similar to: Coercing a character zoo to a numeric

Dear R-helpers, Please have a look at the following. f1 is the same as f2 except that it has some values replaced by NA. But it's corresponding file is slightly bigger than the file containing f2. Could someone please tell me if this is an anomaly ? > load("file1") > ls() [1] "f1" > load("file2") > ls() [1] "f1" "f2" > >

Dear Olivier, Many thanks for your reply. Very cool. You know what I thought ? I thought you had modified the A4 sheet size to compute the paperheight and paperwidth ? I wonder if that's another way of proceeding. Best, Ashim On Sat, Nov 18, 2023 at 12:51?PM Olivier Crouzet <olivier.crouzet at univ-nantes.fr> wrote: > > Dear Ashim, > > these are documented in the LaTeX

Dear All, I want a transformation which will make the spread of the response at all combinations of 2 factors the same. See for example : boxplot(breaks ~ tension * wool, warpbreaks) The closest I can do is : spreadLevelPlot(breaks ~tension , warpbreaks) spreadLevelPlot(breaks ~ wool , warpbreaks) I want to do : spreadLevelPlot(breaks ~tension * wool, warpbreaks) But I get : >

Dear All, we need to do : library(car) for the spreadLevelPlot function I forgot to say that. Apologies, Ashim On Sun, Jan 7, 2018 at 10:37 AM, Ashim Kapoor <ashimkapoor at gmail.com> wrote: > Dear All, > > I want a transformation which will make the spread of the response at all > combinations > of 2 factors the same. > > See for example : > >

Dear All, Here is a reprex: > x<- 1:100 > x[-which(x>100)] integer(0) In words, I am finding out which indices correspond to values in x which are greater than 100 ( there are no such items ) . Then I remove those indices. I should get back the x that I started with since there are no items in x which are bigger than 100 . Instead, it is returning an empty vector. Why is this ?

And indeed again (I did not understand your previous question exactly at first), the 'hard-coded' definition of a landscape a4 sheet would therefore be: paperwidth=29.7cm, paperheight=21cm Olivier. On Sat, 18 Nov 2023 13:20:49 +0530 Ashim Kapoor <ashimkapoor at gmail.com> wrote: > Dear Olivier, > > Many thanks for your reply. > > Very cool. > > You know

Dear R-helpers, Please look at the following minimal code. \documentclass[a4paper]{article} \begin{document} <<>>== library(zoo) a<-zoo(1:4,order.by=Sys.time()+1:4) str(a) @ \end{document} When I do R CMD Sweave,followed by pdflatex ,and view the final pdf, the letter surrounding the phrase zoo, a ***zoo*** series are messed up. They are messed up even inside Emacs now that I

Dear Olivier, Many thanks for your reply. This works well for me. How did you come up with the pagewidth / pageheight numbers? I do understand that their ratio = 16:9, but how did you choose these numbers? Best Regards, Ashim On Fri, Nov 17, 2023 at 9:25?PM Olivier Crouzet <olivier.crouzet at univ-nantes.fr> wrote: > > Dear Ashim, > > I don't think the aspectratio is

Dear all, I have posted a query which has received a response but that is not working on my computer. Here is the query: https://stackoverflow.com/questions/77387434/pdf-from-rmarkdown-landscape-and-aspectratio-169 Can someone please help me ? Best Regards, Ashim

Dear Ashim, these are documented in the LaTeX 'geometry' package (see for example on CTAN: https://ctan.org/pkg/geometry). As I added in my response on Stackoverflow, several parts in the RMarkdown header actually concern information that are processed by LaTeX to actually generate the PDF, among which the 'geometry' line. For someone who is used to working with LaTeX, it is

Dear All, I am trying to use the function ses from the forecast package. >From its help I have : Usage: ses(y, h = 10, level = c(80, 95), fan = FALSE, initial = c("optimal", "simple"), alpha = NULL, lambda = NULL, biasadj = FALSE, x = y, ...) My query is that if I do not mention the initial value will its default value be "optimal". A MWE would be

(Sorry, sent the message before I finished it) Hello, everyone I have a long script that uses zoo objects. In this script I used simple moving averages and these I can very efficiently calculate with filter() functions. Now, I have to use special "exponential" moving averages, and the only way I could write the code was with a for-loop, which makes everything extremely slow. I don't

Dear All, I had done an automatic upgrade of my Debian 10 system which had also upgraded R. I reinstalled the stargazer package and the error went away. Query : Do I need to reinstall all packages with each upgrade of R ? Best, Ashim On Thu, Jun 8, 2023 at 11:11?AM Ashim Kapoor <ashimkapoor at gmail.com> wrote: > > Dear All, > > Here is my reproducible example: > >

Hello, everyone I have a long script that uses zoo objects. In this script I used simple moving averages and these I can very efficiently calculate with filter() functions. Now, I have to use special "exponential" moving averages, and the only way I could write the code was with a for-loop, which makes everything extremely slow. I don't know how to optimize the code, but I need to

On Thu, 08 Jun 2023, Ashim Kapoor writes: > Dear All, > > I had done an automatic upgrade of my Debian 10 system which had also > upgraded R. > > I reinstalled the stargazer package and the error went away. > > Query : Do I need to reinstall all packages with each upgrade of R ? > > Best, > Ashim Your session info says "stargazer 5.2.2". What version

Dear all, I have an error message, when I try to convert a zoo object (called test) to ts (on R 2.4.0, Package zoo version 1.2-1, Windows XP) > test 1994-05-10 1994-06-09 1994-07-09 0.0024943889 0.0024881824 0.0006955831 > str(test) atomic [1:3] 0.002494 0.002488 0.000696 - attr(*, "index")=Class 'Date' num [1:3] 8895 8925 8955 > is.regular(test) [1] TRUE

Dear Ashim, I?ll address your questions briefly but they?re really not appropriate for this list, which is for questions about using R, not general statistical questions. (1) The relevant distribution is within cells of the wool x tension cross-classification because it?s the deviations from the cell means that are supposed to be normally distributed with equal variance. In the warpbreaks data

My problem is the following : I create 2 zoo objects and then I try to subset one of them using logic. indicesthatpass is a vector of trues and falses but when I send it into bckret, it returns an empty bckret. Obviously it has something to do with bckret being a zoo object and if I do the same subsctripting off of coredata(bckret), I'm confident it will work. But, I need to keep the minute

Look at which(x>100) This is a zero-length vector. The negative of nothing is nothing, not a list of all possible index values. Do you want x[ !( x > 100 ) ] ? On April 18, 2018 6:13:30 AM CDT, Ashim Kapoor <ashimkapoor at gmail.com> wrote: >Dear All, > >Here is a reprex: > >> x<- 1:100 >> x[-which(x>100)] >integer(0) > >In words, I am

Dear Ashim, Try spreadLevelPlot(breaks ~ interaction(tension, wool), data=warpbreaks) . I hope this helps, John ----------------------------- John Fox, Professor Emeritus McMaster University Hamilton, Ontario, Canada Web: socialsciences.mcmaster.ca/jfox/ > -----Original Message----- > From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Ashim > Kapoor > Sent: