rank

I'd like to calculate the rank of each index within a vector, e.g: x <- c(0.82324952352792, 0.11953364405781, 0.588659686036408, 0.41683742380701, 0.11452184105292, 0.438547774450853, 0.586471405345947, 0.943002870306373, 0.28184655145742, 0.722095313714817) calcRank <- function(x){ sorted <- x[order(x)] ranks <- sapply(x, function(x) which(sorted==x)) return(ranks) } calcRank(x) > calcRank(x) [1] 9 2 7 4 1 5 6 10 3 8 Is there a better way to do this? Why not just: rank(x) # ..... ? # [1] 9 2 7 4 1 5 6 10 3 8 match is what you want: match(x, sort(x)) 来源： https://stackoverflow.com/questions

In this post it is stated that order(order(x)) is the same as rank(X) . While some experiments corroborate this... #why is order(order(x)) == rank(x)? x <- c(0.2, 0.5, 0.1) x ## [1] 0.2 0.5 0.1 order(x) ## [1] 3 1 2 rank(x) ## [1] 2 3 1 order(order(x)) ## [1] 2 3 1 I fail to see how this can be proved and, even better, intuited. Related: rank and order in R First off look at what happens with an integer sequence formed from a permutation of 1:10: > set.seed(123); x <- sample(10) > x [1] 3 8 4 7 6 1 10 9 2 5 > order(x) [1] 6 9 1 3 10 5 4 2 8 7 > order(order(x)) [1] 3 8 4 7 6 1 10 9 2 5 > rank(x

I read the documentation in R for wilcox.test() and want to determine: How R computes wilcox.test() The docs say that when the number of samples is small, it does the test exactly (instead of using a normal approximation) -- what tables does it use to do this exactly? wilcox.test.default is "hidden" in the stats package's namespace. That's why you need to do getAnywhere("wilcox.test.default") or stats:::wilcox.test.default to view it. Package exactRankTests can do exact wilcox rank sum tests. 来源： https://stackoverflow.com/questions/12395110/how-does-r-do-exact-wilcox-rank-sum-tests