similar to: Basic question about: <<- and method start with dot.

Dear list, When I use VennDiagram package, I got a error as follow: venn.plot <- draw.pairwise.venn( area1 = 3186, area2 = 325, cross.area = 5880); Error in draw.pairwise.venn(area1 = 3186, area2 = 325, cross.area = 588) : Impossible: cross section area too large. Does anyone have suggestion? Thank you.

Dear list, Is there a elegant way to name multiple ordered plot as A, B, C and D? Just put A, B, C and D on the top left corner of each plot. Usually I do it in Illustrator, but I think in R should there is a way. Could you please provide a example? Thank you very much in advance.

Dear All, I am trying to use the do.call command to avoid tedious loops in R when I need to call repetitively a function. I am experiencing a problem when the arguments of a function are given with a vector (I need to express then this way to be able later on to integrate them numerically with the adapt package). Consider the small script: remove(list=ls()) #library(adapt) # first a list of the

Dear R Community, I am new to R, and have a question that I suspect may be quite simple but is proving a formidable roadblock for me. I have a large data set that includes water-quality measurements collected over many 24-hour periods. The date and time of sample collection are in a combined Date/Time field in the format yyyy-mm-dd hh:mm:ss. I need to be able to subset the data for analysis of

Dear All, I'd like to know if it is possible to create animations with R. To be specific, I attach a code I am using for my research to plot some analytical results in 3D using the lattice package. It is not necessary to go through the code. Simply, it plots some 3D density profiles at two different times selected by the user. I wonder if it is possible to use the data generated for different

Hi list, I want to plot two plot in the same figure. I set par(new=TRUE). But it does not work. library(lattice) myPanel <- function(x,...) { panel.histogram(x,alpha=0.4,...) ltext(0.4,1.5,paste("Mean=","0.05",digit=2)),cex=0.8) ltext(0.8,1.5,paste("s.d.=","0.06",digit=2)),cex=0.8) } histogram(sh2,

Hi all, How to plot as the coordinate as in my attachment? I want to trim the coordinate and one of plot as the figure in attachment. Does any one have such example? Thanks. -------------- next part -------------- A non-text attachment was scrubbed... Name: Screen shot 2011-01-17 at 11.22.20 AM.png Type: image/png Size: 23908 bytes Desc: not available URL:

Dear list, How can I detect a package is installed or not? If not, then install it. For example, in a script, I want to check the package DESeq is installed or not. If not, then I will using this script to install it. source("http://www.bioconductor.org/biocLite.R") biocLite("DESeq") The pseudo script would be like this: try: library("DESeq") catch:

Dear list, I want to run two independent calculation (A, B) and then combined their results to do a plot in R. Before I just run calculation A first, then calculation B. It will take some times. I just have one server with multiple core. I noticed that there are some nice package in R for parallel calculation, such as parallel, multicore and snow. Also there is a book, O?Reilly ? ?Parallel R,

Dear list, I have a dataframe with two column as fellow. > head(dat) V1 V2 0.15624 0.94567 0.26039 0.66442 0.16629 0.97822 0.23474 0.72079 0.11037 0.83760 0.14969 0.91312 I want to get the column V2 mean value based on the bin of column of V1. I write the code as fellow. It works, but I think this is not the elegant way. Any suggestions?

I have a csv file, heartrate excercise medicine Time 12 0 0 4:30 am 23 5 0 5:00 am 34 6 0 5:30 am 56 3 0 6:00 am 24 12

OK, I need help!! I've been searching, but I don't understand the logic of some this dataframe addressing syntax. What is this type of code called? test [["v3"]] [is.na(test[["v2"]])] <-10 #choose column v3 where column v2 is == 4 and replace with 10 and where is it documented? The code below works for what I want to do (find the non-missing value in a row),

Not sure where to even start on this.... I'm hoping there's some debugging I can do... I have a loop that cycles through several different data sets (same structure, different info), performing randomForest growth and predictions... saving out the predictions for later study... I get about 5 hours in (9%... of the planned iterations.. yikes!) and R just freezes. This happens in

Hi Jeremy, list, On 06/26/2016 12:11 AM, Jeremy Allison wrote: > We should probably have something in the server that logs > this as an official "event". Can someone log a RFE bug in > the bugzilla so we don't forget this request ? I created this bug: https://bugzilla.samba.org/show_bug.cgi?id=11998 I hope it is (approximately) what you mean. :-) Best regards, MJ

On Tue, 2017-09-19 at 17:02 +0200, L.P.H. van Belle via samba wrote: > Hai Mark, > > I see the bugreport for this is still untouched. > https://bugzilla.samba.org/show_bug.cgi?id=11998 I've closed that bug now. Extensive work has been done to add this feature to Samba 4.7, due out this week: https://wiki.samba.org/index.php/Setting_up_Audit_Logging Two new debug classes,

Description of Problem We are rapidly approaching the time of year when some will transition from standard time (ST) to daylight savings time (DST) and others will make the opposite transition. These vernal and autumnal transitions have important implications for those with Microsoft systems and use utilities that compare file timestamps on different filesystem types or with filesystems on other

https://bugzilla.netfilter.org/show_bug.cgi?id=1077 Bug ID: 1077 Summary: New traffic reduces conntrack timeout Product: netfilter/iptables Version: unspecified Hardware: All OS: All Status: NEW Severity: normal Priority: P5 Component: nf_conntrack Assignee: netfilter-buglog

Hi, If I have the following data.frame >y time val 1 08:00:05.834 1 2 08:03:13.345 2 3 08:10:12.443 3 > and the following function which converts the time string to the number of milliseconds since midnight > str_to_millis function( s ) { a <- as.numeric( unlist( strsplit(s,":",fixed="TRUE") ) ) m <- a[1]*3600000 + a[2]*60000 + a[3]*1000 }

One variable contains values (1.30 - one hour and thirty minutes, 1.2 (which is supposed to be 1.20 - one hour and twenty minutes)). I would like to convert to a minute variable so 1.2 is converted to 80 minutes. How?

Dear List, Does anybody no how to compare mean survival times for two (more) groups in R? What test statistics should I use? Thank you very much! Joe [[alternative HTML version deleted]]