matlab常量匿名函数仅返回一个值，而不是数组 [英] matlab constant anonymous function returns only one value instead of an array

问题描述

我已经在网上搜寻了两个早晨，却什么也没发现，希望能对您有所帮助.

I've been searching the net for a couple of mornings and found nothing, hope you can help.

我有这样的匿名功能

f = @(x,y) [sin(2*pi*x).*cos(2*pi*y), cos(2*pi*x).*sin(2*pi*y)];

需要对点数组进行评估，例如

that needs to be evaluated on an array of points, something like

x = 0:0.1:1;
y = 0:0.1:1;
w = f(x',y');

现在，在上面的示例中，一切正常，结果w是一个11x2矩阵，在每一行中都有正确的值f(x(i)，y(i)). 当我将函数更改为具有恒定值时，就会出现问题:

Now, in the above example everything works fine, the result w is a 11x2 matrix with in each row the correct value f(x(i), y(i)). The problem comes when I change my function to have constant values:

f = @(x,y) [0, 1];

现在，即使像以前一样使用数组输入，我也只能得到一个1x2数组，例如w = [0,1]; 当然，我希望具有与以前相同的结构，即11x2矩阵.

Now, even with array inputs like before, I only get out a 1x2 array like w = [0,1]; while of course I want to have the same structure as before, i.e. a 11x2 matrix.

我不知道Matlab为什么要这么做...

I have no idea why Matlab is doing this...

编辑1 抱歉，我以为我在原始问题中写的内容很清楚，但是我看到一些人在问，所以这里有一个澄清:我想要的是再次拥有一个11x2矩阵，因为我正在向函数提供数组具有11个元素. 这意味着我希望输出与第一个示例完全相同，只是其中的值发生了变化:一个矩阵，该矩阵包含11行2列，第一列只有0值，第二列只有1值，因为对于所有x(i)和y(i)的答案应该是向量[0,1]. 这意味着我希望拥有:

EDIT 1 Sorry, I thought it was pretty clear from what I wrote in the original question, but I see some of you asking, so here is a clarification: what I want is to have again a 11x2 matrix, since I am feeding the function with arrays with 11 elements. This means I expect to have an output exactly like in the first example, just with changed values in it: a matrix with 11 rows and 2 columns, with only values 0 in the first column and only values 1 in the second, since for all x(i) and y(i) the answer should be the vector [0,1]. It means I expect to have:

w = [0 1
     0 1
     0 1
     ...
     0 1]

对我来说似乎很自然...

seems pretty natural to me...

推荐答案

您正在定义一个函数f = @(x,y) [0, 1];，该函数具有输入参数x，y和输出[0,1].您还希望发生什么?

You are defining a function f = @(x,y) [0, 1]; which has the input parameters x,y and the output [0,1]. What else do you expect to happen?

更新:

这应符合您的描述

g=@(x,y)[zeros(size(x)),ones(size(y))]
g(x',y')

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