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.
>
> ought to do the whole thing in one shot and package it nicely into a named
> vector while it's about it.
>
> S Ellison
>
>
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