similar to: help

Dear r-help, Could you help me to find the function which create an empty matrix. I use matrix(), but it gives *a single value that is NA and length of this matrix is 1.* ** *Best Regards* [[alternative HTML version deleted]]

Dear r-help, I create a graph of my baysian network. I use the package igraph. The names of vertex are within the circle, I would leave them outside the circle? > E(g)$color <- "black" > tkplot(g, ,vertex.label=names,layout=layout.kamada.kawai, edge.color=E(g)$color) Best Regards [[alternative HTML version deleted]]

Dear r-help, I create a package. When I installed this package (I use this command : R CMD check namepackage),I find an error: * checking whether package 'namepackage' can be installed ... ERROR Installation failed. Could you help me to find solution for this error. Best Regards [[alternative HTML version deleted]]

Dear r-help, Could you help me to find a solution for this error: Il y a eu 50 avis ou plus (utilisez warnings() pour voir les 50 premiers) > warnings() Messages d'avis : 1: In if ((data[pa, k] == df[, j]) & (data[ch, k] == i)) { ... : la condition a une longueur > 1 et seul le premier élément est utilisé 2: In if ((data[pa, k] == df[, j]) & (data[ch, k] == i)) { ... : la

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))

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

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

Hi all , I have one query. i have list of some .cel files. in my program i have to mention the path of these .cel files part of my program is, rna.data<-exprs(justRMA(filenames=file.names, celfile.path=*datadir*, sampleNames=sample.names, phenoData=pheno.data, cdfname=cleancdfname(hg18_Affymetrix U133A))) in the place of "datadir" i have to mention the character string of the

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",

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",

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 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. ###################################################### ###########

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, im a student so still very new to R. Hope someone could help me out here =) They are 3 slug control products, bustaslug, product X and Y. Im ask to explore the data by plot() and tapply(). But when i try to plot or use the tapply command, it tells me that the x and y lengths differ. so im thinking its because the data is unbalanced? Is there a simple command that i could use to fix this?

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

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,

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 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

Hello, This should be pretty simple but I cannot get it right. Please point to the right code. Thanks. > last <- read.csv(file.path(dataDir,"plot1.csv"), as.is=T,stringsAsFactors = FALSE) > last date r_wvht 1 8/6/2008 0.9766667 2 8/8/2008 0.7733333 3 8/11/2008 1.4833333 4 8/13/2008 1.5766667 5 8/14/2008 1.3900000 6 8/18/2008 0.7800000 7 8/20/2008