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

[Rui Barradas]

Hello,

Yes, or even simpler is to assume that the first group starts at the 
first element of x, a reasonable assumption.

cumsum(c(TRUE, diff(x) > 15))


Hope this helps,

Rui Barradas

Às 10:36 de 19/02/20, PIKAL Petr escreveu:
> 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]]
>>>
>>> ______________________________________________
>>> R-help using r-project.org mailing list -- To UNSUBSCRIBE and more, see
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>>> PLEASE do read the posting guide http://www.R-project.org/posting-
>> guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>
>> ______________________________________________
>> R-help using 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.
>>
>> ______________________________________________
>> R-help using 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.