R help - Mar 2015 - How to speed up a double loop?

Hi Petr,

Thanks for your reply,

Actually it's not what I'm looking for. The aim is not simply to remove
each
value > 15. 

In my loop, I consider the first numeric value of my column as
"correct".
Then, I want to test the second value. If the absolute difference with the
previous correct one is <15, it's a new correct one, but if it's
>15, then
it's a wrong one. 
If it's a wrong one, it has to test the third one to check if it's still
>15
from the last correct value (first one).
The value becomes correct again when the difference with the last correct
one goes under 15 (and so, this value is the new "correct" one, and so
one
for the rest of the column).

My loop is already doing the trick, but I just want to speed it up (or maybe
another faster way to do the job).
Hope it's more understandable right now! 





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Usually there is some trick with diff or role or cumsum that can make something
like this work, but all I can come up with is using Rcpp to brute force it. That
works best if you create a package, but it can work okay without it if this is a
one-off analysis.

Note that demolishing data as in your example makes most statisticians
uncomfortable.. particularly in the name of removing outliers. I would recommend
creating a quality check column and then perhaps subsetting into a new data
frame if you felt it was necessary.
---------------------------------------------------------------------------
Jeff Newmiller                        The     .....       .....  Go Live...
DCN:<jdnewmil at dcn.davis.ca.us>        Basics: ##.#.       ##.#.  Live
Go...
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Sent from my phone. Please excuse my brevity.

On March 2, 2015 6:12:44 AM PST, jeff6868 <geoffrey_klein at
etu.u-bourgogne.fr> wrote:
>Hi Petr,
>
>Thanks for your reply,
>
>Actually it's not what I'm looking for. The aim is not simply to
remove
>each
>value > 15. 
>
>In my loop, I consider the first numeric value of my column as
>"correct".
>Then, I want to test the second value. If the absolute difference with
>the
>previous correct one is <15, it's a new correct one, but if it's
>15,
>then
>it's a wrong one. 
>If it's a wrong one, it has to test the third one to check if it's
>still >15
>from the last correct value (first one).
>The value becomes correct again when the difference with the last
>correct
>one goes under 15 (and so, this value is the new "correct" one,
and so
>one
>for the rest of the column).
>
>My loop is already doing the trick, but I just want to speed it up (or
>maybe
>another faster way to do the job).
>Hope it's more understandable right now! 
>
>
>
>
>
>--
>View this message in context:
>http://r.789695.n4.nabble.com/How-to-speed-up-a-double-loop-tp4704054p4704061.html
>Sent from the R help mailing list archive at Nabble.com.
>
>______________________________________________
>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.

I tried another faster way which seems to do the trick right now:

    myts
<-data.frame(x=c(10,2,50,40,NA,NA,0,50,1,2,0,0,NA,50,0,15,3,5,4,20,0,0,25,22,0,1,100),z=NA)
    
    test <- function(x){
    st1 <- numeric(length(x))
    temp <- st1[1]
    for (i in 2:(length(x))){ 
        if((!is.na(x[i])) & (!is.na(x[i-1]))& (abs((x[i])-(temp)) >=
15)){
    st1[i] <- 1
    } } 
    return(st1)
    }

    myts[,2] <- apply(as.data.frame(myts[,1]),2,test)  
    myts[,2] <- as.numeric(myts[,2])

Thanks anyway for your help!



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Hi

Removal outliers is quite tricky business and especially automatic removal of
outliers. I have in some cases different approach.

If I know that some series is stationary and there are some dropouts in
measurement (somebody disconnected it, process stopped for whatever reasons) I
take large enough sequence, compute median (which shall be close to most of the
points, set some arbitrary value for threshold and remove measurements which are
behind this threshold. I then check my choice by visual inspection of plot.

So in your case

myts <- data.frame(x=c(1,2,50,40,30,40,100,1,50,1,2,3,3,5,4),y=NA)
selec <- abs(myts$x-median(myts$x))>15
myts$y[selec] <- myts$x[selec]
myts$x[selec] <- NA
> myts
    x   y
1   1  NA
2   2  NA
3  NA  50
4  NA  40
5  NA  30
6  NA  40
7  NA 100
8   1  NA
9  NA  50
10  1  NA
11  2  NA
12  3  NA
13  3  NA
14  5  NA
15  4  NA

So you either reconsider your outlier definition or you has to stick with
Jeff's solution.

Cheers
Petr
> -----Original Message-----
> From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of
> jeff6868
> Sent: Monday, March 02, 2015 3:13 PM
> To: r-help at r-project.org
> Subject: Re: [R] How to speed up a double loop?
>
> Hi Petr,
>
> Thanks for your reply,
>
> Actually it's not what I'm looking for. The aim is not simply to
remove
> each value > 15.
>
> In my loop, I consider the first numeric value of my column as
> "correct".
> Then, I want to test the second value. If the absolute difference with
> the previous correct one is <15, it's a new correct one, but if
it's
> >15, then it's a wrong one.
> If it's a wrong one, it has to test the third one to check if it's
> still >15 from the last correct value (first one).
> The value becomes correct again when the difference with the last
> correct one goes under 15 (and so, this value is the new
"correct" one,
> and so one for the rest of the column).
>
> My loop is already doing the trick, but I just want to speed it up (or
> maybe another faster way to do the job).
> Hope it's more understandable right now!
>
>
>
>
>
> --
> View this message in context: http://r.789695.n4.nabble.com/How-to-
> speed-up-a-double-loop-tp4704054p4704061.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> 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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