R devel - Feb 2020 - How to index the occasions in a vector repeatedly under condition 1? if not, it will give a new index.

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 at adelaide.edu.au<mailto:lijun.zhao at
adelaide.edu.au> or lijun.zhao at sahmri.com<mailto:lijun.zhao at
sahmri.com>



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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 at 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 at adelaide.edu.au<mailto:lijun.zhao at
adelaide.edu.au> or
> lijun.zhao at sahmri.com<mailto:lijun.zhao at sahmri.com>
>
>
>
>         [[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.
>
	[[alternative HTML version deleted]]

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 at tibco.com>
Sent: Thursday, 20 February 2020 1:53 AM
To: Lijun Zhao <lijun.zhao at adelaide.edu.au>
Cc: r-help at 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<http://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<http://tibco.com>


On Wed, Feb 19, 2020 at 7:08 AM Lijun Zhao <lijun.zhao at
adelaide.edu.au<mailto:lijun.zhao at 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 at adelaide.edu.au<mailto:lijun.zhao at
adelaide.edu.au><mailto:lijun.zhao at adelaide.edu.au<mailto:lijun.zhao
at adelaide.edu.au>> or lijun.zhao at sahmri.com<mailto:lijun.zhao at
sahmri.com><mailto:lijun.zhao at sahmri.com<mailto:lijun.zhao at
sahmri.com>>



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______________________________________________
R-help at r-project.org<mailto: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.

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