similar to: Colors in barplot

Hi there, I have a barplot and the labels at the x-axis are strings, which are rotated by 90?. But now the sub-title of the barplot is in between these labels, which does not look very nice... Could anybody help me finding the parameter-setting to prevent this? par(mai=c(1.5,1,1,0.6)) barplot(data, las=2, ylim = c(0,plotMax+1), ylab = "Percentage", main = filename, sub = subtitle)

Hello all, I have this simple barplot code: ifn <- "id.dat" dat <- read.table(ifn) ofn <- "id.png" bitmap(ofn, type = "png256", width = 30, height = 30, pointsize = 30, bg = "white",res=50) par(mar=c(5, 5, 3, 2),lwd=5) par(cex.main=1.6,cex.lab=1.6,cex.axis=1.6) names(dat)<-c("NumberOfPeople","Average")

Hi, as a bloody R beginner I failed to solve the probably simple problem to create a barplot of the following data read from a file Year A B C 2000 4 3 0 2001 2 1 3 2002 1 2 5 The Barplot should look like 5 | C 4 | A C 3 | AB C C 2 | AB A C BC 1 |

hello, I would like to create the following barplot: I have 4 different data sets (same length + stddev for each data point) data1 sd1 data2 sd2 data3 sd3 data4 sd4 now, I'd like to plot in the following way: data1[1],data2[1],data3[1],data4[1] with it's sd-values side-by-side at one x-axis label (named "position 1") and each bar in different colors.

Hello, I tried to fit a poisson distribution but looking at the function fitdistr() it does not optimize lambda but simply estimates the mean of the data and returns it as lambda. I'm a bit confused because I was expecting an optimization of this parameter to gain a good fit... If I would use mle() of stats4 package or mle2() of bbmle package, I would have to write the function by myself

Hi there, I though there has been a possibility to force the output on the console with one element per line. Instead of this: > 1:10 [1] 1 2 3 4 5 6 7 8 9 10 something like this > 1:10 [1] 1 [2] 2 [3] 3 [4] 4 [5] 5 [6] 6 [7] 7 [8] 8 [9] 9 [10] 10 Can anybody help? Antje

Hello, I hope this question is not too stupid. I would like to know how to update levels after subsetting data from a data.frame. df <- data.frame(factor(c("a","a","c","b","b")), c(4,5,6,7,8), c(9,1,2,3,4)) names(df) <- c("X1","X2","X3") my.sub <- subset(df, X1 == "a" | X1 == "b")

Hello, I'm looking for a solution for the following problem: I have two vectors V1 <- c("apple","honey","milk","bread","butter") V2 <- c("bread","milk") now, I would like to know for each element in V1 if it's equal to one of the elements in V2 I could do: which(V1 == V2[1] | V1 == V2[2]) but what if I

Hi folks, I've just discovered that the following code leads to boxplot (surprisingly to me). Can anybody explain to me why? Is this documented somewhere? I've never consider this option before. x <- rnorm(300) l <- c(rep("label1",100), rep("label2",50), rep("label3",150)) df <- data.frame(as.factor(l), x) plot(df) Thank you! Antje

Hi all! One quick question: How do I get the standard colors used by barplot? I have some stacked bars and would like to add a horizontal legend via legend() but I don't know how to find the colors for the fills of the legend points. Thanks! Werner

Hello, I was wondering if I can plot two curves I get from "density(data)" into one plot. I want to compare both. With the following commad, I just get one curve plotted: plot( density(mydata) ) Sorry for this stupid question but I could not find a solution until now... Antje

Hi there, I have a question concerning the behaviour of the colouring with levelplot. (I hope, I manage to explain) If I give the parameters "at" and "col.regions" like this: at <- c(1,2,3,4,5,6) col.regions <- c("blue","blue","blue","yellow","yellow","yellow") Which color would have the value 3.5? I would

Hi all, Quick question: What function can I use to draw a line in the margin of a plot? segments() and lines() both stop at the margin. In case the answer depends on exactly what I'm trying to do, see below. I'm using R v. 2.8.1 on Windows XP. Cheers, Alan I'm trying to make a horizontal barplot with a column of numbers on the right side. I'd like to put a line between the

Hello, unfortunately, I don't know a better subject. I would like to be very flexible in how to process my data. Assume the following dataset: par1 <- seq(0,1,length.out = 100) par2 <- seq(1,100) fac1 <- factor(rep(c("group1", "group2"), each = 50)) fac2 <- factor(rep(c("group3", "group4", "group5", "group6"), each =

----Original Message----- From: Martyn Plummer [mailto:plummerm at iarc.fr] Sent: Thursday, January 12, 2017 3:36 AM To: Weiner, Michael <weinerm at ccf.org>; r-sig-fedora at r-project.org Subject: Re: [R-sig-Fedora] rgl >One possibility is that you have the 32-bit version of mesa-libGL-devel installed. This will provide the headers but not the 64-bit l>ibrary which is what you need

Does anyone have any experience retrieving latitutde and longitude for an address from the Google Maps API? I'd like to have an R script that submits a street address, city, state, and zip code and returns the coordinates. So far, I've been submitting the coordinates from another program, then loading the coordinates in R and merging them back into the data frame I want to use.

Hello, I tried to use mle to fit a distribution(zero-inflated negbin for count data). My call is very simple: mle(ll) ll() takes the three parameters, I'd like to be estimated (size, mu and prob). But within the ll() function I have to judge if the current parameter-set gives a nice fit or not. So I have to apply them to observation data. But how does the method know about my observed

Hello, I have a question how to reshape a given matrix to a data frame. # ---------------------------------- > a <- matrix(1:25, nrow=5) > a [,1] [,2] [,3] [,4] [,5] [1,] 1 6 11 16 21 [2,] 2 7 12 17 22 [3,] 3 8 13 18 23 [4,] 4 9 14 19 24 [5,] 5 10 15 20 25 > colnames(a) <- LETTERS[1:5] > rownames(a) <-

A few day ago, I was looking for an answer to my question but didn't get one. Anybody who can help now? Hello, I tried to use mle to fit a distribution(zero-inflated negbin for count data). My call is very simple: mle(ll) ll() takes the three parameters, I'd like to be estimated (size, mu and prob). But within the ll() function I have to judge if the current parameter-set gives a nice

Hi there, as I'm not sure to understand the coloring levelplot uses, I'm looking for another easy way to create a heatmap like this: library(lattice) mat <- matrix(seq(1,5, length.out = 12), nrow = 3) mat[1,2] <- 3.5 my.at <- seq(0.5,5.5, length.out = 6) my.col.regions <- rainbow(5) graph <- levelplot(t(mat[nrow(mat):1, ] ), at = my.at, col.regions = my.col.regions)