need help with chisq
Thank you Stephanie. This is perfect!
On May 10, 12:34?pm, Stephanie Kovalchik <sko... at 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)$p.value
}
apply(X,1,f.chisq)
Quoting JC <jerome.co... at gmail.com>:
I am very new to R. I have some data from a CVS stored in vdata with 4 columns labeled: X08, Y08, X09, Y09.
I have created two new "columns" like so:
Z08 <- (vdata$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!
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______________________________________________ R-h... at 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.