| REGRESS Question [message #31982] |
Wed, 04 September 2002 14:21  |
David Fanning
Messages: 11724
Registered: August 2001
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Senior Member |
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Folks,
I have a client who has asked me to create a pixel density
function between two images and then perform a regression
analysis on the resulting distribution. No problem doing all
this, but she finds that the results of my regression analysis
differ from the same analysis performed in other statistics
packages. In fact, three different packages give the same
answer, and then there is IDL. :-(
For example, if the other packages calculate a "goodness
of fit" of 0.95, IDL might report 0.97.
Here is my question. Are there any known problems with REGRESS?
Or, can I assume that this problem comes from my own mathematical
ignorance?
Cheers,
David
--
David W. Fanning, Ph.D.
Fanning Software Consulting, Inc.
Phone: 970-221-0438, E-mail: david@dfanning.com
Coyote's Guide to IDL Programming: http://www.dfanning.com/
Toll-Free IDL Book Orders: 1-888-461-0155
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| Re: REGRESS Question [message #32046 is a reply to message #31982] |
Fri, 06 September 2002 00:26  |
Mike Alport
Messages: 8
Registered: December 1998
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Junior Member |
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I think Bill may have a point - either R or R^2 is sometimes used as a
measure of "Goodness of Fit". One way to check this would be to compare this
quantity from both programs to eg 5 dec places and see if one is the SQRT of
the other.
Mike
"Bill" <wclodius@lanl.gov> wrote in message
news:3D7794C1.17DD18AB@lanl.gov...
>
>
> David Fanning wrote:
>
>> Folks,
>>
>> I have a client who has asked me to create a pixel density
>> function between two images and then perform a regression
>> analysis on the resulting distribution. No problem doing all
>> this, but she finds that the results of my regression analysis
>> differ from the same analysis performed in other statistics
>> packages. In fact, three different packages give the same
>> answer, and then there is IDL. :-(
>>
>> For example, if the other packages calculate a "goodness
>> of fit" of 0.95, IDL might report 0.97.
>>
>> Here is my question. Are there any known problems with REGRESS?
>> Or, can I assume that this problem comes from my own mathematical
>> ignorance?
>>
>> Cheers,
>>
>> David
>
> <snip>
>
> Almost any package can have problems, but the original REGRESS in
> Bevington has stood the test of time. IDL's version works for me, but it
> is possible that the introduced some problems. One thing that bothers
> me is tha 0.95 is to a good approximation 0.97^2. Could you be fitting
> the square root of the customer's data.
>
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| Re: REGRESS Question [message #32058 is a reply to message #31982] |
Thu, 05 September 2002 11:31 |
julia[1]
Messages: 1
Registered: September 2002
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Junior Member |
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David --
I ran into problems with the regress routine a few years ago, trying
to regress large amounts of data. The problem is that regress.pro
calls the total routine, which is called in floating point precision.
I had more obvious problems than the 2% difference in goodness of fit
that your reporting, but I found I had to modify regress.pro to call
total in double precision.
Julia
David Fanning <david@dfanning.com> wrote in message news:<MPG.17e027561a59e932989994@news.frii.com>...
> Folks,
>
> I have a client who has asked me to create a pixel density
> function between two images and then perform a regression
> analysis on the resulting distribution. No problem doing all
> this, but she finds that the results of my regression analysis
> differ from the same analysis performed in other statistics
> packages. In fact, three different packages give the same
> answer, and then there is IDL. :-(
>
> For example, if the other packages calculate a "goodness
> of fit" of 0.95, IDL might report 0.97.
>
> Here is my question. Are there any known problems with REGRESS?
> Or, can I assume that this problem comes from my own mathematical
> ignorance?
>
> Cheers,
>
> David
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| Re: REGRESS Question [message #32061 is a reply to message #31982] |
Thu, 05 September 2002 10:31 |
William Clodius
Messages: 30
Registered: December 1996
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Member |
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David Fanning wrote:
> Folks,
>
> I have a client who has asked me to create a pixel density
> function between two images and then perform a regression
> analysis on the resulting distribution. No problem doing all
> this, but she finds that the results of my regression analysis
> differ from the same analysis performed in other statistics
> packages. In fact, three different packages give the same
> answer, and then there is IDL. :-(
>
> For example, if the other packages calculate a "goodness
> of fit" of 0.95, IDL might report 0.97.
