similar to: access sublists by a variable

Hello: I need to take a species-sample matrix and transpose it to the format used by PC-ORD for analysis. Unfortunately, the number of species is very large (>5000), and so this operation cannot be performed simply in an application like Excel, which has a 255 column limit. So, I wrote relatively simple code in R that I hoped would do this (appended below). But there are glitches. The

Since upgrading to rails 1.0 I keep getting a parse error in database.yml. I have tried multiple configurations of both mysql and postgres and continue getting this error message when trying to generate a model or scaffold. -Andrew

Dear all, I would like to ask your help concering two R lists. If I did everything should have the same structure (that means the same number of sublists, and their sublists also the same number of sublists). What would change between the two lists is the contents of each element in the lists. Could you please help me understand how I can do that in R? I would like to thank you in advance for

Hi, I've installed samba 4.5.3 as a domain controller and provisioned a domain. Now I'm integrating the win7 machines as clients. If the clients are in the network, all is ok. But if I use a notebook in a different network (for example at home) the win7 logon takes a very long time up to 2-3 minutes. If I disconnect all network connections, the login is as fast as in the domain network.

Hi there, I am wondering if it is possible to do an xyplot with two y-axes. I'd like to print two parameters in a time series but they both have different scales. Which parameter in xyplot can I add to achieve this result? Thanks a lot for any help. Antje [[alternative HTML version deleted]]

Hi all, I'd like to know, if I can solve this with a shorter command: a <- rnorm(100) which(a > -0.5 & a < 0.5) # would give me all indices of numbers greater than -0.5 and smaller than +0.5 I have something similar with a dataframe and it produces sometimes quite long commands... I'd like to have something like: which(within.interval(a, -0.5, 0.5)) Is there anything I

Hello, I tried to fit a poisson distribution but looking at the function fitdistr() it does not optimize lambda but simply estimates the mean of the data and returns it as lambda. I'm a bit confused because I was expecting an optimization of this parameter to gain a good fit... If I would use mle() of stats4 package or mle2() of bbmle package, I would have to write the function by myself

Hello, I'm looking for a solution for the following problem: I have two vectors V1 <- c("apple","honey","milk","bread","butter") V2 <- c("bread","milk") now, I would like to know for each element in V1 if it's equal to one of the elements in V2 I could do: which(V1 == V2[1] | V1 == V2[2]) but what if I

Hello, I have a question how to reshape a given matrix to a data frame. # ---------------------------------- > a <- matrix(1:25, nrow=5) > a [,1] [,2] [,3] [,4] [,5] [1,] 1 6 11 16 21 [2,] 2 7 12 17 22 [3,] 3 8 13 18 23 [4,] 4 9 14 19 24 [5,] 5 10 15 20 25 > colnames(a) <- LETTERS[1:5] > rownames(a) <-

Hello, I'm not very experienced with lattice and I was wondering whether I get get some hints from you how to create a pure heatmap (using levelplot), without any axis, title, legend, margin at all... I just want to see the coloured squares, nothing else. Any suggestions? Antje

Hello, I hope this question is not too stupid. I would like to know how to update levels after subsetting data from a data.frame. df <- data.frame(factor(c("a","a","c","b","b")), c(4,5,6,7,8), c(9,1,2,3,4)) names(df) <- c("X1","X2","X3") my.sub <- subset(df, X1 == "a" | X1 == "b")

Hi there, as I'm not sure to understand the coloring levelplot uses, I'm looking for another easy way to create a heatmap like this: library(lattice) mat <- matrix(seq(1,5, length.out = 12), nrow = 3) mat[1,2] <- 3.5 my.at <- seq(0.5,5.5, length.out = 6) my.col.regions <- rainbow(5) graph <- levelplot(t(mat[nrow(mat):1, ] ), at = my.at, col.regions = my.col.regions)

Hi there, assume that you have data with different sampling like d1 <- rnorm(100) d2 <- rnorm(150) now, I'd like to create two boxplots in one graph but each plot located at the sampling number at the x-axis. This, I can do with "at" l <- list(d1,d2) boxplot(l, at=c(length(d1), length(d2)), xlim=c(0,200) ) but both plots are very thin and I'd like to increase the

Hei R Users, i have the following dataframe: Datetime Temperature and many more collumns 1 2008-6-1 00:00:00 5 2 2008-6-1 02:00:00 5 3 2008-6-1 03:00:00 6 4 2008-6-1 03:00:00 0 5 2008-6-1 04:00:00 6 6 2008-6-1 04:00:00 0 7 2008-6-1 05:00:00 7 8 2008-6-1 06:00:00

Hi there, I though there has been a possibility to force the output on the console with one element per line. Instead of this: > 1:10 [1] 1 2 3 4 5 6 7 8 9 10 something like this > 1:10 [1] 1 [2] 2 [3] 3 [4] 4 [5] 5 [6] 6 [7] 7 [8] 8 [9] 9 [10] 10 Can anybody help? Antje

Hi folks, I've just discovered that the following code leads to boxplot (surprisingly to me). Can anybody explain to me why? Is this documented somewhere? I've never consider this option before. x <- rnorm(300) l <- c(rep("label1",100), rep("label2",50), rep("label3",150)) df <- data.frame(as.factor(l), x) plot(df) Thank you! Antje

Hi there, I have a question, which I thought is very easy to solve, but somehow I can't find a solution. Probably someone could help me quickly? Here it is: I have two matrices: a [,1] [,2] [,3] [1,] 1 4 9 [2,] 2 6 10 [3,] 3 6 11 [4,] 4 8 12 b [,1] [,2] [1,] 1 4 [2,] 2 5 [3,] 3 6 Now I want to find

Hi there, I try to rewrite some Java-code with R. It deals with reading XML files. I started with the XML package. In Java, I had a very useful method which gave me a node by using: name of the node index of appearance start point: global (false) / local (true) So, I could do something like this. setCurrentChildNode("data", 0); getValueOfElement("val",1,true); -->

Hi there, I hope this question is not as stupid as the one before ... I tried to shorten my histogram (because the distribution is quite skewed and I simply don't want to see the long tail but still use the histogram plot). How can I do something like this? (The example does not work but I don't know why...) data <- rnorm(100) # as example, of course this is not skewed... h <-

Hello, I'm running R with windows. Could anybody help me how to install the package biOps (http://cran.mirroring.de/src/contrib/Descriptions/biOps.html) ? I cannot find it at any mirror provided by the GUI; I just found it as "tar.gz" which cannot be installed with the installation method... Maybe, I simply do something wrong? Thanks for any hint! Antje