similar to: Error bar:Problem Fixed

Hi there, Using the code below I generated the plot attached. The error bars at day = -3, 2 and 4 appear are larger. I was thinking if there is a way I could make all the error bars to be of the same size. I don't know if that makes sense. If not, then, is there a way I can plot only these 3 largest errorbars at day axis (day = -3, 2 and 4) and omit the rest of the error bars. I would be much

Good day all. This is to thank all those who have helped in fixing this problem. Starting with a text book was indeed a problem, however, that gave me a clue of what I was looking for. This, with your contributions added to other materials I got on the net, put me on the right track. Thank you so much. Warmest regards Ogbos On 31 January 2010 14:07, S Ellison <S.Ellison@lgc.co.uk> wrote:

Hi Ogbos, Here is your sample data plotted in roughly the same way as the image. You can get the hours to start at the bottom, but it will require more code. swe$hour<-as.numeric(sapply(strsplit(swe$time,":"),"[",1)) swe$FD<-sample(1:2,nrow(swe),TRUE) library(plotrix) clock24.plot(swe$count,swe$hour,rp.type="s",radial.lim=c(3000,4010),

?s 01:43 de 29/03/2024, Ogbos Okike escreveu: > Dear Rui, > Thanks again for resolving this. I have already started using the version > that works for me. > > But to clarify the second part, please let me paste the what I did and the > error message: > >> set.seed(2024) >> data <- data.frame( > + Date = sample(seq(Sys.Date() - 5, Sys.Date(), by = "1

Dear Rui, Thanks again for resolving this. I have already started using the version that works for me. But to clarify the second part, please let me paste the what I did and the error message: > set.seed(2024) > data <- data.frame( + Date = sample(seq(Sys.Date() - 5, Sys.Date(), by = "1 days"), 100L, + TRUE), + count = sample(10L, 100L, TRUE) + ) > > # coerce

Hi R masters! I need a Help with plot confidence intervals for one equation. I use library gplots and plotCI command in this script: require(gplots) ano <-1980:2002 rf<-exp(91.37162-0.04720281*ano) ciw.f<-sqrt(1.766073e-08) plotCI(ano,rf,uiw=ciw.f) But in the graph not shown the errors bar and I have this error msg zero-length arrow is of indeterminate angle and so skipped

Dear list members, I would like to create two series of plotted mean values and error bars, yet with different locations along the x-axis. Plotting of first series using plotCI with the standard arguments goes without any problem. However I do not succeed to add the second series in the same plot, which should be horizontally shifted from the first series along the x-axis. The “add=TRUE”

Dear All, This problem is over. Clock24.plot did the job. Thanks to all those who assisted me. Ogbos On Apr 22, 2016 8:34 PM, "Ogbos Okike" <giftedlife2014 at gmail.com> wrote: > Dear All, > One hand. Many thanks!! The code run as soon as I loaded lubridate. > > Please can you guide me on how to relate this code to my actual data. > My actual data is looking like:

Hello people, A while back I wanted to plot boxplots with interactions that will have a dot for the mean of the sample + bars for the SE. After searching for some code, I found something that did it for one level, but couldn't find something that will allow for interactions the way the original boxplot does. After playing with the original code, I found a way for allowing the boxplot code to

I've shared a document with you: plot.png http://docs.google.com/leaf?id=0B_J5_yv8GrERYWRhZjU4NDItMmMyOS00ZGIxLWE2ZDUtMDFiYTY3MzliMzhk&hl=en&invite=CI2a3aYC It's not an attachment -- it's stored online at Google Docs. To open this document, just click the link above. [[alternative HTML version deleted]]

Hello, I am trying to calculate and plot mean and confidence intervall for a set of data. This is the code that I am currently using: means <- sapply(data, mean, na.rm=TRUE) n <- sapply(data,length) stdev <- sqrt(sapply(data, var, na.rm=TRUE)) ciw <- qt(0.98, n) * stdev / sqrt(n) par(mgp=c(2,0.6,0), las=2, fin=c(7,3), mai=c(1,0.5,0.2,0.2), cex=0.8) plotCI(x=means, uiw=ciw,

Dear All, Many thanks for bailing me out. Ogbos On Apr 18, 2016 9:07 PM, "David Winsemius" <dwinsemius at comcast.net> wrote: > > > On Apr 18, 2016, at 10:44 AM, Ogbos Okike <giftedlife2014 at gmail.com> > wrote: > > > > Dear ALL, > > Thank you so much for your contributions. > > I have made some progress. Below is a simple script I

Hi Ogbos, You can just use ISOdate. If you pass more values, it will process them: ISOdate(2018,01,22) [1] "2018-01-22 12:00:00 GMT" > ISOdate(2018,01,22,18,17) [1] "2018-01-22 18:17:00 GMT" Add something like: if(is.null(data$hour),data$hour<-12 then pass data$hour as it will default to the same value as if you hadn't passed it. Jim On Mon, Jan 22, 2018 at 6:01

Looks like you forgot to load the lubridate package library(lubridate) You are calling functions days(), hours(), minutes(), seconds(), and hour() which all come from that package. ------------------------------------- David L Carlson Department of Anthropology Texas A&M University College Station, TX 77840-4352 -----Original Message----- From: R-help [mailto:r-help-bounces at

Dear Christophe, Yes. It was helpful indeed. The four steps you outlined successfully helped me to install the newest version of R. I have been fiddling with this for days and I am quite grateful to all who contributed. I formally posted to r-help@r-project.org before somebody suggested posting to r-sig-debian@r-project.org. Best regards Ogbos 2011/5/31 Christophe Bonenfant

Dear Christophe, Yes. It was helpful indeed. The four steps you outlined successfully helped me to install the newest version of R. I have been fiddling with this for days and I am quite grateful to all who contributed. I formally posted to r-help@r-project.org before somebody suggested posting to r-sig-debian@r-project.org. Best regards Ogbos 2011/5/31 Christophe Bonenfant

Dear list, Thanks for your time. I was surprised. While I was still waiting for your response, I entered R and discovered that it has been updated!!!! I hurried logged out and went home since I had overstayed battling with it. I wonder how the update happened with the numerous errors that were reported. I wish I will be able to find out, otherwise, I am off with my analysis with the new version :)

Hi, A terrific resource for this type of issue (and pretty much anything related to R) is http://rseek.org/ I'm sure I use it at least daily. Check out ... http://rseek.org/?q=pointsToRaster The first hit is about pointsToRaster() - it has been replaced by raster::rasterize() Cheers, Ben > On Apr 30, 2016, at 4:28 AM, Ogbos Okike <giftedlife2014 at gmail.com> wrote: > >

Hi Uwe, I am pleased to inform you that this problem has, following your hint, been resolved. I installed 64-bit version of R and I was alright. Thanks again for your time. Ogbos 2009/6/24 Uwe Ligges <ligges@statistik.tu-dortmund.de> > > > ogbos okike wrote: > >> Hi, >> I have a data of size 981.1MB(1028707715) and I intend to calculate the >> length of the

Hi Peter, Many thanks. It worked. Regards Ogbos On 1 August 2011 14:05, Peter Ehlers <ehlers@ucalgary.ca> wrote: > On 2011-08-01 03:32, ogbos okike wrote: > >> Dear All, >> I am trying to put 10^-8 st km^-2day^-1 on x-axis of my plot. I tried >> using >> : ylab = expression(paste("st / ", plain(km)^2, " / day")) to see if I can >> at