similar to: Another question on gplot

Hi, suppose I have following codes : library(zoo); library(ggplot2) dat <- matrix(rnorm(500*2), 500); dat <- zooreg(dat, start = as.Date("01/01/01", "%m/%d/%y"), frequency=1); plot(dat) head(dat); month.no <- format(index(dat), "%m"); dat1 <- cbind(coredata(dat), as.numeric(month.no)) x <- dat1[,1]; y <- dat1[,2]; z <- dat1[,3] Now I draw a

Hi, i get no success change the title of the "fill" (colour) legend and the defintion of "levels". Have anybody a hint how i can do this. df <- data.frame(variable=sample(c("A","B","C"),1000,replace=T,prob=c(0.22,0.28,0.5)),group=gl(2,500)) p <- ggplot(df, aes(x = variable)) p + geom_histogram(aes(y= ..count.. /

I'm learning ggplot and am a little confused. Sometimes discrete scales work like I'd expect, and sometimes they don't. For example... This works exactly like one would expect: df<-data.frame(names=c("Bob","Mary","Joe","Bob","Bob")) ggplot(df,aes(names))+geom_histogram() But this yields an error:

Hola Soy Leticia, quería hacer una consulta al grupo a ver si me podéis ayudar. Tengo unos datos de proteínas, el data frame es: 5 columnas iniciales que tienen información sobre las proteínas y después las columnas con las intensidades, de las cuales no se cuantas condiciones tienen ni cuántas réplicas por condición. Estoy intentando hacer un cálculo del coeficiente de variación por grupo y

Hi all, I was trying to draw a stacked density plot like that : library(ggplot2); library(plyr) dat <- cbind(rnorm(300), rep(c(1,2), each=150)) ggplot() + geom_density(aes(x=dat[,1], fill=factor(dat[,2]), position="stack")) + xlab("") + ylab("") + scale_colour_manual(name = "Pallet", labels = c("X", "Y")) Here

Dear All I am trying to graph a proportion and CI95% by a factor with ooplot (any other better solution ?) It works well until I try to add the confidence interval. this is the error message and and a description of the data: > dat1 PointEst TT1 1 3.6 TT2 2 5.0 TT3 3 5.8 TT4 4 11.5 TT5 5 7.5 TT5 6 8.7 TT7 7 17.4 > dat2

I am learning ggplot2 commands and I have figured out how to create histograms and density curves but I am not sure how to add a density curve on top of a histogram. Here are the two graphs that I created. ## Histogram t<-rnorm(500) w<-qplot(t, main="Normal Random Sample", fill=I("blue"), colour=I("black"), geom="histogram") w ##Density Curve

Hello all-- I'm attempting to produce overlaid histograms with partially transparent columns. Whether this display will end up being useful, I can't say. But I do want to get it right. I've already got one solution (shown below), but I tried some other versions and had questions about my results. (Note: I'm using a quartz device, so transparency shows up correctly. You might

Muchas fracias por la rapida contastacion Javier. Si, ya me he dado cuenta que a veces es mas limpiar y formatear que calcular. Lo que yo necesito hacer no es crear un data.frame, es crear una lista por cada grupo/condicion de muestras. Y queria hacerlo en un for loop, para que no tuviese que indicar cuantas listas necesito si no que cogiese esa informacion del usuario (y) Creo que no me

Hola Leticia Acomodar datos para analizar suele llevar más trabajo que el análisis. Yo prefiero en un data.frame colocar todos los datos, porque siempre es más fácil filtrar que agregar. Cuándo tengo el data.frame con todo, realizo un plot (datos), y muchas veces en la gráfica veo cosas que me llaman la atención, de esta forma busco algo de antemano no sospechaba. No es necesario que conozcas

Hi all, How can I overlay these two histograms? ggplot(gg, aes(gg$Alz, fill = gg$veg)) + geom_histogram(alpha = 0.2) ggplot(tt, aes(tt$Cont, fill = tt$veg)) + geom_histogram(alpha = 0.2) thanks for any help! Elahe

Hallo I am not familiar with any of packages you use (except of MASS and ggplot2) and the code is too complicated without any hint where the error could come from and what is the message you get. I wonder if anybody would like to go through your whole code. 1. data seems to be read correctly ICUData <- read.csv(file = "ICUData.csv", stringsAsFactors = TRUE) ICUData.neuro <-

Hi, I have a histogram.But I need seperate colours for fixed range of values.for eg. between 2-3 on the x axis a colour.3-4 another colour etc. and the legend has to say what each colour is.How can this be done in geom_histogram? -- Rajesh.J [[alternative HTML version deleted]]

I'm trying to overlay two histograms using transparency to enable viewing of multiple distributions on a single scale. So far ggplot2 seems to do what I want. However I'm having a problem generating the legend coloring appropriate to each distribution in the plot. Here is a test case to show my best (failed) effort so far: library(ggplot2) x <- data.frame(X=rnorm(1000, mean=0)) y

Dear ggplot2 users, is there an easy/elegant way to suppress zero count bars in histograms with logarithmic y axis ? One (made up) example would be qplot(exp(rnorm(1000))) + geom_histogram(colour = "cornsilk", fill = "darkblue") + scale_x_sqrt() + scale_y_log10() Thanks! Markus [[alternative HTML version deleted]]

Sarah's solutions are good, and here's another, even more basic: tmp1 <- unique(dat1$B) tmp2 <- seq_along(tmp1) dat1$C <- tmp2[ match( dat1$B, tmp1) ] > dat1 N B C 1 1 29_log 1 2 2 29_log 1 3 3 29_log 1 4 4 27_cat 2 5 5 27_cat 2 6 6 1_log 3 7 7 1_log 3 8 8 1_log 3 9 9 1_log 3 10 10 1_log 3 11 11 3_cat 4 12 12 3_cat 4 As a single line

I need to split a given matrix in a sequential order. Let my matrix is : > dat <- cbind(sample(c(100,200), 10, T), sample(c(50,100, 150, 180), 10, > T), sample(seq(20, 200, by=20), 10, T)); dat [,1] [,2] [,3] [1,] 200 100 80 [2,] 100 180 80 [3,] 200 150 180 [4,] 200 50 140 [5,] 100 150 60 [6,] 100 50 60 [7,] 100 100 100 [8,] 200 150 100 [9,]

Buenas tardes a todos, tengo un problema con R: ejecuto el mismo script en el ordenador del trabajo y en mi portátil con los mismos datos y obtengo resultados diferentes (siendo los correctos, los obtenidos en el trabajo): rm(list=ls()) directorio<-"C:\\Documents and Settings\\BDs\\" library(RODBC) library(car) library(gdata) ### DATOS A ###

Hi, Here's one way to approach it, using the coercion of factor to numeric. Note that I changed your data.frame() statement to avoid coercing strings to factors, just to make it simpler to set the levels. dat1 <-data.frame(N=seq(1, 12,1), B=c("29_log","29_log", "29_log", "27_cat", "27_cat", "1_log", "1_log",

Hi Sarah, Thank you so much!! I got your good ideas. Ding -----Original Message----- From: Sarah Goslee [mailto:sarah.goslee at gmail.com] Sent: Friday, May 11, 2018 11:40 AM To: Ding, Yuan Chun Cc: r-help mailing list Subject: Re: [R] add one variable to a data frame [Attention: This email came from an external source. Do not open attachments or click on links from unknown senders or