similar to: Time Series of Histograms

Hello, I have a multi-year dataset (see below) with date, a data value and a flag for the data value. I want to find the monthly median for each month in this dataset and then plot it. If anyone has suggestions they would be greatly apperciated. It should be noted that there are some dates with no values and they should be removed. Thanks Emily > print ( str(data$flow$daily) )

I am running this small program: x <- factor(c("A","B","A","C")) y <- c(1,2,3,4) w <-data.frame(x,y) if (w$x=="A"){ w$z=1 } w And I obtain: x y z 1 A 1 1 2 B 2 1 3 A 3 1 4 C 4 1 And not x y z 1 A 1 1 2 B 2 NA 3 A 3 1 4 C 4 NA Like I should obtain. What am I doing wrong? Please notice that I get a warning approximately saying -

Hi Im working on my master thesis, I need to get a plot currents distributed over a surface. I have the data listed in a dataframe with the x coordinates, y coordinates and the value. Im using the persp function but I’ve not found out how to put the data into the z1 variable es <-read.table("RDatalog_pass_dev_1_wafers_1_8V.txt",header=T,sep="\t") x1 <-

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I am trying to make horizontal violin plots. I have tried both vioplot and simple.violinplot, but both of them seem to not be willing to take the horizontal option. Is this correct, or am I just bungling it somehow? For instance, for vioplot (from the example shown, with the horizontal modification): > vioplot(bimodal,uniform,normal, horizontal=TRUE) Error in median(data) : need numeric data

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Dear R Users, I would like to know is there any package to create a plot like this? http://dl.dropbox.com/u/5409929/cs1160521f01.gif X axis is categorical. And the positions of the points are corresponding to the frequency. (similar to violinplot) Thank you. Regards, CH -- CH Chan

Dear all, I have a function f(x) which return a list as result. $T1 [1] 0.03376190 $T2 [1] 0.04725 $T3 [1] 0.3796071 $T4 [1] 0.3713452 $T5 [1] 0.4523651 $T6 [1] 0.4575873 I now find the result for a vector of x values at one time. I want to store the reuslt for each xi value in a column of a matrix x <- seq(0,1, by=0.1) result <- matrix(0, nrow=6, ncol=length(x)) for (i in

Hello. I am creating an R package that I'd like to submit to CRAN (OS Windows XP). How do I distinguish among 'public' functions, e.g., those that are intended to be called by users of the package and for which I am providing documentation & examples, and 'private' functions, which are used internally by the 'public' functions, but for which I do not wish to

Hello, I have build a syntax to find out if a given substring is included in a larger string that works like this: d1$V1[regexpr("some text = 9",d1$V2)>0] <- 9 and this works all right till "some text" contains standard ASCII set. However, it does not work when accents are included as the following: d1$V1[regexpr("some t?xt = 9",d1$V2)>0] <- 9 I have

In The American Statistician vol 42 (1988) pages 257 - 280, Yoav Benjamini investigates some variations on the box plot, including vaseplots, which maek the width of each box vary proportionally to he estimated density at a particular point. Has anyone implemented these in R ? Thanks as always Peter

Hello, I am using the function simple.violinplot from the package UsingR. I have some outliers in my dataset so that the distribution has very long tails. As a result, the y-axis of the output of simple.violinplot extends to very large values. I would like to zoom on the y-axis with a command such as ylim=c(a,b), as in boxplot(x,ylim=c(a,b)). However, doing simple.violinplot(x,ylim=c(a,b)) does

Hello R-helpers, I would like to calculate least square means after having built a GLM with quasipoisson errors. In my model the dependent variable is continuous, I have one continuous independent variable and one categorical independent variable (that is the variable for which I would like to calculate the least square means). I've looked around for the command to calculate the least