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

[PIKAL Petr]

Hi

You could get similar result with using diff function Rui suggested

c(1,cumsum((diff(x)>15))+1)
[1] 1 1 1 1 1 1 2 3 3

Cheers
Petr

> -----Original Message-----
> From: R-help <r-help-bounces using r-project.org> On Behalf Of Rui Barradas
> Sent: Wednesday, February 19, 2020 8:13 AM
> To: Lijun Zhao <lijunzhao0606 using gmail.com>; 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.
> 
> Hello,
> 
> First of all, a note about your reproducible example.
> 
> When you write diff <- x - lag(x) there are two things to be said.
> 
> 1. There is a base R function named 'diff', it is better to use another name.
> 
> diff(x)
> #[1]   0   0   0   8   0 577  69   0
> 
> 2. There are also several functions named 'lag', one of them in base package
> stats.
> 
> x - lag(x)
> #[1] 0 0 0 0 0 0 0 0 0
> #attr(,"tsp")
> #[1] 0 8 1
> 
> This is not the one you are using.
> 
> x - dplyr::lag(x)
> #[1]  NA   0   0   0   8   0 577  69   0
> 
> That's the one. When you have a package loaded in your session, please start
> your scripts with library(<pkgname>), in this case library(dplyr).
> 
> 
> Now for the question's problem. I will use a different name, 'd', not
> 'diff'. And qualify the function name with the package name prefix.
> 
> The main problem is the NA in the first element of 'd', without it
> cumsum(d > 15) would be enough. This works because the logical values
> FALSE/TRUE are coded as 0/1 and their cumulative sum goes up every time
> a TRUE is found.
> 
> d <- x - dplyr::lag(x)
> cumsum(is.na(d) | d > 15)
> #[1] 1 1 1 1 1 1 2 3 3
> 
> 
> Hope this helps,
> 
> Rui Barradas
> 
> 
> Às 06:56 de 19/02/20, Lijun Zhao escreveu:
> > Dear All,
> >
> > could you please help me how to get the output from 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 occassions in x repeatedly if the diff>15? if not, it will
> > give a new index
> >
> > i want the output be like y
> >
> > y<-c(1,1,1,1,1,1,2,3,3)
> >
> >
> > thanks,
> >
> >
> > Lijun
> >
> > 	[[alternative HTML version deleted]]
> >
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> > and provide commented, minimal, self-contained, reproducible code.
> >
> 
> ______________________________________________
> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
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