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Hi there, I've looked through the very helpful advice about adding fitted lines to plots in the r-help archive, and can't find a post where someone has offered a solution for my specific problem. I need to plot logistic regression fits from three differently-sized data subsets on a plot of the entire dataset. A description and code are below: I have an unbalanced dataset

Hi, I am trying to print out my confusion matrix after having created my random forest. I have put in this command: fit<-randomForest(MMS_ENABLED_HANDSET~.,data=dat,ntree=500,mtry=14, na.action=na.omit,confusion=TRUE) but I can't get it to give me the confusion matrix, anyone know how this works? Thansk! Ruben [[alternative HTML version deleted]]

I really like these Trellis graphics but how do I get this code to skip the missing? logreg<-read.csv("logreg.csv", header=TRUE, sep=",", na.string=" ") attach(logreg) bwplot(yesno~bc_pcb_tot |varlist, data=logreg, main="Box Cox PCB transformation", auto.key=TRUE, fontfamily = "HersheySans" ) Dean Sonneborn M.S. Public Health Sciences *

Hi Bert, I am sorry to bother you on weekend. I am still struggling on defining a correct function. I first defined the function RFS (see below), then run it by provide the two argument. m52.2cluster <-RFS(inputfile =allinfo_m52, N=2 ) I do not get error message, but no figure displays on screen. I do not know what is going on. Can you help me a little more on this issue? Thank you,

Hi Richard, Thank you so much!! I understand the problem now, I assign a name to the "ggsurvplot" object and then add print(fig) at bottom of function definition, now figure gets printed on screen. Ding # function to generate RFS curves RFS <- function( inputfile, N ) { cluster<- survfit(Surv(RFS_days2, OV_Had_a_Recurrence_CODE) ~ clusters, data =

I have been trying to run a logistic regression using a number of studies. Below is the syntax, error message & data. Any advice regarding what I am doing wrong or solutions are appreciated, regards Bob Green > logreg <- read.csv("c:\\logregtest.csv",header=T) > attach(logreg) > names(logreg) [1] "medyear" "where" "who"

Dear R Experts, Is there a way in R to add an error bar (say in the y direction) for each data point? Thanks Ming Chow -- __________________________________________________________________ -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help",

Dear R Experts, Is there a way in R to add an error bar (say in the y direction) for each data point? Thanks Ming Chow -- __________________________________________________________________ -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help",

FAQ 7.22 You must print a ggplot object, for example with print(m52.2cluster) For the FAQ, run the line system.file("../../doc/FAQ") in R on your computer. Open up the resulting filepath in your favorite editor and scroll down to 7.22 On Sun, Jan 14, 2018 at 4:21 PM, Ding, Yuan Chun <ycding at coh.org> wrote: > Hi Bert, > > I am sorry to bother you on weekend. >

That is certainly OK, but you can also just use print(ggsurvplot(...)) as your final statement. out <- RFS( ...) would then return the ggsurvplot object *and* graph it. Any good R tutorial or a web search will provide more details on function returns, which you might find useful. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and

Thank you, your suggestion is simpler and logically better. I had impression that the last object in a function gets returned, so I did not add the print function at the bottom line of the function definition. Returning an object and graph the object are different process, I am a beginner for writing R function and need to find a good guide source about writing R functions. If you know a good

Hi, I'm sorry to send email to everyone on this list, but I have a simple question which is bothering me and I can't seem to figure out the correct answer. I just downloaded R and I'm trying to reproduce some simple analysis I've done on other packages. In particular, if I have a vector of experimental values and a vector of their uncertainties, is there a way to plot the

Hi, I was trying to make a density plot with 13 samples. To distinguish each sample, it would be good if each color is as different as possible from the other colors. I could use the built in function, but that does not do more than 8 colors and then goes back to recycling the cols. If I use a palette, then it is really difficult to distinguish between the colors. So, is there a way that I can

I propose to add a function that allows to display colors selected by a text pattern or by color vectors in a plot. Wolfram Fischer #--- showcolors.R showcolors <- function( col = "red" , index = NULL , pie = TRUE , lwd = 6 , cex = 1.0 , main = NULL , sub = NULL , ... ){ n.colors <- length( col ) if( n.colors > 1 ){ main <- deparse( substitute( col ) )

Hi All. I'd like to change the default plotting colors used to construct a scatterplot with regression line in the CAR package. So, scatterplot(y~pred,smooth=FALSE, xlab="X", ylab="Y", lwd=2) If I change the palette (e.g., palette(ranbow(6)), I can change the color of the lines and points. However, the axes and labels remain in black (i.e., the first color in the

Perhaps this is not a bug but rather my misunderstanding of lty and col. I would like to generate a legend for plots as follows: par(mfcol = c(2, 1), mar = c(5.1, 6.1, 4.1, 2.1)) matplot(t(matrix(1:8,8,20)), type="l") par(mfg = c(2, 1, 2, 1)) box(col = 0) # Rbug workaround. Thanks Martin legend((par()$usr)[1:2], (par()$usr)[3:4], as.character(1:8), lty=1:8, col=1:8,

Perhaps this is not a bug but rather my misunderstanding of lty and col. I would like to generate a legend for plots as follows: par(mfcol = c(2, 1), mar = c(5.1, 6.1, 4.1, 2.1)) matplot(t(matrix(1:8,8,20)), type="l") par(mfg = c(2, 1, 2, 1)) box(col = 0) # Rbug workaround. Thanks Martin legend((par()$usr)[1:2], (par()$usr)[3:4], as.character(1:8), lty=1:8, col=1:8,

Hi there: recently i try to use LogicReg package for a tree model(logistics fit ) . i list my code and error below: > dim(model.dat) [1] 48000 745 > fit1 <- logreg(resp = model.dat[,745], bin=model.dat[, 9:700], type = 3, select = 3, ntrees = c(1,2), nleaves=c(1,7), ) Insufficient declaration LGCn1MAX in logreg() is 20000 LGCn1MAX should be at least 48000 Please fix and

Gracias, Javier. Está muy bueno el post, pero en efecto refiere a legend y no a los names.arg Adjunto un archivo para poder correr el gráfico. Creo que eso ayuda a precisar el problema. Las etiquetas que aparecen al costado izquierdo de las barras son las que quiere pintar de negro y azul. También va a necesitar este objeto para que la sintaxis corra sin errores: library(RColorBrewer) #Para

Dear R-users: How does one calculate in R the odds ratios for a CATEGORICAL predictor variable that has 4 levels. I see r-help inquiries regarding odds ratios for what looked like a continuous predictor variable. I was wondering how to get the pairwise odds ratios for comparisons of levels of a categorical predictor variable. I can't seem to get the correct output using: >