寻找曲线上的最佳权衡点 [英] Finding the best trade-off point on a curve
问题描述
假设我有一些数据，我想为其拟合参数化模型.我的目标是为这个模型参数找到最佳值.
我正在使用 .基本上，对于曲线上的每个点 p，我们找到最大距离 d 的点:
快速找到弯头的方法是从曲线的第一个点到最后一个点画一条线，然后找到距离最远的数据点那条线.
这当然在一定程度上取决于您在直线平坦部分的点数，但如果您每次测试相同数量的参数，结果应该是合理的.
<预> <代码>曲线= [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 4.17854.178 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 4.03884.0385 4.0385 4.0383 4.038 4.0379 4.0375 4.0364 4.0353 4.0344]；%# 获取所有点的坐标nPoints = 长度(曲线)；allCoord = [1:nPoints;curve]';%'# SO 格式%# 拉出第一点firstPoint = allCoord(1,:);%# 获取第一个和最后一个点之间的向量 - 这是线lineVec = allCoord(end,:) - firstPoint;%# 归一化线向量lineVecN = lineVec/sqrt(sum(lineVec.^2));%# 求每个点到直线的距离:%#所有点和第一个点之间的向量vecFromFirst = bsxfun(@minus, allCoord, firstPoint);%# 为了计算到线的距离，我们将 vecFromFirst 一分为二%# 分量，一个平行于直线，一个垂直于直线%# 然后，我们取垂直于线的部分的范数，然后%# 获取距离.%# 我们通过将 vecFromFirst 投影到直线上找到平行于线的向量%# 行.垂直向量是 vecFromFirst - vecFromFirstParallel%# 我们通过向量的标量积来投影 vecFromFirst%# 指向直线方向的单位向量(这给了我们%# vecFromFirst 投影到线上的长度).要是我们%# 将标量乘积乘以单位向量，我们有 vecFromFirstParallelscalarProduct = dot(vecFromFirst, repmat(lineVecN,nPoints,1), 2);vecFromFirstParallel = scalarProduct * lineVecN;vecToLine = vecFromFirst - vecFromFirstParallel;%# 到线的距离是 vecToLine 的范数distToLine = sqrt(sum(vecToLine.^2,2));%# 绘制到线的距离图('名称'，'曲线到线的距离')，绘图(distToLine)%# 现在你只需要找到最大值[maxDist,idxOfBestPoint] = max(distToLine);％# 阴谋图，图(曲线)坚持，稍等情节(allCoord(idxOfBestPoint，1)，allCoord(idxOfBestPoint，2)，'或')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.
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).
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):
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.
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.
Any thoughts on how to implement this, or better ideas?
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)
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:
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.
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')
这篇关于寻找曲线上的最佳权衡点的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
