Calculating Kendall's tau
On Apr 2, 2015, at 7:06 AM, S Ellison wrote:
I am analyzing trend using Mann-kendall test for 31 independent sample, each sample have 34 years dataset. I supposed to find Kendall ?tau? for each sample. The data is arranged in column wise (I attached the data).To find Kendall tau, I wrote R script as: ... Anyone can tell me how can I get orderly displayed ?tau? value?
Usually, in R, a hypothesis test returns an object, and you can extract an individual element of that object. MannKendall seems to be no exception. Looking at the help page, a MannKendall test returns... " A list with class Kendall. tau Kendall?s tau statistic sl two-sided p-value S Kendall Score D denominator, tau=S/D varS variance of S" To get just tau, say something like MannKendalltau[i]<-MannKendall(y[,i])$tau But your code is a bit of a mess.... MannKendalltau<- numeric(nc) simply makes MannKendalltau a single integer equal to nc; that doesn't look sensible when the next thing you do is treat MannKendalltau as a vector. R's been kind to you and extended MannKendalltau when you tried to add things to later, non-existent, elements, but it clearly wasn't the right thing to do. Look up ?numeric, and then look up ?vector for next time you want to set up an empty vector. Second, since MannKendall(y[,i]) ) returns a list object of class Kendall, MannKendalltau[i]<-MannKendall(y[,i]) assigns a whole object containing 5 values to each new element of your MannKendalltau. So your result is a list of lists. Finally, you don?t need a loop at all. On a data frame, sapply would work nicely, so (although I've not tested it) something like sapply(desta[,2:nc], 2, function(x) ManKendall(x)$tau)
That looks more line an apply call. The second argument to sapply needs to be a function or function name.
David.
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> ought to do the whole thing in one shot and package it nicely into a named vector while it's about it.
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> S Ellison
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