similar to: Downloading file with lapply

Dear Bert and all, I am really sorry. This is the full code: dat<-read.csv("example.csv")attach(dat) sdate<-as.Date(Date, format="%Y-%m-%d") #change the colour of the strip background my.settings <- canonical.theme(color=FALSE) my.settings[['strip.background']]$col <- "blue" dat_key <- list(space="right", ??????????????? lines =

Not reproducible, as we have neither "example.csv" nor "Date" . Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Mon, Sep 11, 2017 at 2:38 PM, Maria Lathouri <mlathouri at yahoo.gr> wrote: > Dear

1. Not reproducible since my.settings, dat_key, sdate not provided. 2. Why did you not try something like ylim = c(-5,80) ? Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Mon, Sep 11, 2017 at 12:33 PM, Maria Lathouri via

I am a bit worried I am reinventing the wheel. Isn't there a calendar system in R? I have written a function to add months to a date and return the number of days resulting. I am newish to R so I am hoping there is a package that can do this sort of date arithmetic for me... Worik DaysInMonths <- function(s,d){ ## Days in d months from s sdate <- MSTD(s) ## Get day, month

Dear all I am trying to make a plot in xyplot lattice by groups. I would like to show "0" in y axis but I don't want to be aligned with the x axis. I want to be a little bit above. I have tried many options but I don't get what I want. I also tried ylim=c(0, 80) but both 0 and 80 are fully aligned with the x-axis and the upper boundary of the plot, respectively: xyplot(upper.zn +

Hi Maria Rule 1 make sure your data is in the right format dat <- source("G:/1/savedat.txt") > dat $value Location Date Zn2 upper.zn lower.zn 1 upstream 2016-04-27 29.92477 55.59800 13.912207 2 spill 2016-04-27 12.84040 22.07006 6.964934 3 downstream 2016-04-27 22.49673 41.60901 11.739109 4 upstream 2016-06-28 23.98425 45.60219 10.690640 5

Dear all, Thank you very much for the help. ylim=c(-5, 80) worked.? Regarding the reproducible example, I used dput () and saved the file as txt. It is not the first time that I used this way and normally it works. Because when I try to attach a .csv file with the data, most of the time it doesn't go through.? I will know for the next time.? Many thanks. Regards,Maria ???? 2:11 ?.?.

Dear R Helpers, I am having difficulty with strptime. I wish to find the differences between two vectors of times. I have apparently no difficulty to convert the vectors to the appropriate format using strptime. But, then difftime does not calculate all the differences. Here is the code and output:- dob=strptime(as.character(datx$BDT),'%d-%b-%y'); dob$year=dob$year-100

Dear Maria The file you attached to your first e-mail did come through but I think some people on the list must have missed it. Michael On 12/09/2017 12:54, Maria Lathouri via R-help wrote: > Dear all, > Thank you very much for the help. ylim=c(-5, 80) worked. > Regarding the reproducible example, I used dput () and saved the file as txt. It is not the first time that I used this way

It would depend on which one of the 9 quantile definitions you are using. The discontinuous ones aren't invertible, and the continuous ones won't be either, if there are ties in the data. This said, it should just be a matter of setting up the inverse of a piecewise linear function. To set ideas, try x <- rnorm(5) curve(quantile(x,p), xname="p") The breakpoints for the

Hi, I see different results in png() and pdf() for polypath() on Windows when using the "winding" rule ## overlapping, both clock-wise x <- cbind(c(.1, .1, .6, .6, NA, .4, .4, .9, .9), c(.1, .6, .6, .1, NA, .4, .9, .9, .4)) pfun <- function() { plot(x) polypath(x * 0.8 + 0.2, rule = "winding", col = "#BEBEBE80") polypath(x, rule =

My x axis is a series of daily dates (e.g., 01/01/2000, 01/02/2000, etc.) from 2000 to end of 2008. The default only gives me 4 ticks. I want more. Why doesn't this work? sdate<-as.POSIXct(strptime(date,format="%m/%d/%Y")) plot(ppt~sdate,type="l",ylim=c(0,47),col=1,lwd=1,pch=16,ylab="Salinity, psu",xlab="Year",las=1,main="Duck Key

Newbie here~ I've spent a ton of time on this, but have to admit I am stuck. I've tried various combinations of strptime and the package CHRON, but still can't get there. My simple task is that I want to plot salinity (S1 or S2) on y-axis and time on x-axis, but I want to use a lay-understandable representation of time (not Julian). If I use the following command, I can make a

Hi Just to clarify, I think this IS a problem with grid.path() as well as polypath(). For the example you give, grid.path() diverts to drawing a polygon (because there is no 'id' specified), and the NAs in 'x' generate two separate polygons, which get drawn one on top of the other. The correct analogy to the polypath() example is ... x2 <- matrix(x[!is.na(x)], ncol=2)

On Thu, 4 Aug 2016 at 17:53 Michael Sumner <mdsumner at gmail.com> wrote: > On Thu, 4 Aug 2016 at 11:17 Paul Murrell <paul at stat.auckland.ac.nz> wrote: > >> Hi >> >> Just to clarify, I think this IS a problem with grid.path() as well as >> polypath(). >> >> > Hi, oh dear - sorry about that > > I appreciate the deeper explanation, I

Dear list, I've been using plotmeans {gplots} a lot before, and found the "connect" argument to be quite useful. I've moved to ggplot for several reasons, but would still like to connect lines conditionally, somehow. Is it possible to do? Small example: df <- data.frame(cat=LETTERS[1:4],num=rnorm(4)) ggplot(df, aes(x=cat, y=num)) + geom_point() +

Hi everyone, I have the following problem: I have three variables, 'group', 'city' and 'pressure' There is an interaction effect between group and city and I'd like to show this in an interaction plot: interaction.plot(group, city, pressure, type="b", col= c(1:2), leg.bty="o", leg.bg="blue", lwd=1,

Dear List, This may be related to this email thread initiated by Ben Bolker last month: https://stat.ethz.ch/pipermail/r-devel/2011-July/061630.html In answering this Question on StackOverflow http://stackoverflow.com/q/7195628/429846 I noticed that `methods()` was not listing some S3 methods for `plot()` provided by the mgcv package. At the time I wanted to check the development version of R as

David, thanks for the response. In your response the quantile function (if I see correctly) runs on the columns versus I need to run it on the rows, which is an easy fix, but that is not exactly what I had in mind... essentially we can remove t() from my original code to make "res" look like this: res<-apply(z, 1, quantile, probs=c(0.3)) but after all maybe I did not explain

Hello, I am having a problem using sqldf. I'm trying to choose a subset of observations from a data set based on the date and maximum value of a variable by date. Here is the code I am using: test<-sqldf("select distinct * from bextuse group by sdate having bext=max(bext)",method="raw"); The result I get back is a data frame with 0 rows and 0 columns. I have tried