Greg's solution is most elegant (I think). This is more of an illustrative approach:
set.seed(1) #to replicate identical sampling if you use this code x=sample(1:100,replace=T) x=rev(sort(x)) #reverse order sum(x)/2 # what is the mean of x: 2613.5 cumsum(x) # the cumulative sums for x=1:i all.greater=which(cumsum(x)>sum(x)/2) # which cumulative sums are greater than your critical value first=min(all.greater) #what is the first cumulative sum that is greater than your critical value: 32 x[first] # what is x at "first": 72 #or integrating the above: x[min(which(cumsum(x)>sum(x)/2))] Cheers, Daniel ------------------------- cuncta stricte discussurus ------------------------- -----Ursprüngliche Nachricht----- Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im Auftrag von Jeremy Leipzig Gesendet: Tuesday, November 25, 2008 12:45 PM An: r-help@r-project.org Betreff: Re: [R] calculating an N50 > Given a set of integers of different values how do I calculate the > minimum number of the largest of integers that are required, when > summed, to equal 50% of the total sum of the the set? > Actually I need the value of the smallest member such that the sum of all members equal or greater to that is 50% of the total sum of the set ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. ______________________________________________ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
