search for: newdim

...bind(Admitted,Rejected)~Gender+Dept,data=berkeley,family="binomial") setCoefTemplate(est.se=c(est = "($est:#)($se:#)")) mtable(berk0,berk1,berk2, + coef.style="est.se", + summary.stats=c("Deviance","AIC","N")) Error in dim(ans) <- newdims : dims [product 1] do not match the length of object [2] Thank you in advance. -- Sebasti?n Daza sebastian.daza at gmail.com

...d,Rejected)~Gender+Dept,data=berkeley,family="binomial") setCoefTemplate(est.se=c(est = "($est:#)($se:#)")) mtable(berk0,berk1,berk2, + coef.style="est.se", + summary.stats=c("Deviance","AIC","N")) Error in dim(ans) <- newdims : dims [product 1] do not match the length of object [2] Thank you in advance. -- Sebasti?n Daza sebastian.daza at gmail.com

...ire data set, but only have PRVDR_NUM == 050108. This is where I'm having trouble. I've tried code like this: newinpatient <- subset(oldinpatient, oldinpatient$PRVDR_NUM == 050108) #OR newinpatient <- oldinpatient[oldinpatient$PRVDR_NUM == 050108, ] #OR providernum <- data.frame(newdim(PRVDR_NUM = c(050108)) newinpatient <- merge(providernum, oldinpatient) With checking "class" at one point, I gathered that R interprets PRVDR_NUM as a factor, not a number .. so I've understood a potential reason why I would have errors (with code above). So, I then tried someth...

?? > z <- array(1:24,dim=2:4) > all.equal(f(z,3),f2(z,3)) [1] "Attributes: < Component ?dim?: Mean relative difference: 0.4444444 >" [2] "Mean relative difference: 0.6109091" In fact, > dim(f(z,3)) [1] 2 3 4 > dim(f2(z,3)) [1] 3 4 2 Have I made some sort of stupid error here? Or have I misunderstood what was wanted? Cheers, Bert Bert Gunter

Here is an alternative approach using apply(). Note that with apply() you are reversing rows or columns not indices of rows or columns so apply(junk, 2, rev) reverses the values in each column not the column indices. We actually need to use rev() on everything but the index we are interested in reversing: f2 <- function(a, wh) { dims <- seq_len(length(dim(a))) dims <-