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...e are a couple of *applys (including mApply in Hmisc) but I haven't found out how to do it straightforward. Applying to row indices works, but looks like a poor hack to me: sens <- function(test, gold) { if (any(gold==1)) { sum(test[which(gold==1)]/sum(which(gold==1))) } else NA } numtest <- 6 numsubj <- 20 newtest <- array(rbinom(numtest*numsubj, 1, .5), dim=c(numsubj, numtest)) goldstandard <- array(rbinom(numtest*numsubj, 1, .5), dim=c(numsubj, numtest)) t(sapply(1:nrow(newtest), function(i) { sens(newtest[i,], goldstandard[i,])})) Is there any shortcut...

Hi, I am wondering if anyone can suggest how to test the equality of 2 proportions. The caveat here is that the 2 proportions were calculated from the same number of samples using 2 different tests. So essentially we are comparing 2 accuracy rates from same, say 100, samples. I think this is like a paired test, but don't know if really we need to consider the "paired" nature of the

...be of same length!") } else if(any(testvec, na.rm=TRUE)) { applvec[min(which(testvec)) : length(applvec)] <- NA } applvec } fillafter <- function(applvec, testvec=NULL) { if (is.null(testvec)) testvec <- applvec fillfrom(applvec, c(FALSE, testvec[-length(testvec)])) } numtest <- 6 numsubj <- 20 newtest <- array(rbinom(numtest*numsubj, 1, .5), dim=c(numsubj, numtest)) goldstandard <- array(rbinom(numtest*numsubj, 1, .5), dim=c(numsubj, numtest)) newtest.NA <- t(sapply(1:nrow(newtest), function(i) { fillafter(newtest[i,], goldstandard[i,]==1)}))...

...er. Is there any reason, perhaps to do with random number generation in R or the nature of the normal distribution simulated by the rnorm function that could explain this depletion? Here are two key parts of my code to show what functions I'm working with: #Calculating the p values while(i<numtests){ Group1<-rnorm(6,-0.0065,0.0837) Group2<-rnorm(6,-0.0065,0.0837) PV<-t.test(Group1,Group2)$p.value pscoresvector<-c(PV,pscoresvector) i<-i+1 } #Binning the results freqtbl1<-binning(pscoresvector,breaks=bins) Thanks in advance for any insights, Andrew [[alternative HTML ver...

...KEY; by A B C; run; I tried using "aggregate" to find the maximum TS for each combination of A/B/C, but it's slow. I also tried "by" but it's also slow. My current (faster) solution is: DF$abc<-paste(DF$A,DF$B,DF$C,sep="") abclist<-unique(DF$ABC) numtest<-length(abclist) maxTS<-rep(0,numtest) for(i in 1:numtest){ maxTS[i]<-max(DF$TS[DF$abc==abclist[i]],na.rm=TRUE) } maxTSdf<-data.frame(device=I(abc),maxTS=maxTS ) DF<-merge(DF,maxTSdf,by="abc",all.x=TRUE) DF<-Df[DF$TS==DF$maxTS,,drop=TRUE] DF$maxTS<-NULL This...

...ec)) : length(applvec)] <- NA > > } > > applvec > > } > > > > fillafter <- function(applvec, testvec=NULL) { > > if (is.null(testvec)) testvec <- applvec > > fillfrom(applvec, c(FALSE, testvec[-length(testvec)])) > > } > > > > numtest <- 6 > > numsubj <- 20 > > > > newtest <- array(rbinom(numtest*numsubj, 1, .5), > > dim=c(numsubj, numtest)) > > goldstandard <- array(rbinom(numtest*numsubj, 1, .5), > > dim=c(numsubj, numtest)) > > > > newtest.NA <- t(sapply(1:n...