similar to: panel failure in xyplot

Hello I want to draw a xyplot. Its dots will have three colors: red for meat, green for vegetable, and blue for both. I used the code below but could not make the dot in the same group show the same color. Please kindly advise how to modify it. Thank you. code library (lattice) diet.code <- c("Herbivore", "Omnivore", "Carnivore") Diet.colors <-

Hello This is Elaine. I am using package lattice to generate boxplots. Using Richard's code, the display was almost perfect except the outlier shape. Based on the following code, the outliers are vertical lines. However, I want the outliers to be empty circles. Please kindly help how to modify the code to change the outlier shapes. Thank you. code package (lattice) dataN <-

Hello, I am making a boxplot of 13 boxes. I tried to color the box using 13 colors but failed. Only red and brown were displayed. Green, blue, and grey disappeared. Please kindly advise modification after checking the code below. Thank you in advance. Elaine R code # data input dataN <-read.csv("H:/a_mig_distance_B_NB/R_data/Mig_bird_586_20120925.csv",header=T, row.names=1)

Hello I have a data composing three columns: body weight, body length, and sex. I want to draw a box plot, X-axis for body weight, Y-axis for body length. In the boxplot, I want to compare the body length between two sexes of three groups of body weight: <50, 51-100, >101 kg. (For instance, from left to right, the first boxplot: sex =1, body weight <50 the second boxplot: sex

Hello, I want to draw a boxplot using 13 colors for 13 boxes. Each box represents a type of diet of birds. Y axis is the breeding range of the birds. I checked the previous r-help and found a possible solution. However, it did not work by showing "error in ncol(Diet_B)," Diet_B not found. Please kindly help with modification and thank you. Also, if anything is unclear, please let me

Can somebody please take a look at this and tell me whats going wrong? It seems to be parsing wronly around the 'if' statement and gives me a directory listing. Thanks in advance Tom N.B. datan is an invented dataset xvals<-c(1,0.4,0.2) datan<-data.frame(s1=c(3,4,5),s2=c(5,5,5),s3=c(21,55,34),s4=c(5,3,2)) datan$sint<-NA datan$sgrad<-NA for(icount in 1:dim(datan)[1]) {

Hello I want to draw a plot using the code below. Further, I want to add labels (White, Yellow, Red) to x-axis. Please kindly advise how to add the categorical labels, by modifying the code. Thank you. Elaine Code dataN <-read.csv("H:/skin_color.csv",header=T, row.names=1) dim(dataN) dataN[1,] str(dataN) obs.group<-dataN$skin_color

Hello, I want to know if two quadratic regressions are significantly different. I was advised to make the test using step 1 bootstrapping both quadratic regressions and get their slope coefficients. (Let's call the slope coefficient *â*^1 and *â*^2) step 2 use the slope difference *â*^1-*â*^2 and bootstrap the slope coefficent step 3 find out the sampling distribution above and

Hi, I'm having difficulty with getting a table to show with multiple rows and columns. Below is the commands that I've typed in and errors that I am getting. Thank you. Laura Table trying to enter: Diet: Binger-yes: Binger-No: Total: None 24 134 158 Healthy 9 52 61 Unhealthy 23 72 95 Dangerous 12 15 27 >

Otras variantes con y sin paquetes adicionales... > sapply(split(datIn$Gain, as.factor(datIn$Diet)), mean) d1 d2 d3 280 278 312 > by(datIn$Gain, datIn$Diet, mean) datIn$Diet: d1 [1] 280 -------------------------------------------------------------- datIn$Diet: d2 [1] 278 -------------------------------------------------------------- datIn$Diet: d3 [1] 312 > > library(dplyr) >

Me gusta la respuesta uqe has dado, pero si por ejemplo, alguno de los datos tiene datos faltantes, entonces devuelve NA. He probado con: sapply(split(datos$uno, as.factor(datos$dos)), mean(na.rm=TRUE)) pero da fallo. ¿Cómo se podría hacer para que devolviera además la media obviando los NA y que contara el numero de NA por categoria? > Date: Wed, 28 Oct 2015 00:13:45 +0100 > From: cof

according to the documentation of the cast function in the reshape function, I would expect this bit of code from the examples to calculate marginal means over only the 'diet' variable. #Chick weight example names(ChickWeight) <- tolower(names(ChickWeight)) chick_m <- melt(ChickWeight, id=2:4, na.rm=TRUE) cast(chick_m, diet + chick ~ time, mean, margins="diet") But,

Dear all, This question may be too basic quesition for this list, but if someone has time to answer I will be happy. I have tried to find out, but haven't found a consice answer. As an example I use "Pinheiro, J. C. & Bates, D. M. 2000. Mixed-effects models in S and S-PLUS. Springer, New York." page 225, where rats are fed by 3 different diets over time, which body mass has

Hi, In the examples from the ReShape package there is a simple example of using melt followed by cast that produces a smallish amount of output about the chicks database. Here's the code: library(reshape) names(ChickWeight) <- tolower(names(ChickWeight)) chick_m <- melt(ChickWeight, id=2:4, na.rm=TRUE) DietResults <- cast(chick_m, diet + chick ~ time) DietResults My challenge

Hi all , I have a dataframe (Hypertension) with following headers :- > Hypertension ID Hypertension(before drug A) Hypertension(On drug A) On drug B? Healthy diet? 1 160 90 True True 2 190

I am trying to apply a permuation-based MANOVA (Anderson 2001) to a set of morphological data from three ecomorphs of fish reared under two different conditions and measured at two points during ontogeny. I will supply a distance matrix based on Procrustes distances calculated outside of vegan. I have not found an example of a design such as this for adonis. However, I have designed my factors

Dear users, I think my questions are pretty simple, but I got lost in the hundreds of par() and plot() arguments and plot functions, so I don't know in which direction I should go. Here is my sample dataset: test <- structure(list(DIET = structure(c(1L, 1L, 1L, 1L, 1L, 3L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 2L, 2L, 4L, 4L, 4L, 4L, 4L), .Label = c("G", "GG", "L",

I am running the following code on the coagulation data and I am getting an error. Please let me know if I am missing anything from my code. coag<- matrix( scan("//Samba3/nair/R/blood.dat", sep=","), 24, 3, byrow=TRUE) colnames(coag) <- c("time","diet","order") coag <- as.data.frame(coag) oneway.test(time ~ diet, data=coag,

Estimados Cuando existia epicalc, había una manera muy fácil de determinar la media de una variable (en esta caso Gain) por grupos, en este caso (Diet). ?Como se puede hacer ahora? Diet Gain 1 d1 270 2 d1 300 3 d1 280 4 d1 280 5 d1 270 6 d2 290 7 d2 250 8 d2 280 9 d2 290 10 d2 280 11 d3 290 12 d3 340 13 d3 330 14 d3 300 15 d3 300

In /usr/share/doc/rsync/README.gz: BUG REPORTS ----------- If you have web access then please look at... Well what if you don't have web access? You don't mention what to do then. "Send bugs to rsync@lists.samba.org" probably. OK. Will do. Here's one now: $ rsync -avz debian.linux.org.tw:var/lib/apt/diet.lists var/lib/apt/diet.lists receiving file list ...