similar to: Tables without names

Hi Hadley and everyone, here's a patch for ggplot2 that fixes the behavior of opts(legend.position={left,top,bottom}). If you try the following code in an unmodified ggplot2 options(warn = -1) suppressPackageStartupMessages(library("ggplot2")) data <- data.frame( x = c(1, 2, 3, 4, 5, 6), y = c(2, 3, 4, 3, 4, 5), colour = c(TRUE, TRUE, TRUE, FALSE, FALSE, FALSE))

I have two vectors, a and b. b is a text file. I want to find in b those elements of a which occur at the beginning of the line in b. I have the following code, but it only returns a value for the first value in a, but I want both. Any ideas please. a = c(2,3) b = NULL b[1] = "aaa 2 aaa" b[2] = "2 aaa" b[3] = "3 aaa" b[4] = "aaa 3 aaa"

Dear all, I am trying to plot a matrix I have? as an image str(matrixToPlot) ?num [1:21, 1:66] 0 0 0 0 0 0 0 0 0 0 . ?that contains only 0s and 1s, where the xlabel will be Labeled as str(xLabel) ?num [1:66] 1e+09 1e+09 1e+09 1e+09 1e+09 ... and the yLabels will be labeled as str(yLabel) ?num [1:21] -88 -87 -86 -85 -84 -83 -82 -81 -80 -79 ... I have found on the internet that I can do

I recently traced a bug of mine to the fact that cumsum(s)[length(s)] is not always exactly equal to sum(s). For example, x<-1/(12:14) sum(x) - cumsum(x)[3] => 2.8e-17 Floating-point addition is of course not exact, and in particular is not associative, so there are various possible reasons for this. Perhaps sum uses clever summing tricks to get more accurate results? In some

Hi I am very new to R and statistical programming in general. I am trying to reorder data from a .csv file. I have managed to import the data and create a zero matrix. I am now trying to fill the matrix. There seems to be some problem with this section of my code. I have highlighted the dodgy code in red. Please help if possible. ###################################################### ###########

Good day all, I'm trying to plot a continuous line plot, which is followed by two forecast points eg. one forecast point is 12 months out, and another 24 months out from the last date of the line plot. In my attempts so far, the second plot (the forecast points) is scaled against a new axis scale, thus the two plots are not directly comparable (I need the forecast points to be scaled

To test two environments for object equality (Lisp EQ), I can use 'identity': > e1 <- environment(local(function()x)) > e2 <- environment(local(function()x)) > identical(e1,e2) # compares object identity [1] FALSE > identical(as.list(e1),as.list(e2)) # compares values as name->value mapping [1] TRUE # (is there a

Tested in: R version 2.8.1 (2008-12-22) / Windows Recursive default argument references normally give nice clear errors. In the first set of examples, you get the error: Error in ... : promise already under evaluation: recursive default argument reference or earlier problems? (function(a = a) a ) () (function(a = a) c(a) ) () (function(a = a) a[1] ) () (function(a = a)

What is the recommended class for time of day (independent of calendar date)? And what is the recommended way to get the time of day from a POSIXct object? (Not a string representation, but a computable representation.) I have looked in the man page for DateTimeClasses, in the Time Series Analysis Task View and in Spector's Data Manipulation book but haven't found these. Clearly I can

Dear users, I'm trying to produce a 3d bar plot but the x and y dimensions have categorical data -- so I only want 3 points on each axis. So I try: require(scatterplot3d) mymat<-data.frame( x=c(1,1,1,2,2,2,3,3,3), y=c(1,2,3,1,2,3,1,2,3), z=c(1,2,3,4,5,6,7,8,9)) scatterplot3d(mymat, type="h", lwd=5, pch=" ", xlab="xlabel",

Hi! How to find out which plot function is used when i call plot(hist.default(1:10,plot=F)) and all works fine ? The reason why I would like to know it is that after loading some self written R functions > plot(hist.default(1:10,plot=F)) Error in xy.coords(x, y, xlabel, ylabel, log) : x and y lengths differ > traceback() 5: stop("x and y lengths differ") 4: xy.coords(x, y,

On Tue, Feb 10, 2009 at 10:11 AM, Duncan Murdoch <murdoch at stats.uwo.ca> wrote: > Stavros Macrakis wrote: >> On Tue, Feb 10, 2009 at 8:31 AM, Duncan Murdoch <murdoch at stats.uwo.ca>wrote: >>> The evaluator recognizes the context of usage and will get the >>> function for a function call.... >> Can you point me to chapter and verse in the language

Dear All: good morning I am trying to graph the function y=f(x)=1/x over the interval (-5,5). But I am getting an error message. Please see below. I am getting the error message: *Error in xy.coords(x, y, xlabel, ylabel, log) : * * 'x' and 'y' lengths differ* x x <- seq(-5, 5, 0.01) y < 1/x plot(x,y, type='l', xlim=c(-5, 5), ylim=c(-5, 5), xlab = "x",

I don't usually do much with graphs in R, and this is my first time adding a line of best fit. Hopefully this is an easy problem to solve. I'm looking at a variable called soloKills along the range 5:28. Here are all my commands, in script form: range=5:28 graph=soloKills title="Solo kill/death data" xlabel="Number of deaths/1 game" ylabel="Mean number of kills/1

i want to plot the histogram and the curve in the same graph.if i have a set of data ,i plot the histogram and also want to see what distribution it was.So i want to plot the curve to know what distribution it like. -- View this message in context: http://www.nabble.com/how-to-plot-the-histogram-and-the-curve-in--the-same-graph-tp20082506p20082506.html Sent from the R help mailing list archive at

Dear List, I am trying to modify the xlab and ylab for a current figure that was plotted by a package, I searched through the low level plotting command and they do not seem to contain how to do this (the only way is to use xlab, ylab as arguments in "plot" command, which I can not do since the plot is plotted using some other package, not by my own script). Is there any command for

Dear R users, Please can you help me with a relatively straightforward problem that I am struggling with? I am simply trying to plot a baseline survivor and hazard function for a simple data set of lung cancer survival where `futime' is follow up time in months and status is 1=dead and 0=alive. Using the survival package: lung.wbs <- survreg( Surv(futime, status)~ 1, data=lung,

HI, sorry but i don't understand how to make a function with as.function() formula<-"2+3*x" formu<-as.symbol(formula) > formu 2+3*x formul<-as.function(alist(x=,formu)) curve(formul,1,5,col="blue") Error in xy.coords(x, y, xlabel, ylabel, log) : x and y lengths differ > typeof(formul) [1] "closure" and not plot the curve function, Why?

Hi i want do a line graph. My y axis contains numeric values. My x axis contains non numeric statements. This is what i want the graph to look like. When i try to plot this graph on R it comes up with the following error message: "Error in plot.window(...) : need finite 'xlim' values In addition: Warning messages: 1: In xy.coords(x, y, xlabel, ylabel, log) : NAs introduced by

What is the 'idiomatic' way of writing a vectorized switch statement? That is, I would like to write, e.g., vswitch( c('a','x','b','a'), a= 1:4, b=11:14, 100 ) => c(1, 100, 13, 4 ) equivalent to ifelse( c('a','x','b','a') ==