similar to: Password-protect script files

Displaying 20 results from an estimated 2000 matches similar to: "Password-protect script files"

2007 Sep 27
3
Expressing number in percentage
I am wondering if there is any procedure to write a particular value in Percentage format, still maintaining it's numeric character. for example I want to write '.33' as '33%' regards, thanks in advance --------------------------------- [[alternative HTML version deleted]]
2007 Oct 18
5
R-squared value for linear regression passing through origin using lm()
Hi, A have small technical question about the calculation of R-squared using lm(). In a study case with experimental values, it seems more logical to force the regression line to pass through origin with lm(y ~ x +0). However, R-squared values are higher in this case than when I compute the linear regression with lm(y ~ x). It seems to be surprising to me: is this result normal ? Is there
2007 Sep 17
1
Importing a dataset
? stato filtrato un testo allegato il cui set di caratteri non era indicato... Nome: non disponibile Url: https://stat.ethz.ch/pipermail/r-help/attachments/20070917/f6d8d606/attachment.pl
2018 Nov 29
2
Unexpected argument-matching when some are missing
On Thu, Nov 29, 2018 at 1:10 PM S Ellison <S.Ellison at lgcgroup.com> wrote: > > > > > plot(x=1:10, y=) > > > plot(x=1:10, y=, 10:1) > > > > > > In both cases, 'y=' is ignored. In the first, the plot is for y=NULL (so not > > 'missing' y) > > > In the second case, 10:1 is positionally matched to y despite the
2007 Aug 17
2
for plots
Hi, All, I am a beginner for R. Now I have installed R 2.5.1 in Window environment. After I run a program such as "gam" I would like to display a plot for the object. The following is an example. When I did this, only the last plot was presented on my screen. How can I get a plot before the last plot? I mean if the object has several plots how can I get those? "gam.object <-
2007 Aug 06
4
Function for trim blanks from a string(s)?
I feel like an idiot posting this because every language I've ever seen has a string function that trims blanks off strings (off the front or back or both). Ideally, it would process whole data frames/matrices etc but I don't even see one that processes a single string. But I've searched and I don't even see that. There's a strtrim function but it does something completely
2007 Sep 25
3
Legend
I have following syntax for putting a legend : legend("bottom", fill=c("red","blue"), legend=expression(p==0.30, p==0.50), bty="n") However what I want is that : the value "0.30" should be a value of a variable instead of a constant, so that I can put the name of this variable and in legend it's value will be displayed. Can anyone tell me how
2004 Oct 29
4
Installation problems with R.classes bundle
Firstly hi to everyone on the list, I am new to this list ans also R so please forgive the simplicity of my questions over the next few months. I have version R 1.9.1 and want to perform z-scoring for a benchmarking procedure. I have tried to install Henrik Bengtsson's R.classes bundle (http://www.maths.lth.se/help/R/R.classes). I have downloaded the .zip file and placed the relevant folders
2007 Aug 03
2
Opening a script with the R editor via file association (on Windows)
Is there an easy way to open an R script file in the R Editor found in more recent versions of R on Windows via a file association? I am looking for functionality akin to how the ".ssc" file extension works for S-Plus: upon double-clicking on a ".R" file, Rgui.exe starts up and loads the script file in the R Editor. As far as I can tell, Rgui.exe does not have a command line
2007 Oct 10
2
how to generate and evaluate a design using Algdesign
Hi, I have some problems when using AlgDesign->optFederov() generating designs. I have 6 variables, all factors. 3^2 and 4^4, I want to have a design that can take care of main effects and two interactions within 2 pair of variables v3-v4 and v5-v6, the following is the code ################ require(AlgDesign) set.seed(1) levels = c(v1=3,v2=3, v3=4,v4=4,v5=4,v6=4)
2014 Oct 24
1
package checking apparently ok but R-forge version does not build
Dear r developers, I'm writing a set of new functions for an existing R package on R-forge (called COGARCH). I wrote the new R code (but still no documentation), and updated the NAMESPACE file. I installed it from my local repository and everything seems to work. I checked the local repository and the check did not produces errors, but 2 warnings and 2 notes (I attach the log) one of which
2019 Jul 26
1
R evolution suggestion request
Good evening R devel mailing list. Just wonder if there is a way to ask for an evolution of the R language. In affirmative case, let me know. I use R for now many years, and still like it very much. It evolves quite smoothly and improves well. Nevertheless, there is one topic that I still miss in R language. It is pure native R annotation, I mean ? la java. I wish to have an @ operator
2015 Oct 06
5
authorship and citation
> The former co-author contributed, so he is still author and probably copyright > holder and has to be listed among the authors, otherwise it would be a CRAN > policy violation ... It's a bit of a philosophical question right now, but at some point in a developing package's life - particularly one that starts small but is subsequently refactored in growth - there may be no code
2017 Aug 08
0
Latin hypercube sampling from a non-uniform distribution
> However, my variable is simulated from the cumulative distribution function > of the Poisson distribution. Then I am afraid I don't know what you're trying to achieve. Or why. However, the principle holds; write a function that maps [0,1] to the 'pattern' you want, do that and apply it to the result from randomLHS. It happens that for generating data that follow a given
2017 Aug 04
2
Latin hypercube sampling from a non-uniform distribution
Hello, I am performing a sensitivity analysis using a Latin Hypercube sampling. However, I have difficulty to draw a Hypercube sample for one variable. I?ve generated this variable from a Poisson distribution as follows: set.seed(5) mortality_probability <- round(ppois(seq(0, 7, by = 1), lambda = 0.9), 2) barplot(mortality_probability, names.arg = seq(0, 7, by = 1), xlab = "Age
2016 Apr 06
0
Problem with <= (less than or equal): not giving the expected result
> Apparently, abs(1 - 0.95) is not equal to 0.05, which I find however quite > disturbing. It's normal.* See R FAQ 7.31 in the html help system. S Ellison *... and common to all computers that use binary. ******************************************************************* This email and any attachments are confidential. Any use, copying or disclosure other than by the intended
2008 May 09
2
Regarding anova result
  Hi, I fitted tree growth data with Chapman-Richards growth function using nls. summary(fit.nls) Formula: Parameters: Estimate Std. Error t value Pr Signif. codes: 0 ''***'' 0.001 ''**'' 0.01 ''*'' 0.05 ''.'' 0.1 '' '' 1 Residual standard error: 1.879 on 713 degrees of freedom Algorithm
2018 Nov 30
2
Unexpected argument-matching when some are missing
But the main point is where arguments are mixed together: > debugonce(plot.default) > plot(x=1:10, y=, 'l') ... Browse[2]> missing(y) [1] FALSE Browse[2]> y [1] "l" Browse[2]> type [1] "p" I think that's what I fall over mostly: that named, empty arguments behave entirely different from omitting them (", ,") And I definitely agree we need
2018 Aug 01
1
RFC: make as.difftime more consistent or convenient
Hello! you, Emil Bode <emil.bode at dans.knaw.nl>, wrote on Tuesday, July 31, 2018 1:55 PM: > Some of the changes you're proposing could be made (with effort), but note that you're not > restricted to providing strings with a format. > What you're trying to do can be accomplished with as.difftime(12, units='weeks'), see also > ?as.difftime > > Or if
2017 Aug 07
2
Latin hypercube sampling from a non-uniform distribution
Thanks for your answer. However, my variable is simulated from the cumulative distribution function of the Poisson distribution. So, the pattern obtained from the function "qpois" is not the same as the observed pattern (i.e., obtained from the function "ppois") set.seed(5) mortality_probability <- round(ppois(seq(0, 7, by = 1), lambda = 0.9), 2)