Thank you Rui!
that works as I want it... :)
/Johannes
On Fri, Sep 28, 2012 at 12:30 PM, Rui Barradas <ruipbarra...@sapo.pt> wrote:
> Hello,
>
> Try the following.
>
>
> f <- function(x, y, ...,
> alternative = c("two.sided", "less", "greater"), exact = NULL){
> #w <- getOption("warn")
> #options(warn = -1) # ignore warnings
> p <- ks.test(x, y, ..., alternative = alternative, exact =
> exact)$p.value
> #options(warn = w)
> p
> }
>
> n <- 1e1
> dat <- data.frame(X=rnorm(n), Y=runif(n), Z=rchisq(n, df=3))
>
> apply(dat, 2, function(x) apply(dat, 2, function(y) f(x, y)))
>
> Hope this helps,
>
> Rui Barradas
> Em 28-09-2012 11:10, Johannes Radinger escreveu:
>>
>> Hi,
>>
>> I have a dataframe with multiple (appr. 20) columns containing
>> vectors of different values (different distributions).
>> Now I'd like to create a crosstable
>> where I compare the distribution of each vector (df-column) with
>> each other. For the comparison I want to use the ks.test().
>> The result should contain as row and column names the column names
>> of the input dataframe and the cells should be populated with
>> the p-value of the ks.test for each pairwise analysis.
>>
>> My data.frame looks like:
>> df <- data.frame(X=rnorm(1000,2),Y=rnorm(1000,1),Z=rnorm(1000,2))
>>
>> And the test for one single case is:
>> ks <- ks.test(df$X,df$Z)
>>
>> where the p value is:
>> ks[2]
>>
>> How can I create an automatized way of this pairwise analysis?
>> Any suggestions? I guess that is a quite common analysis (probably with
>> other tests).
>>
>> cheers,
>> Johannes
>>
>> ______________________________________________
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>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
>
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