similar to: Legend in plots

Dear R-helper, I wonder if anyone can help me. I am trying to convert a dataset to a format recognizable by a software onboard a research vessel but I am having problems with some steps. I have a data frame as follows: Conc Lat Lon Depth Point0000 56.25 -5.65 70 Point0001 56.55 -5.35 85 Point0002 56.25 -5.65 65 Point0003 56.37

Dear R users, I am trying to build a barplot2 graph however I can't find a way of defining the scale for the x-axis. I would like to show in my x-axis only the numbers 0, 25, 50, 75 etc. (so far R is giving me a random scale hard to interpret and it doens't look nice...). Could anyone advise me on how to do this please, it would be a great help! Thank you. Below I show the code I have

Dear R users, I have a very simple problem but I can't find the function in R to deal with it. I need to split (or decompose) one line into many lines using one field as a reference. I have a table with the following format: A B Frequency 23 3 2 24 2 5 25 1 3 And need to split each line into several lines according to the frequency to achieve something like this: A B

Dear R users, I need to find the coordinates for the point (midpoint) located half way between two pairs of coordinates (lon1,lat1 and lon2,lat2) assuming a straight line between them. What would be the best way? I tried to find an answer in the help archives but without success. I would greatly appreciate any help. df<- data.frame(lon1=c(-4.568,-4.3980), lat1=c(59.235,56.369),

Hello, thanks for the help. Yes And the commands to discovery iscsi results ok Such iscsiadm --mode discovery --type sendtargets And scsiadm -m node -T And the disks appear on pvdisplay 2018-03-14 16:23 GMT-03:00 Marcelo Roccasalva < marcelo-centos at irrigacion.gov.ar>: > On Wed, Mar 14, 2018 at 4:08 PM, marcos sr <msr.mailing at gmail.com> wrote: > > > >

Dear all, I am still a R apprentice... Apologies for the basic question. I am trying to reshape a dataframe based on the order of two variables (a character variable and a numerical variable). To simplify it, consider the following dataframe > df<-data.frame(id=c("b","b","a","a","a"),ord=c(2,1,1,3,2)) id ord 1 b 2 2 b 1 3 a 1 4

Thanks for consideration I use this command for update my repo rsync -avzH --delete-after centos.mirror.angkasa.id::centos /home/centos/ --log-file=/var/log/rsync/rsync.log.$(date +%Y%m%d%H%m%S) Best Regard Mohsen On Mon, Jul 10, 2017 at 11:32 PM, Marcelo Roccasalva < marcelo-centos at irrigacion.gov.ar> wrote: > On Mon, Jul 10, 2017 at 3:49 PM, mohsen Abbaspour >

Hello all, As an example, consider the following dataframe > df<-data.frame(id=c("b","b","a","a","a"),ord=c(2,1,1,3,2)) > dates<-as.Date(c("02/27/92", "02/27/92", "01/14/92", "02/28/92", "02/01/92"),"%m/%d/%y") > df$dates<-dates which was ordered with >

Hello all, Can someone tell me why the following mixture of two log-normal distributions does not get truncated? What puzzles me is that the function works almost always, but for certain combinations (like the one below), it does not. # R code example library(distr) mix<-UnivarMixingDistribution(Lnorm(3.2,0.5),Lnorm(5.4,0.6),mixCoeff=c(0.3,0.7))

Hello. I am using R version 2.1.1 on Windows 2000. I am using a par(mfrow=c(2,2)) statement to produce 4 plots on one screen. I want a single horizontal legend to appear at the top of the four plots. My code is something like this: par(mfrow=c(2,2)) plot(x,y1) lines(x,y2) lines(x,y3) plot(x,z1) lines(x,z2) lines(x,z3) plot(x,t1) lines(x,t2) lines(x,t3) plot(x,w1) lines(x,w2) lines(x,w3)

Hi, I want to get a histogram with the legend for my data. I drew a normal density curve and kernel density curve in the histogram, and I also label mean and median in the X axis. From the code, I got two legend: One shows "Normal Density" and "Kernel Density" and their corresponding lines, the other shows "Mean = value" and "Median = value" and their

Hi all, I would like to do some QSAR analysis (quantitative structure activity relationship). I need to use some Partial Least Squares (PLS) regression, but I have not seen this option on the R-project. Is it possible to do this kind of regression on R? thank you in advance best regards, olivier [[alternate HTML version deleted]]

Hi Have this helper method inside a sites_helper.rb module def distritos [ ['0', 'Todos'], ['1', 'Aveiro'], ['2', 'Beja'], ['3', 'Braga'], ['4', 'Brangança'], ['5', 'Castelo Branco'], ['6', 'Coimbra'], ['7',

I am trying to plot data with multiple logical and physical groups using R. Below is a sample of the kind of data I am working with and the desired output. We have a jmp script that can do the same thing, but at ~$200 a year the licensing is counterproductive. Thanks for any help. Data input: Loaded from a CSV file. Physical_1 Logical_1 Logical_2 Data_1 Data_2 x Y Z2

Hi List, Am trying to customize a legend in trellis: Draws 2x5 lines in 5 colors and 2 linetypes. I would like to add two more items to the legend showing the key for the line types above the colored legend. Any suggestions welcome - thanks Herry ############################# #Following example code: library(gregmisc) trellis.device(bg="white") i1=0 i2=-1.89767506 i3=-1.17087085

Hello guys, Does anyone know if it is possible to authenticate an ftp server (proftpd or vsftpd) based LDAP Samba 4? Thanks -- Celso Vianna BSD User: 51318 http://www.bsdcounter.org Palmas/TO

Dear all, I want to update an lm (or glm) object by changing the response variable and I want to do so inside a function. Doing the update outside of a function is straight forward: x <- 1:5 y <- c(1,2,3,3,6) mm <- lm(y~x) y2 <- c(1,3,3,4,6) mm2<- update(mm, y2 ~ .) But I want to make the update inside a function (where the new response variable is declared inside the

Hello, I am trying to do a fit a loglikelihood function with Multivariate distribution via copulas with fitMdvc. The problem is that it doesn't recognize that my beta is a vector of km parameter and when I try to run it it say that the length of my initial values is not the same as the parameter. Can somebody guide me where my mistake is. Thanks, Elisa. #################################

Hi everybody! Is there any package or functions to make Partial Least Squares analysis with R? Thanks a lot Luis -------------------------------------------- sapo.pt/kitadsl -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or

I am getting this message after run: ruby script/server and click in "About your application environment" in the browser. -- Posted via http://www.ruby-forum.com/. -- You received this message because you are subscribed to the Google Groups "Ruby on Rails: Talk" group. To post to this group, send email to rubyonrails-talk-/JYPxA39Uh5TLH3MbocFF+G/Ez6ZCGd0@public.gmane.org To