R help - Aug 2011 - how to start R script editor by default

Hi All,
 
1) Is it possible to set the options such that R opens a new script editor
every time I start the R and 2) specify the size of windows.
 
Thanks for the suggestion and Best regards,
 
Krishna

 

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On 08/29/2011 08:03 PM, SNV Krishna wrote:
> Hi All,
>
> 1) Is it possible to set the options such that R opens a new script editor
> every time I start the R and 2) specify the size of windows.
>
Hi Krishna,
You can start an editor like this:

system("my_editor",wait=FALSE)

where "my_editor" is the name of your favorite editor. Adding this
line
to your .First function will start that editor when you start R. Getting 
a particular window size depends upon whether you can specify the size 
on the command line. Say you're using NEdit. You could do something like 
this:

cat("How many rows, Krishna?")
rows<-scan(n=1)
cat("How many columns, Krishna?")
columns<-scan(n=1)
system(
  paste("nedit
-rows",rows,"-columns",columns,collapse=" "),
  wait=FALSE)

Jim

I did a simple little simulation of a binary variable in a two armed trial. I
was quite surprised by the number of p-values delivered by the fisher.test
function which was >1(!). Of course, under the null hypothesis you expect a
fair number of outcomes with the same number of event in both arms but still?

Is there some silly error in my crude code?

##########################################################################################

niter<-50000
ra<-rbinom(niter,100,.05)

rb<-rbinom(niter,100,.05)

pval<-rep(NA,niter)


for (i in 1:niter){
apa<-matrix(c(100-ra[i],ra[i],100-rb[i],rb[i]),byrow=T,ncol=2)
pval[i]<-fisher.test(apa)$p.value

}


cbind(ra,rb,pval)[pval < 0.06 & pval > 0.04,]
hist(pval,probability=T)
summary(pval)

table(pval< 0.05)/niter

sum(pval>1)/niter

################################################################################################



Patrik ?hagen
Biostatistiker
Enheten f?r effekt och s?kerhet 4

Box 26, 751 03 Uppsala
Bes?ksadress: Dag Hammarskj?ldsv?g 42
Telefon: 018 - 17 49 24, v?xel: 018 - 17 46 00
Fax: 018 - 54 85 66
patrik.ohagen at mpa.se
www.lakemedelsverket.se

On 29-Aug-11 11:44:28, ?hagen Patrik wrote:
> I did a simple little simulation of a binary variable in a two armed
> trial. I was quite surprised by the number of p-values delivered by the
> fisher.test function which was >1(!). Of course, under the null
> hypothesis you expect a fair number of outcomes with the same number of
> event in both arms but still?
> 
> Is there some silly error in my crude code?
> 
>#########################################################################
>################
> 
> niter<-50000
> ra<-rbinom(niter,100,.05)
> 
> rb<-rbinom(niter,100,.05)
> 
> pval<-rep(NA,niter)
> 
> 
> for (i in 1:niter){
> apa<-matrix(c(100-ra[i],ra[i],100-rb[i],rb[i]),byrow=T,ncol=2)
> pval[i]<-fisher.test(apa)$p.value
> 
> }
> 
> 
> cbind(ra,rb,pval)[pval < 0.06 & pval > 0.04,]
> hist(pval,probability=T)
> summary(pval)
> 
> table(pval< 0.05)/niter
> 
> sum(pval>1)/niter
> 
>########################################################
> Patrik ?hagen
After reading your posting, and being puzzled by "I was
quite surprised by the number of p-values delivered by the
fisher.test function which was >1(!).", I ran your code.
In each of three runs I got sum(pval>1) = 0.

It would indeed be surprising to get any pval > 1, since
the Fisher P-value is the sum of a subset of the probabilities
possible for the table. This subset may sometimes be all of
them, but even so (unless there was an unusual rounding error)
pone should not see pval > 1.

There are certainly many P-values equal to 1. On my third sun
(of 50000) I get

  sum(pval==1)
  # [1] 11520

  sum(pval==1)/niter
   [1 ] 0.2304


The probability of pval=1 in an interation is *at least* the
probability (in your setup of the tables) that ra[i] = rb[i],
which would be the probability that two independent binomial
samples of size 100 with p=0.05  should give the same result,
which is

sum((dbinom((0:100),100,0.05))^2)   which = 0.1307316

*"at least"* because, depending on the configuration of the
random 2x2 table, there are other possibilities for the Fisher
P-value to equal 1.

So the large number of P-values equal to 1 (as can be clearly
seen from the histogram) is not a surprise.

I am, therefore wondering if you really observed any pval > 1?
Did you confuse "pvale == 1" with "pval > 1" in your
posting?
If you really did get any "pval > 1", how many did you get?

Ted.

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Date: 29-Aug-11                                       Time: 15:47:12
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