在MATLAB中使用@()调用函数 [英] Call functions with @() in MATLAB

问题描述

我试图在以下代码片段中找出@(t)的目的:

I'm trying to figure out what is the purpose of @(t) in the following code snippet:

 [theta] = ...
     fmincg (@(t)(lrCostFunction(t, X, (y == c), lambda)), ...
             initial_theta, options);

lrCostFunction:

lrCostFunction:

function [J, grad] = lrCostFunction(theta, X, y, lambda)
%LRCOSTFUNCTION Compute cost and gradient for logistic regression with 
%regularization
%   J = LRCOSTFUNCTION(theta, X, y, lambda) computes the cost of using
%   theta as the parameter for regularized logistic regression and the
%   gradient of the cost w.r.t. to the parameters. 

和选项:

options = optimset('GradObj', 'on', 'MaxIter', 50);

我希望能得到一些解释.预先感谢

I'd appreciate some explanation. Thanks in advance

推荐答案

让我回答您的问题，重点是

Let me answer your question focusing on anonymous function itself.

以下功能，在单独的.m文件中定义

The following function, defined in a separate .m file

function y = foo(x, a, b)
    y = x^(a-b);
end

等同于在主脚本中定义匿名函数

is equivalent to defining an anonymous function in the main script

bar = @(x, a, b) x^(a-b);

当主脚本调用函数foo(5, 1, 2)时，Matlab在工作目录中搜索，然后读取并执行文件foo.m中的代码.相反，当您运行bar(5, 1, 2)行时，Matlab会调用函数句柄" 并将其视为函数(尽管其功能受一行代码限制-您无法轻松执行switch或for之类的操作).

When your main script calls function foo(5, 1, 2), Matlab searches in working directory, then reads and executes code within file foo.m. Contrarily, when you run a line bar(5, 1, 2), Matlab calls a "function handle" and treat it as a function (though its power is limited by a single line of code - you can't perform things like switch or for easily).

有时，我们需要将一些功能包装到易于使用的功能中.考虑一种情况，我们希望对foo进行1000次评估，但只有输入x会发生变化，而a和b保持不变.当然可以在for循环中编写foo(x, 1, 2)，但是您也可以在进入循环之前包装该函数.

Sometimes we need to wrap some function into an easier-to-use one. Consider a case where we want to evaluate foo 1000 times, but only input x changes, while a and b remains same. It's of course OK to write foo(x, 1, 2) in the for loop, but you can also wrap the function before going into the loop.

a = 1;
b = 2;
foobar = @(x) foo(x, a, b); 

当调用foobar(5)时，Matlab首先调用函数句柄foobar，将5作为其唯一输入.该函数句柄有一条指令:调用名为foo的另一个函数(或函数句柄(如果已定义)). foo的参数是:x，当用户调用foobar(x)时定义.在执行功能句柄定义代码之前首先定义的a和b.

When you call foobar(5), Matlab first invokes the function handle foobar, taking 5 as its only input. That function handle has one instruction: call another function (or function handle, if you define it as so) named foo. The arguments of foo are: x, which is defined when user calls foobar(x); a and b, which have been defined in the first place BEFORE the function handle definition code is executed.

在您的情况下，fmincg仅接受仅具有一个输入参数的函数作为其第一个参数.但是lrCostFunction需要四个. fmincg不知道如何治疗x，y或lambda(我也不是).因此，将成本函数包装为一般优化器可以理解的形式是您的工作.这还需要您预先分配x，y，c和lambda.

In your case, fmincg only accepts, as its first argument, a function that only has one input argument. But lrCostFunction takes four. fmincg doesn't know how to treat x, y, or lambda (I don't either). So it's your job to wrap the cost function into the form that a general optimizer can understand. That also requires you assign x, y, c and lambda in advance.

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