similar to: ggplot2 question: keeping the order as in the input data

Hello! I am using ggplot2: ggplot(myplotdata, aes(x=att_levels, y=WTP)) + geom_bar(stat="identity",fill="dark orange",colour="black", alpha = 1,position = "identity") + geom_text(aes(label=WTP),colour="black",size=4,hjust=1.1,position='dodge') + coord_flip() + xlab("") +

Hi, I want to plot grouped bars to compare 95% confidence interval estimates from two models. Each bar represents a 95% confidence interval estimate of a coefficient from one of the two models. Each group represents confidence interval estimates of the same coefficient from the two models. I think such a bar plot will nicely present whether 95% confidence interval estimates of the same

hi there.. i got a problem with ggplot2. here my example: library (ggplot2) v1 <- c(1,2,3,3,4) v2 <- c(4,3,1,1,9) v3 <- c(3,5,7,2,9) gender <- c("m","f","m","f","f") d.data <- data.frame (v1, v2, v3, gender) d.data x <- names (d.data[1:3]) y <- mean (d.data[1:3]) pl <- ggplot (data=d.data, aes (x=x,y=y)) pl

Hi there, I'm trying to find out the command to stop clipping to plot region in ggplot. I have a bar chart (axis flipped) with labels on the bars, but the labels are clipped at the plot region box. I know it's possible to turn this off for base and lattice, but how about ggplot? par(xpd=NA) The ggplot2 manual doesn't seem to mention it, and a search for ggplot and

Hi, I wanted to generate a plot which is almost like the plot generated by the following codes. category <- paste("Geographical Category", 1:10) grp1 <- rnorm(10, mean=10, sd=10) grp2 <- rnorm(10, mean=20, sd=10) grp3 <- rnorm(10, mean=15, sd=10) grp4 <- rnorm(10, mean=12, sd=10) mydat <- data.frame(category,grp1,grp2,grp3,grp4) dat.m <- melt(mydat) p <-

I'm having problems with this example, it is posted with reproduceable code below, both with the normal 0-6 scale and the desired 3-6 scale (with bars removed). How can I get the graph to have the desired 3-6 scale without removing the bars. Thanks! #Data

Hello! I have my data frame "mydata" (below) and data frame "reference" - that contains all the dates I would like to be present in the final data frame. I am trying to merge them so that the the result data frame contains all 8 dates in both subgroups (i.e., Group1 should have 8 rows and Group2 too). But when I merge it it's not coming out this way. Any hint would be

I think you should be more suspicious of yourself, Dimitri. A letter T variable can easily arise in the problem domain when you are not thinking of logical values at all, at which point your cavalier use of T as a synonym for TRUE can suddenly become a bug. -- Sent from my phone. Please excuse my brevity. On July 27, 2017 8:18:03 AM PDT, Dimitri Liakhovitski <dimitri.liakhovitski at

Dear R-list, maybe some of you could point me in the right direction: Are you aware of any FREE Fortran or Java libraries/actual pieces of code that are VERY efficient (time-wise) in running the regular linear least-squares multiple regression? More specifically, I have to run small regression models (between 1 and 15 predictors) on samples of up to N=700 but thousands and thousands of them. I

Hello! I have a column in my data frame that I have to split: I have to distill the numbers from the text. Below is my example and my solution. x<-data.frame(x=c("aaa1_bbb1_ccc3","aaa2_bbb3_ccc2","aaa3_bbb2_ccc1")) x library(stringr) out<-as.data.frame(str_split_fixed(x$x,"aaa",2)) out2<-as.data.frame(str_split_fixed(out$V2,"_bbb",2))

Just a thought: Did you try na.rm = TRUE in case you have an object named "T" in scope? -- Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Thu, Jul 27, 2017 at 7:49 AM, Dimitri Liakhovitski <dimitri.liakhovitski at

Hello! I am trying to understand how ggplot2's geom_bar treats NAs. The help file says: library(ggplot2) ?geom_bar na.rm: If FALSE, the default, missing values are removed with a warning. If TRUE, missing values are silently removed. I am trying it out: md <- data.frame(a = c(letters[1:5], letters[1:4], letters[1:3], rep(NA, 3))) str(md); levels(md$a) ggplot(data = md, mapping = aes(x =

Hello! s<-"start"; e<-"end" middle<-as.character(c(1,2,3)) I would like to get the following result: "start 123 end" or "start 1 2 3 end" or "start 1,2,3 end" How can I avoide this (undesired) result: paste(s,middle,e,sep=" ") Thank you! -- Dimitri Liakhovitski Ninah Consulting www.ninah.com

?hanks for the advice, Jeff. Will keep it in mind. But I am anal - I shy away from using letters and words that "look familiar" to me in R (such as mean, sd, T, etc.) But still, it's a good advice. On Thu, Jul 27, 2017 at 11:53 AM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote: > I think you should be more suspicious of yourself, Dimitri. A letter T > variable can

To clarify: my question is not about "who could I exclude NAs from being counted" - I know how to do that. My question is: Why na.rm = T is not working for geom_bar in this case? On Thu, Jul 27, 2017 at 8:24 AM, Dimitri Liakhovitski < dimitri.liakhovitski at gmail.com> wrote: > Hello! > > I am trying to understand how ggplot2's geom_bar treats NAs. > The help file

Hello! I am using function "curve" to create a line graph. I was wondering, if it's possible to "turn off" the default tick marks and introduce those tick marks in specific locations. For example, currently in my X axis tick marks are (automatically) at 10, 11, 12, 13 but I want them to be in 5 specific locations, like 9.89, 10.34, etc. Any hint would be greatly

Dear R-ers! I have x as a variable in a data frame x. x<-data.frame(x=c("aa.bb","cc.dd.ee")) x$x<-as.character(x$x) x I am sorry for such a simple question - but how can I replace all periods in x$x with spaces? sub('.', ' ', x$x) - removes all letters to the left of each period... Thanks a lot for your advice! -- Dimitri Liakhovitski Ninah.com

Hello everyone! I have a list X with 3 elements, each of which is a data frame, for example: a<-data.frame(a=1,b=2,c=3) b<-data.frame(a=c(4,7),b=c(5,8),c=c(6,9)) c<-data.frame(a=c(10,13,16),b=c(11,14,17),c=c(12,15,18)) X<-list() X[[1]]<-a X[[2]]<-b X[[3]]<-c (X) How can I most effectively transform X into a data frame with columns a, b, and c? I would love to find a generic

Hello! I wrote a function that returns a data frame. Nowhere in the function do I say print(my.data.frame), but when I run the function - the data frame is printed on the console. Is there any way to suppress it? Thank you! -- Dimitri Liakhovitski Ninah Consulting www.ninah.com

Hi John (and other authors of ChoiceModelR package), I am experiencing a weird thing when using the function choicemodelr under R3.0.1. Before I updated to R3.0, I had used choicemodelr unde R2.15. It was always as fast or faster than exactly the same task in Sawtooth software. Now, I've tried to run it under R3.0.1. It seems to be doing the job. But something is slowing it down dramatically.