similar to: ggplot2: Adding points to a density plot

Hi, Please consider the example below. How can I manage to overlay the points the way I want in the second case? Thanks, Joh library(ggplot2) # Modify data to match "real" case myDiamonds <- diamonds myDiamonds[["clarity"]] <- as.character(myDiamonds[["clarity"]]) myDiamonds[myDiamonds[["clarity"]]=="I1","clarity"] <- 1

Hi, Is there such a thing? If no: is it easily simulated? thanks, Joh

ggplot2 ------------------------------------------------------------ ggplot2 is a plotting system for R, based on the grammar of graphics, which tries to take the good parts of base and lattice graphics and avoid bad parts. It takes care of many of the fiddly details that make plotting a hassle (like drawing legends) as well as providing a powerful model of graphics that makes it easy to produce

ggplot2 ------------------------------------------------------------ ggplot2 is a plotting system for R, based on the grammar of graphics, which tries to take the good parts of base and lattice graphics and avoid bad parts. It takes care of many of the fiddly details that make plotting a hassle (like drawing legends) as well as providing a powerful model of graphics that makes it easy to produce

Hi all, When running R interactively, I have the problem as following: > library(ggplot2) Loading required package: reshape Loading required package: plyr Attaching package: 'reshape' The following object(s) are masked from 'package:plyr': round_any Loading required package: grid Loading required package: proto > data(VADeaths) > pg <- ggplot(melt(VADeaths),

Why is class(spectrum[["Ion"]]) after this "factor"? spectrum <- cbind(spectrum,Ion=rep("", nrow(spectrum)),Deviation.AMU=rep(0.0, nrow(spectrum))) slowly going crazy ... Joh

>From: Johannes Graumann <johannes_graumann at web.de> >Date: 2007/12/18 Tue PM 04:40:37 CST >To: r-help at stat.math.ethz.ch >Subject: [R] All anchored series from a vector? lapply(1:length(myvector) function(.length) { c(myvector[1}:myvector[.length]) }) but test it because i didn't. >Hi all, > >What may be a smart, efficient way to get the following result:

Hi, Any ideas on how to efficiently convert > list(c(1,2,3),c(4,5,6)) to > data.frame(OriginalListIndex=c(1,1,1,2,2,2),Item=c(1,2,3,4,5,6)) Thanks for any hints, Joh

Dear R users, First just want to say thank you to all for developing such a wonderful software and packages. I need to produce a wind rose plot. Tried with packages circular and plotrix and couldn't quite get what I want. Moved to package ggplot2 and it's going great. However stuck in how to move axis labels. I am using the wind rose from the help to learn how to do what I need (code

Hello, I'd like to sitch to a monochrome/bw color-palette for the filling of geom_bar-bars (produced via "qplot" as in the example below). Hours of googling didn't yield anything useful, so I thought, I'd just ask ... Thanks, Joh library(ggplot2) qplot(factor(cyl), data=mtcars, geom="bar", fill=factor(cyl))

Dear R-help listers, I want to add 95% confidential interval as vertical lines to x axe in density plot. I have found the library(hdrcde) can do this work, but I do not know how to handle functions of this library when I used ggplot2 to draw the graph. Thank you in advance. The data and codes followed: # dummy data factor<-rep(c("Alice","Jone","Mike"),each=100)

Hello, Banging my head against a wall here ... can anyone light the way to a pattern modification that would make the following TRUE? identical( grep( "^Intensity\\s[^HL]", c("Intensity","Intensity L", "Intensity H", "Intensity Rep1")), as.integer(c(1,4))) Thank you for your time. Sincerely, Joh

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

Dear all, I know that ggplot2 documentation is coming along, but at the moment I can't find how to do the following: a) change the title of the legend b) get rid of the closing line at the bottom of the density line. I also observed that the density lines (after limiting the x-scale) extend a little bit into the surrounding of the plot, which can be seen very strong when plotted as pdf.

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]

Dear all, I do not fully understand how ggplot2 handles NAs. See the following example: library(ggplot2) x <- rnorm(150) g <- as.factor(c(rep(c(0,1,NA),50))) mydf <- data.frame(x,g) m <- ggplot(aes(x = x, group = g, color = g), data = mydf) m + geom_density() How do I get rid of the NAs (i.e. the blue colored curve)? I thought ## m <- ggplot(aes(x = x, group = g, color = g,

Hello, I have a list of character vectors like this: sequences <- list( c("M","G","L","W","I","S","F","G","T","P","P","S","Y","T","Y","L","L","I","M",

Hi, Is there any way to trick R CMD check into not throwing this error after I created a dedicated "plot" incarnation for my custom function? * checking S3 generic/method consistency ... WARNING plot: function(x, ...) plot.MQUSpecMatch: function(x, mozlabel, labelcex) Thanks again, Joh

<posted & mailed> Dear all, I'm trying to solve the problem, of how to find clusters of values in a vector that are closer than a given value. Illustrated this might look as follows: vector <- c(0,0.45,1,2,3,3.25,3.33,3.75,4.1,5,6,6.45,7,7.1,8) When using '0.5' as the proximity requirement, the following groups would result: 0,0.45 3,3.25,3.33,3.75,4.1 6,6.45 7,7.1