Thank you Stephanie. This is perfect!
On May 10, 12:34 pm, Stephanie Kovalchik <sko...@ucla.edu> wrote:
> JC,
>
> If each row are the counts for a 2 x 2 contingency table - so for the
> ith contingency table you have counts for row 1 c(Y08[i],Z08[i]) and
> row 2 (Y09[i],Z09[i]) then you could use apply:
>
> X <- cbind(vdata$Y08,vdata$X08-vdata$Y08,vdata$Y09,vdata$X09-vdata$Y98)
>
> f.chisq <- function(x){
> m <- matrix(x,2,2)
> chisq.test(m)$p.value
>
> }
>
> apply(X,1,f.chisq)
>
> Quoting JC <jerome.co...@gmail.com>:
>
>
>
>
>
> > I am very new to R. I have some data from a CVS stored in vdata with 4
> > columns labeled:
> > X08, Y08, X09, Y09.
>
> > I have created two new "columns" like so:
>
> > Z08 <- (vdata$X08-vdata$Y08)
>
> > Z09 <- (vdata$X09-vdata$Y09)
>
> > I would like to use chisq.test for each "row" and output the p-value
> > for each in a stored variable. I don't know how to do it. Can you
> > help?
>
> > so far I have done it for one row (but I want it done automatically
> > for all my data):
>
> > chidata=rbind(c(vdata$Y08[1],Z08[1]),c(vdata$Y09[1],Z09[1]))
> > results <- chisq.test(chidata)
> > results$p.value
>
> > I tried removing the [1] and the c() but that didn't work... Any
> > ideas?
>
> > THANKS!
>
> > ______________________________________________
> > r-h...@r-project.org mailing list
> >https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guidehttp://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
>
> ______________________________________________
> r-h...@r-project.org mailing listhttps://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guidehttp://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
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