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:
? ? ?desta<-read.csv("rainfall.csv", header=T, sep=",")?
? ?require(Kendall)? ? ? ? ? ? ? MK<-function(y) {? ? ? ? ? ? ? ?
?nc<-ncol(y)? ? ? ? ? ? ? ? ?MannKendalltau<- numeric(nc)? ? ? ? ? ? ? ?
?for(i in 2:nc){? ? ? ? ? ? ? ? ? ? ? ? ?
MannKendalltau[i]<-MannKendall(y[,i])?????????? }?????
??????MannKendalltau??? }? ? MK(desta)
The ?displayed result showed ?both ?tau?? and ?2-sided p-value?in unorganized
way. ?But, I want only ?tau? value that is presented in organized ?manner.
Anyone can tell me how can I get orderly displayed ??tau? value? here is my
sample result:?? ? ?[[1]][1] 0
[[2]][1] 0.4352941attr(,"Csingle")[1] TRUE
[[3]][1] 0.5462185attr(,"Csingle")[1] TRUE
[[4]][1] 0.4218487attr(,"Csingle")[1] TRUE....Thank you for your
guidance?
[[alternative HTML version deleted]]
This sounds like homework. Homework is discouraged on this list (but you might get lucky). Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 "Data is not information. Information is not knowledge. And knowledge is certainly not wisdom." Clifford Stoll On Tue, Mar 31, 2015 at 9:08 AM, Desta Yoseph via R-help <r-help at r-project.org> 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: > desta<-read.csv("rainfall.csv", header=T, sep=",") require(Kendall) MK<-function(y) { nc<-ncol(y) MannKendalltau<- numeric(nc) for(i in 2:nc){ MannKendalltau[i]<-MannKendall(y[,i]) } MannKendalltau } MK(desta) > The displayed result showed both ?tau? and ?2-sided p-value?in unorganized way. But, I want only ?tau? value that is presented in organized manner. Anyone can tell me how can I get orderly displayed ?tau? value? here is my sample result: [[1]][1] 0 > [[2]][1] 0.4352941attr(,"Csingle")[1] TRUE > [[3]][1] 0.5462185attr(,"Csingle")[1] TRUE > [[4]][1] 0.4218487attr(,"Csingle")[1] TRUE....Thank you for your guidance > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see > 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.
OK. But always reply to the list (which I am ccing here) so that everyone knows -- and re-submit your OP in **PLAIN TEXT**, not html, as this is a plain text list and html typically garbles everything. Also, reading and following the posting guide (see end of this email) generally improves your chance of getting useful help. -- Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 "Data is not information. Information is not knowledge. And knowledge is certainly not wisdom." Clifford Stoll On Tue, Mar 31, 2015 at 9:24 AM, Desta Yoseph <desta_yo at yahoo.com> wrote:> Dear Bert, > It is not homework. Actually my real work is for 10,360 sample data. But if > some one showed me for 31 sample dataset, i can manage for large sample > data. > hopefully this give you few hint why i really want someone help. > cheers > > > > On Tuesday, March 31, 2015 6:14 PM, Bert Gunter <gunter.berton at gene.com> > wrote: > > > This sounds like homework. Homework is discouraged on this list (but > you might get lucky). > > Cheers, > Bert > > Bert Gunter > Genentech Nonclinical Biostatistics > (650) 467-7374 > > "Data is not information. Information is not knowledge. And knowledge > is certainly not wisdom." > Clifford Stoll > > > > > On Tue, Mar 31, 2015 at 9:08 AM, Desta Yoseph via R-help > <r-help at r-project.org> 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: >> desta<-read.csv("rainfall.csv", header=T, sep=",") >> require(Kendall) MK<-function(y) { nc<-ncol(y) >> MannKendalltau<- numeric(nc) for(i in 2:nc){ >> MannKendalltau[i]<-MannKendall(y[,i]) } MannKendalltau >> } MK(desta) >> The displayed result showed both ?tau? and ?2-sided p-value?in >> unorganized way. But, I want only ?tau? value that is presented in >> organized manner. Anyone can tell me how can I get orderly displayed ?tau? >> value? here is my sample result: [[1]][1] 0 >> [[2]][1] 0.4352941attr(,"Csingle")[1] TRUE >> [[3]][1] 0.5462185attr(,"Csingle")[1] TRUE >> [[4]][1] 0.4218487attr(,"Csingle")[1] TRUE....Thank you for your guidance >> >> [[alternative HTML version deleted]] >> >> ______________________________________________ >> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see >> 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. > >
> 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) ought to do the whole thing in one shot and package it nicely into a named vector while it's about it. S Ellison ******************************************************************* This email and any attachments are confidential. Any use, copying or disclosure other than by the intended recipient is unauthorised. If you have received this message in error, please notify the sender immediately via +44(0)20 8943 7000 or notify postmaster at lgcgroup.com and delete this message and any copies from your computer and network. LGC Limited. Registered in England 2991879. Registered office: Queens Road, Teddington, Middlesex, TW11 0LY, UK
> 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.No, numeric(nc) makes a "numeric" (double precision) vector of length nc filled with zeros. Perhaps you were thinking of as.numeric(nc), which makes a numeric vector of length one containing the value nc. Bill Dunlap TIBCO Software wdunlap tibco.com On Thu, Apr 2, 2015 at 7:06 AM, S Ellison <S.Ellison at lgcgroup.com> 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) > > ought to do the whole thing in one shot and package it nicely into a named > vector while it's about it. > > S Ellison > > > ******************************************************************* > This email and any attachments are confidential. Any u...{{dropped:17}}
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 > > > ******************************************************************* > This email and any attachments are confidential. Any u...{{dropped:19}}