>
> Here is my question. Are there any known problems with REGRESS?
> Or, can I assume that this problem comes from my own mathematical
> ignorance?
>
> Cheers,
>
> David
<snip>
Almost any package can have problems, but the original REGRESS in
Bevington has stood the test of time. IDL's version works for me, but it
is possible that the introduced some problems. One thing that bothers
me is tha 0.95 is to a good approximation 0.97^2. Could you be fitting
the square root of the customer's data.
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| Re: REGRESS Question [message #36721 is a reply to message #31982] |
Wed, 22 October 2003 06:03 |
wmconnolley
Messages: 106
Registered: November 2000
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Senior Member |
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"Christopher Lee" <cl> wrote:
> "Kevin M. Lausten" <kevinlausten@hotmail.com> wrote:
>> I am having difficulty working with the REGRESS function. I continually
>> get values <1 for my slope when doing a regression between two vectors.
>> When I do a regression mapping y to x (slope = regress(x, y, const =
>> const)) and when I do a regression mapping x to y (slope = regress(y, x,
>> const = const) I get a slope<1 for both calculations. Shouldn't the
>> y=mx+b of these two regressions be inverses of each other (leading to
>> one slope>1, and one<1?) Maybe I am misunderstanding regressions?
You've certainly misunderstood some basic maths: the inverse (as in
reciprocal) of -1 is -1, not 1. If the regression of y against x
has a negative slope, then you would expect the regression of x against y
to have too.
OTOH the relation is *not* reciprocal anyway, unless the fit is perfect.
(probably because the fit is asymmetric: y values are assumed uncertain,
x values exact).
-W.
--
William M Connolley | wmc@bas.ac.uk | http://www.antarctica.ac.uk/met/wmc/
Climate Modeller, British Antarctic Survey | Disclaimer: I speak for myself
I'm a .signature virus! copy me into your .signature file & help me spread!
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| Re: REGRESS Question [message #36722 is a reply to message #31982] |
Wed, 22 October 2003 03:44 |
Chris Lee
Messages: 101
Registered: August 2003
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Senior Member |
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In article <932b9720.0310210627.f93c6f2@posting.google.com>, "Kevin M.
Lausten" <kevinlausten@hotmail.com> wrote:
> I am having difficulty working with the REGRESS function. I continually
> get values <1 for my slope when doing a regression between two vectors.
> When I do a regression mapping y to x (slope = regress(x, y, const =
> const)) and when I do a regression mapping x to y (slope = regress(y, x,
> const = const) I get a slope<1 for both calculations. Shouldn't the
> y=mx+b of these two regressions be inverses of each other (leading to
> one slope>1, and one<1?) Maybe I am misunderstanding regressions?
> Thanks,
> kevin
Hi,
If you try the regression with the simplest possible straight line
y = mx +c
where m=1 and c=0 , so
y=x
if you regress with y=f(x), you get a value of 1 (and a constant of 0)
if you regress with x=f(y), you get a value of 1, again.
if the gradient is negative for y=f(x), it has to be negative for x=f(y).
The two equations you are assuming in the regressions are
y= mx + c OR x = (y-c)/m = ny + d
n=1/m, so sign is preserved. (and d=-c/m = -cn)
HTH
Chris.
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| Re: regress question [message #64049 is a reply to message #31982] |
Mon, 01 December 2008 00:38 |
Wout De Nolf
Messages: 194
Registered: October 2008
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Senior Member |
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On Sun, 30 Nov 2008 23:42:38 -0800 (PST), James McCreight
<mccreigh@gmail.com> wrote:
> I have some vague recollection of doing this once within an IDL
> function. A quick look turned up this, looks promising and like
> something i've seen before:
>
> Curvefit( X, Y, Weights, A [, Sigma] [, CHISQ=variable] [, /DOUBLE] [,
> FITA=vector] [, FUNCTION_NAME=string] [, ITER=variable] [,
> ITMAX=value] [, /NODERIVATIVE] [, STATUS={0 | 1 | 2}] [, TOL=value] [,
> YERROR=variable] )
> A
> A vector with as many elements as the number of terms in the user-
> supplied function, containing the initial estimate for each parameter.
> On return, the vector A contains the fitted model parameters.
