similar to: Feed list of vectors to vioplot()?

Hi, I have some data (lots of year,distance pairs), which I can straightforwardly boxplot: dists <- read.table("movedists.dat") with(dists,plot(as.factor(V1),V2)) If I want to plot these data as violin plots using vioplot, this approach doesn't work: > with(dists,vioplot(as.factor(V1),V2)) Error in Summary.factor(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, : min not

I have been using the package vioplot to make boxplot/densityplots. Now I am looking for a variation of this, and I was wondering if someone could give me any tips on how to do what I want. I have nine groups of values, each group containing two sets of numbers which show the difference between two values, such as this: Start dev Stop dev -1 10 5 -2 0

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

Dear all I am trying to make a graphic with the "vioplot" package. I use the following code: library(vioplot) x1 <- GSMrxDL x2 <- WIFI x3 <- UMT vioplot(x1, x2, x3, ylim=c(0, 10), names=c("GSMrxDL", "WIFI", "UMT"), col="gold") title("NIS Strahlung", xlab="Sender", ylab="V/m") Now I want to scale the y-axis

Hi I am using violin plots (type of boxplots) and  I am trying to increase the font size in the plots. It looks like that the violin plots do not work as "normal" plots as the cex parameters are ignored. You can have a loot at the code below require('vioplot') data1<-rnorm(100) data2<-rnorm(10) data3<-rnorm(1000) vioplot(data1,data2,data3,cex.lab=2,cex.axis=2)

Hi,? I have made violin plot with both ggplot2 and vioplot package with same data.? The results and code are as follows.? ################### ?Loading data? data.melt <- dget("https://gubox.box.com/shared/static/rirth0eym114afwyjxwe128sjzipzdym.txt") ################### ?Violin plot by library(vioplot) library(vioplot) data.use_11<- data.melt$value[data.melt$ident == 0]

Hi First post and a relative R newbie.... I am using the vioplot library to produce some violin plots. I have an input CSV with columns off irregular length that contain NAs. I want to strip the NAs out and produce a multiple violin plot automatically labelled using the headers. At the moment I do this Code:? ds1 = read.csv("http://www.lecturematerials.co.uk/data/spelling.csv")

Dear Rgurus, To my knowledge the best way to visualize the distribution of a discrete variable X is plot(table(X)) The problem which I have is the following. I have to discrete variables X and Y which distribution I would like to compare. To overlay the distribution of Y with lines(table(Y)) gives not satisfying results. This is the same in case of using density or histogram. Hence, I am

Dear list, How can I add the mean and standard deviation to each of the boxplots using the example provided in the boxplot function? boxplot(len ~ dose, data = ToothGrowth, boxwex = 0.25, at = 1:3 - 0.2, subset = supp == "VC", col = "yellow", main = "Guinea Pigs' Tooth Growth", xlab = "Vitamin C dose mg",

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

I am creating boxplots from a dataframe and would like to add to the standard output a marker representing the value from a particular row in the dataframe. .....And I apologize if the solution is as trivial as it seems it should be. Thanks for any assistance. Keith

Hi wizards, I need to know how can I generate this numbers I have n numbers N1,N2 , Nn , but Nn = 1- sum(N1+N2+...+Nn-1) and sum(N1+N2+..Nn)=1 and N1,N2,..Nn with 0<N<1 . Does somebody know how to generate it ? Some distribution or algorithm. Thanks in advance. -- Web Page http://www.geocities.com/ricardo_rios_sv/index.html

Does anybody have any insight into how to make this faster? I suspect, that the rounding going on may be an issue, as is the stepping through data frame rows using integers ... If you have the patience to teach a noob, he will highly appreciate it ;0) Joh digit <- 4 for (minute in seq(from=25,to=lrange[2])){ # Extract all data associtaed with the current time (minute) frame <-

Is there any reason that foreign key fields do not show in any of the scaffold views ? project table: id serial primary key, category_id integer, user_id integer, name varchar, morestuff varchar foreign key (category_id) references categories(id), foreign key (user_id) references users(id) scaffold generated views only show name, morestuff fields

Hi Chris, thanks for your answer. It is the first time I decided to encrypt my lvm. I choosed to encrypt the volume group, not every logical volume itself, because in case of doing lvm snapshots in that group they will be encrypted too? And how do I create a new encrypted volume group? Regards Tim Am 6. M?rz 2015 01:58:23 MEZ, schrieb Chris Murphy <lists at colorremedies.com>: >On

Dear all, Can anyone please shed some light onto how to do this? This will give me all "intensity" columsn in my data frame: intensityindeces <- grep("^Intensity",names(dataframe),value=TRUE) This will give me the maximum intensity for the first row: intensityone <- max(dataframe[1,intensityindeces]) What I'm now looking for is how to dfo this for the whole data

Hello R-ologists, Imagine you have a list "list" like so: >list [[1]] [1] "IPI00776145.1" "IPI00776187.1" [[2]] [1] "Something" "IPI00807764.1" "IPI00807887.1" [[3]] [1] "IPI00807764.1" [[4]] [1] "Somethingelse" What I need to achieve is a filtered list "list2" like so: >list2 [[1]] [1]

Hello, I'm new to this and am trying to teach myself some R by plotting biological data. The growth curve in question is supposed to be fitted to the Verhulst equation, which may be transcribed as follows: f(x)=a/(1+((a-0.008)/0.008)*exp(-(b*x))) - for a known population density (0.008) at t(0). I am trying to rework the example from "An Introduction to R" (p. 72) for my case and

I have read and reread the boxplot and the boxplot stats page, and I still cannot understand how and what boxplot shows. I realize that this might be due to me not knowing enough statistics, but anyway... First, how does boxplot determine the size of the box? And is the line inside the box the mean or the median (or something completely different?) And how does it determine how long out the

Dear friends, The user coordiante system in the R graphics is easy to understand as it simply corresponds to the range of values on the axes of the plot. However, sometimes, we want to kown the coordinates of a region in other system, e.g., the normal coordinate system whose origin is the the upper-left cornner of the drawing area which is used to set the hot regions in the HTML image maps.