凸随机值的最小化 [英] Minimization of convex stochastic values

问题描述

我有一个等距变量(假设values=0:1e-3:1).我想从最接近统一随机值(假设x=rand)的values中获取值及其对应的索引.

I have a variable of equidistant values (suppose values=0:1e-3:1). I want to get the value and its correspondent index from values which is closest to a uniformly random value (suppose x=rand).

我可以做[value,vIdx]=min(abs(values-x))，这是我能做的最简单的最小化. 不幸的是，min函数不会从数据的一个属性(即凸的)中获益.我不需要搜索所有索引，因为一旦找到不小于前一个索引的索引，我便找到了全局最小值.话虽如此，我不想用matlab min函数代替一个循环，具体取决于它与我将要开始的值之间的距离，该循环会更慢.有很多方法可以用作黄金分割，但是我不确定使用matlab fmincon是否会比min方法更快.

I could do [value,vIdx]=min(abs(values-x)), which would be the simplest minimization I could do. Unfortunately the min function won't take advantage from one property from the data, that is to be convex. I don't need to search all indexes, because as soon as find an index that is no more lesser than the previous I've found the global minimum. Said that, I don't want to substitute the matlab min function for a loop that would be slower depending on how distant it is from the value I will start. There are many methods that could be used, as the golden section, but I am not sure if using matlab fmincon would be faster than the min method.

有人比使用所述的min方法更快地获得所需值的任何提示/想法吗?有空时我会检查时间表现，但是如果有人知道先验答案很好，请告诉我.

Does anyone have any hints/ideas how to get the required value faster than using the described min method? I'll check the time performance when I have time, but if someone knows a good answer a priori, please let me know.

可能的应用程序:捕捉到最近的图形数据

Possible Application: Snap to nearest graphic data

推荐答案

由于您的点是等距的，因此可以使用值x来找到索引:

Since your points are equidistant you can use the value x to find the index:

vIdx = 1+round(x*(numel(values)-1));

这个想法是，您将间隔[0, 1]分成大小相等的numel(values)-1个间隔.现在，通过将x乘以该数字，您可以将间隔映射到[0, numel(values)-1]，在此处将点映射到整数值.现在，使用round可以得到最接近的一个，通过加1可以得到MATLAB所需的基于1的索引.

The idea is that you are dividing the interval [0, 1] into numel(values)-1 equally sized intervals. Now by multiplying x by that number you map the interval to [0, numel(values)-1] where your points are mapped to integer values. Now using round you get the closest one and by adding 1 you get the one-based index that MATLAB requires.

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