Fitting multiple horizontal lines to data

The changepoint package might give you a way to do this.

-------------------------------------
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352


-----Original Message-----
From: r-help-bounces at r-project.org
[mailto:r-help-bounces at r-project.org] On Behalf Of Sashikanth
Chandrasekaran
Sent: Wednesday, November 6, 2013 11:20 AM
To: r-help at r-project.org
Subject: [R] Fitting multiple horizontal lines to data

I am not trying to fit a horizontal line at every unique value
of y. I am
trying fit the y values with as few horizontal lines by trading
off the
number of horizontal lines with the error. The actual problem I
am trying
to solve is to smooth data in a time series. Here is a realistic
example of
y

y=c(134.45,141.82,143.81,141.81,145,141.61,143.72,145.71,200,175
,140,200,148.77,71.64,111.57,118.15,119.15,112.8,111.64,111.64,1
57.26,143.8,40.19,64.99,64.99,129.98,64.99,65,64.98,64.99)

An example fit for y using multiple horizontal lines (may not be
the best
fit in terms of squared error or another error metric, but I
have included
the y value for concreteness)

1. A horizontal line at approximately y=140 (to fit the first 13
values -
134.45 to 148.77)
2. A horizontal line at approximately y=110 (to fit the next 7
values -
71.64 to 111.64)
3. A horizontal line at approximately y=150 (to fit the next 2
values -
157.26 to 143.8)
4. A horizontal line at approximately y=65 (to fit the last 8
values -
40.19 to 64.99)
-sashi.


______________________________________________
R-help at r-project.org mailing list
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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.

I am not trying to fit a horizontal line at every unique value of y. I am
trying fit the y values with as few horizontal lines by trading off the
number of horizontal lines with the error. The actual problem I am trying
to solve is to smooth data in a time series. Here is a realistic example of
y

y=c(134.45,141.82,143.81,141.81,145,141.61,143.72,145.71,200,175,140,200,148.77,71.64,111.57,118.15,119.15,112.8,111.64,111.64,157.26,143.8,40.19,64.99,64.99,129.98,64.99,65,64.98,64.99)

An example fit for y using multiple horizontal lines (may not be the best
fit in terms of squared error or another error metric, but I have included
the y value for concreteness)

The human brain searches for patterns and often finds them where there is no underlying mechanism. If you are asking for a regime-change method that is statistically based and will replicate your brain-driven pencil-and-paper methods you will probably be disappointed.

plot(y)
abline(h=c(140,110,150,65) )
abline(v=c(13,20,22,30) ,col="red")

1. A horizontal line at approximately y=140 (to fit the first 13 values -
134.45 to 148.77)
2. A horizontal line at approximately y=110 (to fit the next 7 values -
71.64 to 111.64)
3. A horizontal line at approximately y=150 (to fit the next 2 values -
157.26 to 143.8)
4. A horizontal line at approximately y=65 (to fit the last 8 values -
40.19 to 64.99)
-sashi.

If you want a method that is driven by the magnitude of the shift in adjacent values, then this would find some but not all of your proposed breakpoints:

which( abs(diff(y)) >55)

[1] 11 13 22 25 26

You could perhaps refine that set of candidates by requiring that the next value have some other defining feature but I was unable to come up with a simple rule-set that agreed with your candidates . There are packages that do segmented regression but hey are not generall set up to assume all regression coefficients are 0 and that you are only interested in

	[[alternative HTML version deleted]]

______________________________________________
R-help at r-project.org mailing list
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.

David Winsemius
Alameda, CA, USA

Possibly see the
   strucchange
package

Don MacQueen

Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062





On 11/6/13 9:19 AM, "Sashikanth Chandrasekaran"
<sashikanth.chandrasekaran at gmail.com> wrote:

>I am not trying to fit a horizontal line at every unique value of y. I am
>trying fit the y values with as few horizontal lines by trading off the
>number of horizontal lines with the error. The actual problem I am trying
>to solve is to smooth data in a time series. Here is a realistic example
>of
>y
>
>y=c(134.45,141.82,143.81,141.81,145,141.61,143.72,145.71,200,175,140,200,1
>48.77,71.64,111.57,118.15,119.15,112.8,111.64,111.64,157.26,143.8,40.19,64
>.99,64.99,129.98,64.99,65,64.98,64.99)
>
>An example fit for y using multiple horizontal lines (may not be the best
>fit in terms of squared error or another error metric, but I have included
>the y value for concreteness)
>
>1. A horizontal line at approximately y=140 (to fit the first 13 values -
>134.45 to 148.77)
>2. A horizontal line at approximately y=110 (to fit the next 7 values -
>71.64 to 111.64)
>3. A horizontal line at approximately y=150 (to fit the next 2 values -
>157.26 to 143.8)
>4. A horizontal line at approximately y=65 (to fit the last 8 values -
>40.19 to 64.99)
>-sashi.
>
>	[[alternative HTML version deleted]]
>
>______________________________________________
>R-help at r-project.org mailing list
>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.

You already asked this on StackOverflow.
The answer remains the same,  pretty much what David W.  wrote:   this is
not a question about fitting lines to data.  You need to step back and think
about what message you want to deliver to those who will view your graph,
and what the meaning of your dataset is in the first place.




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