寻找最佳平衡点的曲线上 [英] Finding the best trade-off point on a curve
问题描述
说我有一些数据，我想通过它来适应参数化模型。我的目标是找到这个模型参数的最佳值。
Say I had some data, for which I want to fit a parametrized model over it. My goal is to find the best value for this model parameter.
我正在做使用 AIC /的 BIC / MDL 的标准型以奖励款低误差，而且惩罚具有高复杂度模型（我们正在寻找最简单最有说服力的解释，这些数据可以这么说，一拉奥卡姆剃刀）。
I'm doing model selection using a AIC/BIC/MDL type of criterion which rewards models with low error but also penalizes models with high complexity (we're seeking the simplest yet most convincing explanation for this data so to speak, a la Occam's razor).
按照上面的，这是我的东西得到的三种不同的标准的排序的一个例子（两个是被最小化，和一个被最大化）：
Following the above, this is an example of the sort of things I get for three different criteria (two are to be minimized, and one to be maximized):
在视觉上，你可以很容易地看到肘部的形状，你会在该区域选择一个值参数的地方。 问题是，我做的做了大量的实验，我需要一种方法来发现不干预该值。
Visually you can easily see the elbow shape and you would pick a value for the parameter somewhere in that region. The problem is that I'm doing do this for large number of experiments and I need a way to find this value without intervention.
我的第一个直觉是要尽量画一条线从墙角45度角，保持移动，直到它相交的曲线，但说起来容易，做起来难:)此外，它可以错过感兴趣的区域如果曲线有些歪斜。
My first intuition was to try to draw a line at 45 degrees angle from the corner and keep moving it until it intersect the curve, but that's easier said than done :) Also it can miss the region of interest if the curve is somewhat skewed.
这是如何实现这一点，或更好的想法有什么想法？
Any thoughts on how to implement this, or better ideas?
下面是重现上述地块之一所需要的样本：
Here's the samples needed to reproduce one of the plots above:
curve = [8.4663 8.3457 5.4507 5.3275 4.8305 4.7895 4.6889 4.6833 4.6819 4.6542 4.6501 4.6287 4.6162 4.585 4.5535 4.5134 4.474 4.4089 4.3797 4.3494 4.3268 4.3218 4.3206 4.3206 4.3203 4.2975 4.2864 4.2821 4.2544 4.2288 4.2281 4.2265 4.2226 4.2206 4.2146 4.2144 4.2114 4.1923 4.19 4.1894 4.1785 4.178 4.1694 4.1694 4.1694 4.1556 4.1498 4.1498 4.1357 4.1222 4.1222 4.1217 4.1192 4.1178 4.1139 4.1135 4.1125 4.1035 4.1025 4.1023 4.0971 4.0969 4.0915 4.0915 4.0914 4.0836 4.0804 4.0803 4.0722 4.065 4.065 4.0649 4.0644 4.0637 4.0616 4.0616 4.061 4.0572 4.0563 4.056 4.0545 4.0545 4.0522 4.0519 4.0514 4.0484 4.0467 4.0463 4.0422 4.0392 4.0388 4.0385 4.0385 4.0383 4.038 4.0379 4.0375 4.0364 4.0353 4.0344];
plot(1:100, curve)
修改
我接受<给定的解决方案href="http://stackoverflow.com/questions/2018178/finding-the-best-trade-off-point-on-a-curve/2022348#2022348">Jonas.基本上，在曲线上的每一点 P ，我们发现一个与最大距离 D 由下式给出：
EDIT
I accepted the solution given by Jonas. Basically, for each point p on the curve, we find the one with the maximum distance d given by:
推荐答案
发现肘部的一个快速的方法是借鉴第一线，以曲线的最后一个点，然后找到数据点是距离最远该行。
A quick way of finding the elbow is to draw a line from the first to the last point of the curve and then find the data point that is farthest away from that line.
这是当然，一定程度上取决于点你在该行的平坦部分的数目，但如果你每次测试的参数相同的数字，它应该走出相当好的。
This is of course somewhat dependent on the number of points you have in the flat part of the line, but if you test the same number of parameters each time, it should come out reasonably ok.
curve = [8.4663 8.3457 5.4507 5.3275 4.8305 4.7895 4.6889 4.6833 4.6819 4.6542 4.6501 4.6287 4.6162 4.585 4.5535 4.5134 4.474 4.4089 4.3797 4.3494 4.3268 4.3218 4.3206 4.3206 4.3203 4.2975 4.2864 4.2821 4.2544 4.2288 4.2281 4.2265 4.2226 4.2206 4.2146 4.2144 4.2114 4.1923 4.19 4.1894 4.1785 4.178 4.1694 4.1694 4.1694 4.1556 4.1498 4.1498 4.1357 4.1222 4.1222 4.1217 4.1192 4.1178 4.1139 4.1135 4.1125 4.1035 4.1025 4.1023 4.0971 4.0969 4.0915 4.0915 4.0914 4.0836 4.0804 4.0803 4.0722 4.065 4.065 4.0649 4.0644 4.0637 4.0616 4.0616 4.061 4.0572 4.0563 4.056 4.0545 4.0545 4.0522 4.0519 4.0514 4.0484 4.0467 4.0463 4.0422 4.0392 4.0388 4.0385 4.0385 4.0383 4.038 4.0379 4.0375 4.0364 4.0353 4.0344];
%# get coordinates of all the points
nPoints = length(curve);
allCoord = [1:nPoints;curve]'; %'# SO formatting
%# pull out first point
firstPoint = allCoord(1,:);
%# get vector between first and last point - this is the line
lineVec = allCoord(end,:) - firstPoint;
%# normalize the line vector
lineVecN = lineVec / sqrt(sum(lineVec.^2));
%# find the distance from each point to the line:
%# vector between all points and first point
vecFromFirst = bsxfun(@minus, allCoord, firstPoint);
%# To calculate the distance to the line, we split vecFromFirst into two
%# components, one that is parallel to the line and one that is perpendicular
%# Then, we take the norm of the part that is perpendicular to the line and
%# get the distance.
%# We find the vector parallel to the line by projecting vecFromFirst onto
%# the line. The perpendicular vector is vecFromFirst - vecFromFirstParallel
%# We project vecFromFirst by taking the scalar product of the vector with
%# the unit vector that points in the direction of the line (this gives us
%# the length of the projection of vecFromFirst onto the line). If we
%# multiply the scalar product by the unit vector, we have vecFromFirstParallel
scalarProduct = dot(vecFromFirst, repmat(lineVecN,nPoints,1), 2);
vecFromFirstParallel = scalarProduct * lineVecN;
vecToLine = vecFromFirst - vecFromFirstParallel;
%# distance to line is the norm of vecToLine
distToLine = sqrt(sum(vecToLine.^2,2));
%# plot the distance to the line
figure('Name','distance from curve to line'), plot(distToLine)
%# now all you need is to find the maximum
[maxDist,idxOfBestPoint] = max(distToLine);
%# plot
figure, plot(curve)
hold on
plot(allCoord(idxOfBestPoint,1), allCoord(idxOfBestPoint,2), 'or')
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