Re: 3D distance problem
- From: "Phil Sherrod" <phil.sherrod@xxxxxxxxxxxxxxxxxxx>
- Date: Thu, 6 Oct 2005 18:01:41 GMT
On 6-Oct-2005, "alphapoint05" <millerjm@xxxxxxx> wrote:
> I have to keep the actual variables confidential for now, but I could
> frame it this way. Suppose I have 3 cognitive variables. From previous
> analyses, these 3 variables have been applied to 20 different car
> brands with large samples and replication. So, I now have a relatively
> good idea of the average score on each variable and for each car brand.
> I take a 3D plot and plot the 20 car brands based on their 3 variable
> scores. I then take a new sample who all have scores on the 3 variables
> but not in reference to any of these car brands. I now plot that
> subject's score on the 3D plot. I want to know which car brand is
> closest, second closest, etc for this person and then over all people
> in rank-order.
>
> As for the weighting, I know from strong theory that the third variable
> isn't as influential. When plotting in 2D space without that third
> variable, the person is close to Honda. However, suppose the Honda
> score for the third variable is very different from the subject's score
> for the third variable. In this case, the closest car could be very far
> from Honda simply due to the influence of this third variable. So, I'd
> like to find a way, maybe a boundary constraint, that keep selection of
> the nearest car within a range based on the first two variables. It
> would be even better to find a way to determine that constraint based
> on the influence of the third variable rather than on an aribitrary
> decision.
You didn't directly respond to my suggestion that you compute a weighted
distance, so I'm not sure if we are on the same page in terms of a solution.
There are many possible ways to do the weighting. For example, you could use
a simple weighting coefficient for each of the x,y,z distances, or you could
set the weighted distance to a very large number if it exceeded some
threshold value, or you could use a non-linear function like distance^3 or
exp(distance), or you could use a combination of these. It may be possible
to perform a (non)linear regression to determine the optimal weighting
coefficients.
> What do you mean by "exponentially increasing cost"?
if (actual_distance <= threshold) {
weighted_distance = actual_distance;
} else {
weighted_distance = actual_distance + exp(actual_distance -
threshold);
}
The reason I asked about the "big picture" is that there are other types of
models than distance measures to determine the best category for a
prediction. I don't know if distance measures are the best type of
predictive model for your problem.
--
Phil Sherrod
(phil.sherrod 'at' sandh.com)
http://www.dtreg.com (decision tree and SVM predictive modeling)
http://www.nlreg.com (nonlinear regression)
.
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