# [R] fancy linear model and grouping

PIKAL Petr petr.pikal at precheza.cz
Wed Feb 3 08:37:50 CET 2016

```Hi Bill and Gabor

PET approach is quite interesting but as I hardly understand what it really does it seems to me rather complicated to use it properly.

OTOH using Mclust is quite understandable and what is important it provides correct grouping after I used EEV model.

As usual, Rhelp is powerful tool for getting answers if one tries to keep Posting guide rules.

Cheers
Petr

From: William Dunlap [mailto:wdunlap at tibco.com]
Sent: Tuesday, February 02, 2016 6:09 PM
To: PIKAL Petr
Cc: R Help R
Subject: Re: [R] fancy linear model and grouping

Perhaps you can try clustering the output of the Hough transform.
PET::hough() will compute it, given a matrix like gplots::hist2d(1/temp[,1],temp[,2])\$hData.  I do not have much experience here.

Bill Dunlap
TIBCO Software
wdunlap tibco.com<http://tibco.com>

On Tue, Feb 2, 2016 at 3:35 AM, PIKAL Petr <petr.pikal at precheza.cz<mailto:petr.pikal at precheza.cz>> wrote:
Dear all

I have data like this

> dput(temp)

temp <- structure(list(X1 = c(93, 82, NA, 93, 93, 79, 79, 93, 93, 85,
82, 93, 87, 93, 92, NA, 87, 93, 93, 93, 74, 77, 87, 93, 82, 87,
75, 82, 93, 92, 68, 93, 93, 73, NA, 85, 81, 79, 75, 87, 93, NA,
87, 87, 85, 92, 87, 92, 93, 87, 87, NA, 69, 87, 93, 87, 93, 87,
82, 79, 87, 93, 87, 80, 87, 87, 87, 92, 93, 69, 76, 87, 82, 93,
82, NA, 54, 87, 77, 73, 93, 82, 73, 93, 92, 82, 77, 93, 87, 75,
87, 87, 87, 60, 92, 87, 87, NA, 77, 78), X2 = c(224, 624, NA,
224, 224, 642, 642, 224, 224, 599, 622, 224, 239, 224, 225, NA,
239, 224, 224, 224, 688, 657, 239, 224, 624, 239, 672, 254, 224,
225, 499, 224, 224, 692, NA, 599, 627, 642, 677, 239, 224, NA,
239, 239, NA, 375, 239, 375, 224, 239, 239, NA, 299, 239, 224,
239, 224, 239, 621, 642, 239, 224, 239, 638, 239, 239, 239, 225,
224, 299, 672, 239, 618, 224, 620, NA, 626, 239, 657, 693, 224,
624, 693, 224, 225, 621, 657, 224, 239, 673, 239, 239, 239, 569,
224, 239, 239, NA, 657, 651)), .Names = c("X1", "X2"), row.names = c(NA,
-100L), class = "data.frame")
>

You can see there are 3 distinct linear relationships of those 2 variables.

plot(1/temp[,1], temp[,2])

Is there any simple way how to evaluate such data without grouping variable? I know that in case I have proper grouping variable I can evaluate it with lme and get intercepts and/or slopes.

My question is:

Does anybody know about a way/package/function which can give me appropriate grouping of such data or which can give me separate slope/intercept for each set.

I hope I expressed my problem clearly.

Best regards
Petr

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