>
> FITA
> Set this keyword to a vector, with as many elements as A, which
> contains a zero for each fixed parameter, and a non-zero value for
> elements of A to fit. If not supplied, all parameters are taken to be
> non-fixed.
Why using a non-linear least squares fitting algorithm for a linear
problem? Fixing parameters is not all that difficult using the linear
algorithms (i.e. orthogonal decomposition methods like SVD), although
you have to do it yourself.
Suppose y=a.x1+b.x2+c then you find the least squares solution by (X1
and X2 column vectors)
SVDC, [X1,X2,replicate(1,1,n_elements(X1))], W, U, V
result=SVSOL(U, W, V, Y) ; gives LSSol. [a,b,c]
Suppose I want to fix b=3 then you would do this
SVDC, [X1,replicate(1,1,n_elements(X1))], W, U, V
result=SVSOL(U, W, V, Y-3*X2); gives LSSol. [a,c]
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| Re: regress question [message #64050 is a reply to message #31982] |
Sun, 30 November 2008 23:42 |
mccreigh
Messages: 13
Registered: January 2008
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Junior Member |
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I have some vague recollection of doing this once within an IDL
function. A quick look turned up this, looks promising and like
something i've seen before:
Curvefit( X, Y, Weights, A [, Sigma] [, CHISQ=variable] [, /DOUBLE] [,
FITA=vector] [, FUNCTION_NAME=string] [, ITER=variable] [,
ITMAX=value] [, /NODERIVATIVE] [, STATUS={0 | 1 | 2}] [, TOL=value] [,
YERROR=variable] )
A
A vector with as many elements as the number of terms in the user-
supplied function, containing the initial estimate for each parameter.
On return, the vector A contains the fitted model parameters.
FITA
Set this keyword to a vector, with as many elements as A, which
contains a zero for each fixed parameter, and a non-zero value for
elements of A to fit. If not supplied, all parameters are taken to be
non-fixed.
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| Re: regress question [message #64083 is a reply to message #31982] |
Thu, 27 November 2008 07:37 |
Brian Larsen
Messages: 270
Registered: June 2006
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Senior Member |
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Russ,
this has been discussed on this newsgroup for y=mx+b before, I have
turkey on the brain now and not regression but extending this idea to
multiple is probably not too bad (if it turns out to be the right
thing). And if this is not easy to do it is an interesting thread
that is good to remind oneself of.
This is the thread: http://tinyurl.com/2bfhl9
Here's a nice summary: http://tinyurl.com/2aqlgx
Cheers,
Brian
------------------------------------------------------------ --------------
Brian Larsen
Boston University
Center for Space Physics
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| Re: regress question [message #64088 is a reply to message #31982] |
Thu, 27 November 2008 02:47 |
Wout De Nolf
Messages: 194
Registered: October 2008
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Senior Member |
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On Thu, 27 Nov 2008 01:23:06 -0800 (PST), russ <rlayberry@hotmail.com>
wrote:
> Hi
>
> I'm using multiple linear regression using the REGRESS function. This
> gives me
>
> y = c + a1x1 + a2 x2 ...+ anxn
>
> with the coefficents a1,a2 etc.
>
> What I want to do is the above but force the constant to be zero. ie
> find the coeffcients that give the best linear fit whilst the function
> goes through thr origin (which it should do for physical reasons).
>
> Any ideas?
>
> Thanks
>
> Russ
You can create the design-matrix yourself and then use some
factorization like LU, SVD, Cholesky, QR,... (is your linear system
over/under determined?) The example below uses SVD. First it solves a
system not going through the origin by REGRESS and then by SVD.
Finally SVD is used for a system that goes through the origin.
X1 = [1.0, 2.0, 4.0, 8.0, 16.0, 32.0]
X2 = [0.0, 1.0, 2.0, 3.0, 4.0, 5.0]
X = transpose([[X1],[X2]])
Y = 3*X1 - 4*X2 + 5
Yorg = 3*X1 - 4*X2
; Regress
result1=regress(X,Y,const=const)
result1=[reform(result1),const]
; SVD (concat. X with 1's for the const)
SVDC, [X,replicate(1,1,n_elements(Y))], W, U, V
result2=reform(SVSOL(U, W, V, Y))
; SVD (origin)
SVDC, X, W, U, V
result3=reform(SVSOL(U, W, V, Yorg))
print,'Regress: ',result1
print,'SVD: ',result2
print,'SVD(origin): ',result3
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