[R] How to index the occasions in a vector repeatedly under condition 1? if not, it will give a new index.

[William Dunlap]

> 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.zhao using 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 <wdunlap using tibco.com>
> *Sent:* Thursday, 20 February 2020 1:53 AM
> *To:* Lijun Zhao <lijun.zhao using adelaide.edu.au>
> *Cc:* r-help using 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.zhao using 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.zhao using adelaide.edu.au<mailto:lijun.zhao using adelaide.edu.au> or
> lijun.zhao using sahmri.com<mailto:lijun.zhao using sahmri.com>
>
>
>
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>
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