在Matlab中使用nlinfit? [英] Using nlinfit in Matlab?

问题描述

我在Matlab中无法理解和应用nlinfit函数的使用.所以，假设我得到了向量

I'm having trouble understanding and applying the use of nlinfit function in Matlab. So, let's say I'm given vectors

x = [1, 2, 3, 4, 5] 
y = [2.3, 2.1, 1.7, .95, .70] 

，然后要求我将此数据拟合为指数形式(我不知道数字是否会起作用，我将它们补足了)，其中y = A*e^(Bx) + C(A/B/C是常数).

and I'm asked to fit this data to an exponential form (I don't know if the numbers will work, I made them up) where y = A*e^(Bx) + C (A/B/C are constants).

我的理解是nlinfit接受4个参数，两个向量，一个modelfunction在这种情况下应该是我上面的公式，然后是beta0，我一点都不明白.我的问题是如何在nlinft中实现modelfunction，以及如何找到beta0(当仅使用要绘制/拟合的2个向量时)以及应该如何实现?有人可以给我看一个例子，以便我适合自己使用此功能吗?我怀疑我将来会经常使用它，并且真的想学习它.

My understanding is that nlinfit takes 4 arguments, the two vectors, a modelfunction which in this case should be the equation I have above, and then beta0, which I don't understand at all. My question is how do you implement the modelfunction in nlinft, and how do you find beta0 (when only working with 2 vectors you want to plot/fit) and how should it be implemented? Can someone show me an example so that I can apply this function for any fit? I suspect I'll be using this a lot in the future and really want to learn it.

推荐答案

查看文档中的第二个示例: http://www.mathworks.com/help/stats/nlinfit.html

Check out the second example in the docs: http://www.mathworks.com/help/stats/nlinfit.html

基本上，您将函数句柄作为您的modelfunction参数传递.要么在文件中创建一个函数，然后在函数名称前传递一个@，要么创建一个匿名函数，如下所示:

Basically you pass a function handle as your modelfunction parameter. Either make a function in a file and then just pass it the function name with an @ in front or else make an anonymous function like this:

nlinfit(x, y, @(b,x)(b(1).*exp(b(2).*x) + b(3)), beta0)

您会注意到，在上文中，我将所有参数都固定在一个向量中.函数的第一个参数必须是要尝试求解的所有点的矢量(例如，在您的情况下为A，B和C)，第二个必须为x.

You'll notice that in the above I have stuck all your parameters into a single vector. The first parameter of your function must be a vector of all the points you are trying to solve for (i.e. A, B and C in your case) and the second must be x.

就像木片所说的那样，beta0是您的起点，因此，您对A，B和C参数的最佳猜测(不一定是很大的).所以类似[1 1 1]或rand(3,1)的东西，这是非常特定于问题的.您应该玩一些.请记住，这是一个本地搜索功能，因此可能会卡在本地最优值上，因此您的出发点实际上可能非常重要.

As woodchips has said beta0 is your starting point so your best guess (doesn't have to be great) of your A, B and C parameters. so something like [1 1 1] or rand(3,1), it is very problem specific though. You should play around with a few. Just remember that this is a local search function and thus can get stuck on local optima so your starting points can actually be quite important.

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