> all.equal(y, ave(d, cumsum(c(TRUE,is_true(diff(a)!=0))), FUN=function(di)1L+cumsum(is_true(di>15)))) [1] TRUE
Bill Dunlap TIBCO Software wdunlap tibco.com On Wed, Feb 19, 2020 at 7:20 PM Lijun Zhao <lijun.z...@adelaide.edu.au> wrote: > Dear William, > > Thank you so much. > > > > I am quiet new in R. I would like to do this based on another repeated > variables. > > > > For example: > > a<-c(1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2) > > d<-c(NA, 0, 0, 0, 8, 0, 577, 69, 0, NA, 0, 0, 0, 8, 0, 577, 69, 0) > > the outcome I want is : > > y<-c(1, 1, 1, 1, 1, 1, 2, 3, 3, 1, 1, 1 ,1, 1 ,1, 2, 3, 3) > > > > Therefore, I would like to create y based on the variable a. once variable > a has changed, the index will start from 1 again. I wrote a for loop, but > it did not give me what I want. Could you please help me again? > > > > Thanks in advance, > > > > Lijun > > > > > > > > *From:* William Dunlap <wdun...@tibco.com> > *Sent:* Thursday, 20 February 2020 1:53 AM > *To:* Lijun Zhao <lijun.z...@adelaide.edu.au> > *Cc:* r-help@r-project.org > *Subject:* Re: [R] How to index the occasions in a vector repeatedly > under condition 1? if not, it will give a new index. > > > > Use cumsum(logicalVector) to increment a counter at the TRUE positions in > logicalVector. . E.g., > > > > > d <- c(NA, 0, 0, 0, 8, 0, 577, 69, 0) > > > is_true <- function(x) !is.na(x) & x > > 1 + cumsum( is_true(d >= 15) ) > [1] 1 1 1 1 1 1 2 3 3 > > > > Some packages have the equivalent of that is_true function, which maps > FALSE and NA to FALSE and TRUE to TRUE. I don't think core R contains such > a function. > > > Bill Dunlap > TIBCO Software > wdunlap tibco.com > > > > > > On Wed, Feb 19, 2020 at 7:08 AM Lijun Zhao <lijun.z...@adelaide.edu.au> > wrote: > > Dear all, > Could you please help me how to get the output as I described in the > following example? > > x<-c(543, 543, 543, 543, 551 , 551 ,1128 ,1197, 1197) > diff<-x-lag(x) > diff > [1] NA 0 0 0 8 0 577 69 0 > > How to index the occasions in x repeatedly if the diff<15? if diff>=15, it > will give a new index. > I want the output be like y. > > y<-c(1,1,1,1,1,1,2,3,3) > > Thank you so much, > > Lijun Zhao (PhD Candidate) > Nutrition and Metabolism > Level 7 SAHMRI > North Terrace > Adelaide 5005 > Ph : +61 8 8128 4898 > e-mail: lijun.z...@adelaide.edu.au<mailto:lijun.z...@adelaide.edu.au> or > lijun.z...@sahmri.com<mailto:lijun.z...@sahmri.com> > > > > [[alternative HTML version deleted]] > > ______________________________________________ > R-help@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. > > [[alternative HTML version deleted]] ______________________________________________ R-help@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.