similar to: pca in R: Problem Fixed

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),

Hi, I am learning how to do principal component analysis in R. However, since I am family with only a few built-in functions like prcomp, sd, cor, I started manually with examples in text books while trying to use the few functions I know to manipulate what they have in the text. From the example in the text I obtained a data set. Using cor and cov, I calculated the correlation and covariance of

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:

?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 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: > >

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

Great stuff! Both plotrix and plotCI do the same but what count most here is the idea of assigning NA to some of the ciw values. Following your code, I assign "ciw[c(1,2,4,5,6,7,9,11)]<-NA" and that left me with the three error bars I wanted to display. Many thanks and warmest regards Ogbos On 7 April 2010 00:28, Jim Lemon <jim@bitwrit.com.au> wrote: > On 04/07/2010 03:49

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

Hey, Thanks for your great inputs. While "index = apply(mat == 0, MARGIN = 1, any)" gives you an idea of the rows containing zero(s), "index<-data[!apply(data==0,MARGIN=1,any),]" does the actual job of removing the rows with zeros. Kind regards Ogbos On 9 March 2010 15:12, Paul Hiemstra <p.hiemstra@geo.uu.nl> wrote: > Dimitris Rizopoulos wrote: > >>

Dear ALL, Thank you so much for your contributions. I have made some progress. Below is a simple script I gleaned from your kind responses: Sys.setenv(TZ="Etc/GMT") dates <- c("02/27/92", "02/27/92", "01/14/92", "02/28/92", "02/01/92") times <- c("23:0:0", "22:0:0", "01:00:00", "18:0:0",

After doing a PCA using princomp, how do you view how much each component contributes to variance in the dataset. I'm still quite new to the theory of PCA - I have a little idea about eigenvectors and eigenvalues (these determine the variance explained?). Are the eigenvalues related to loadings in R? Thanks, Paul -- View this message in context:

Hey Duncan, Thank so much for the site. Quite helpful. Containing everything and informing me that raster has got no older version. I have also installed the new R version and I am fine. Regards Ogbos On 5 October 2010 15:11, Duncan Murdoch <murdoch.duncan@gmail.com> wrote: > On 05/10/2010 9:00 AM, ogbos okike wrote: > >> Hello, >> The R version on my system is R version

The most important thing is that Date objects by definition do not include time of day. You want to look at ISOdatetime() and as.POSIXct() instead. And beware daylight savings time issues. -pd On 18 Apr 2016, at 15:09 , Ogbos Okike <giftedlife2014 at gmail.com> wrote: > Dear All, > > I have a data set containing year, month, day and counts as shown below: > data <-