 6 6 FRIULI AND > tab id nome 1 1 PIEMONTE 2 2 VALLED'AOSTA 3 3 LOMBARDIA 4 4 TRENTINO 5 25 CAMPANIA 6 28 LAZIO Is there any

Hi R, I have a matrix A, A= [,1] [,2] [1,] a u [2,] b v [3,] c x [4,] d x [5,] e x I want to put the 2nd column of this matrix in a vector without duplicates. i.e., my vector v should be (u, v, x), whose length is 3. Can anybody help me on this? Thanks in advance Shubha. [[alternative HTML version deleted]]

Hi I have a vector x of length n. I am interested in x[1] being different from the other observations (ie x[-1]). My null hypothesis is that x[1] is drawn from a Gaussian distribution of the same mean as observations x[-1], which are assumed to be iid Gaussian. The (unknown) variance of x[1] is assumed to be the same as the variance of x[-1]. This should be an unpaired t-test. But

Dear R gurus, I would like to use the ccf function on two matrices that are each 196000 x 12. Ideally, I want to be able to go row by row for the two matrices using apply for the ccf function and get one 196000 X 1 array output. The apply function though wants only one array, no? Basically, is there a way to use apply when there are two arrays in order to do something like correlation on a row

Dear useRs, I have a few hundred plots that I'd like to export to one document. pdf() isn't an option, because the file created is prohibitively huge (due to scatter plots with many points). So I have to use png() instead, but then I end up with a lot of files (would prefer just one). 1. Is there a way to have pdf() embed images, instead of vector instructions? (What would have to be

Dear people, I've a problem in doing a power calculation. In Fryer and Nicholson (1993), ICES J. mar. Sci. 50: 161-168 page 164 an example is given with the following characteristics T=5, points in time R=5, replicates Var.within=0.1 q=10, a 10% increase per year The degrees of freedom for the test are calculated as Vl=T*R-2=23 and the non-centrality parameter Dl=4.54. Using this they get a