similar to: tapply, levelinformation

Hi all, I have written a R-script under Windows using choose.dir. Now, I have seen that this function is missing at MacOS. Does anybody know an alternative? Antje

Hello, I have some problems with sapply. I wanted to do the following: s <- sapply(filelist, function(x) { if(file.exists(x)) { ... return( list(density(file$V1)$x, density(file$V1)$y)) } else { print(paste("plotDensity ERROR - File does not exist: ",x,sep="")) return( NULL) } }) if(is.null( - ??? - )) { print("no plot") } That

Hello, I'm looking for a solution for the following problem: 1) I have a folder with several csv files; each contains a set of measurement values 2) The measurements of each file belong to a position in a two dimensional matrix (lets say "B02.csv" belongs to position 2,2 3) The size of the matrix is fix 4) I cannot assure to have a csv file for each position 5) Each position

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dear members, i would like to write a variable in a plot title (main="") but i don't know the right syntax:(...i tried a lot of different ways without success. here my example: y=30 z=33 for (i in 10:length(tissue)) { png(filename = tissues[i], width = 1024, height = 768, pointsize = 12, bg = "white") gene.graph("ENSG00000115252", rma.affy, gps=list(1:3,

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

Hallo, I have a question to plotDensity and do not understand what stand behind. In a density plot the x-axis is the signal intensity but for what stands than density on the y-axis? Here I have the values 0.00-0.30 Can anyone discribe it by his own words? I do not understand the help. Thanks, Conny --

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

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")

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

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

Hello, I was wondering if I can plot two curves I get from "density(data)" into one plot. I want to compare both. With the following commad, I just get one curve plotted: plot( density(mydata) ) Sorry for this stupid question but I could not find a solution until now... Antje

Hello, I tried to use mle to fit a distribution(zero-inflated negbin for count data). My call is very simple: mle(ll) ll() takes the three parameters, I'd like to be estimated (size, mu and prob). But within the ll() function I have to judge if the current parameter-set gives a nice fit or not. So I have to apply them to observation data. But how does the method know about my observed

Hi there, I have a question concerning the behaviour of the colouring with levelplot. (I hope, I manage to explain) If I give the parameters "at" and "col.regions" like this: at <- c(1,2,3,4,5,6) col.regions <- c("blue","blue","blue","yellow","yellow","yellow") Which color would have the value 3.5? I would

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, unfortunately, I don't know a better subject. I would like to be very flexible in how to process my data. Assume the following dataset: par1 <- seq(0,1,length.out = 100) par2 <- seq(1,100) fac1 <- factor(rep(c("group1", "group2"), each = 50)) fac2 <- factor(rep(c("group3", "group4", "group5", "group6"), each =

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

A few day ago, I was looking for an answer to my question but didn't get one. Anybody who can help now? Hello, I tried to use mle to fit a distribution(zero-inflated negbin for count data). My call is very simple: mle(ll) ll() takes the three parameters, I'd like to be estimated (size, mu and prob). But within the ll() function I have to judge if the current parameter-set gives a nice

Hello, I guess the solution is rather simple but whatever I tried, I don't manage to get the result as I want to have it: I have several vectors of equal length in a list and I'd like to combine all first elements to a single string, all second elements to a single string, ..., all n-th elements to a single string. # Example code (how it should look like): t1 <- c(1,2,3) t2 <-