On Mar 12, 2012, at 3:07 PM, Emmanuel Levy wrote:
Hi Jeff,
Thanks for your reply and the example.
I'm not sure if it could be applied to the problem I'm facing though,
for two reasons:
(i) my understanding is that the inverse will associate a new Y
coordinate given an absolute X coordinate. However, in the case I'm
working on, the transformation that has to be applied depends on X
*and* on its position relative to the *normal* of the fitted curve.
This means, for instance, that both X and Y will change after
transformation.
(ii) the fitted curve can be described by a spline, but I'm not sure
if inverse of such models can be inferred automatically (I don't know
anything about that).
The procedure I envision is the following: treat the curve "segment by
segment", apply rotation+translation to each segment to bring it on
the
diagonal,
That makes sense. Although the way I am imagining it would be to do a
condition (on x) shift.
and apply the same transformation to all points
corresponding to the same segment (i.e., these are the points that are
close and within the "normal" area covered by the segment).
Does this make sense?
The first part sort of makes sense to me... maybe. You are thinking of
some sort of local transformation that converts a curve to a straight
line by rotation or deformation. Seems like a problem of finding a
transformation of a scalar field. But you then want it extended
outward to affect the regions at some distance from the curve. That's
where might break down or at least becomes non-trivial. Because a
region at a distance could be "in the sights" of the "normal" vector
to the curve (not in the statistical sense but in the vector-field
sense) of more than one segment of the curve. Only if you were going
apply some sort of transformation that did not extend globally would
you be able to do anything other than a y|x (y conditional on x) shift
or expansion contraction
All the best,
Emmanuel
On 12 March 2012 02:15, Jeff Newmiller <jdnew...@dcn.davis.ca.us>
wrote:
It is possible that I do not see what you mean, but it seems like
the following code does what you want. The general version of this
is likely to be rather more difficult to do, but in theory the
inverse function seems like what you are trying to accomplish.
x <- 1:20
y <- x^2 + rnorm(20)
y.lm <- lm( y ~ I(x^2) + x )
plot( x, y )
lines( x, predict( y.lm ) )
qa <- coef(y.lm)["I(x^2)"]
qb <- coef(y.lm)["x"]
qc <- coef(y.lm)["(Intercept)"]
# define inverse of known model
f1 <- function( y ) { ( sqrt( 4*qa*( y -qc ) + qb^2 ) - qb ) /
( 2*qa ) }
f2 <- function( y ) { -( sqrt( 4*qa*( y -qc ) + qb^2 ) + qb ) /
( 2*qa ) }
plot( x, f1( y ) )
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Emmanuel Levy <emmanuel.l...@gmail.com> wrote:
Dear Jeff,
I'm sorry but I do not see what you mean. It'd be great if you could
let me know in more details what you mean whenever you can.
Thanks,
Emmanuel
On 12 March 2012 00:07, Jeff Newmiller <jdnew...@dcn.davis.ca.us>
wrote:
Aren't you just reinventing the inverse of a function?
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Go...
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Sent from my phone. Please excuse my brevity.
Emmanuel Levy <emmanuel.l...@gmail.com> wrote:
Hi,
I am trying to normalize some data. First I fitted a principal
curve
(using the LCPM package), but now I would like to apply a
transformation so that the curve becomes a "straight diagonal
line"
on
the plot. The data used to fit the curve would then be
normalized by
applying the same transformation to it.
A simple solution could be to apply translations only (e.g., as
done
after a fit using loess), but here rotations would have to be
applied
as well. One could visualize this as the "stretching of a curve",
i.e., during such an "unfolding" process, both translations and
rotations would need to be applied.
Before I embark on this (which would require me remembering long
forgotten geometry principles) I was wondering if you can think of
packages or tricks that could make my life simpler?
Thanks for any input,
Emmanuel
Below I provide an example - the black curve is to be "brought"
along
the diagonal, and all data points normal to a "small
segment" (of the
black curve) would undergo the same transformation as it - what
"small" is remains to be defined though.
tmp=rnorm(2000)
X.1 = 5+tmp
Y.1 = 5+ (5*tmp+rnorm(2000))
tmp=rnorm(1000)
X.2 = 9+tmp
Y.2 = 40+ (1.5*tmp+rnorm(1000))
X.3 = 7+ 0.5*runif(500)
Y.3 = 15+20*runif(500)
X = c(X.1,X.2,X.3)
Y = c(Y.1,Y.2,Y.3)
lpc1 = lpc(cbind(X,Y), scaled=FALSE, h=c(1,1) )
plot(lpc1)
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West Hartford, CT
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