英文:
Conditional function for each element of a matrix
问题
我想要定义一个函数 y(x)，如下所示：
如果 x >= 0： y = x^(1/2)
如果 x < 0： y = NaN
我已经尝试了下面的代码，显然不起作用。
y = @(x) (x>=0) * x.^(1/2) - NaN* (x<0);
Test = y([-1,1,4,9])
我的期望是：
y([-1,1,4,9]) = [NaN, 1, 2, 3];
在我有限的理解中，arrayfun 可能会有帮助，但速度可能会很慢。请帮我找到一个解决方案。非常感谢。
编辑：正如Cris Luengo指出的那样，我在问题中不小心没有使用逐元素相乘 .*。请注意，即使使用 .*，我的方法仍然不起作用。
y = @(x) (x>=0) .* x^(1/2) - NaN.* (x<0);
Test = y([-1,1,4,9])
英文:
I want to define a function y(x) as follows:
if x >= 0: y = x^(1/2)
if x< 0: y = NaN
I have tried below code, which clearly does not work.
y = @(x) (x>=0) * x.^(1/2) - NaN* (x<0);
Test = y([-1,1,4,9])
My expectation is that:
y([-1,1,4,9]) = [NaN, 1, 2, 3];
In my limited understanding, arrayfun might help, but it would be very slow. Kindly help me a solution. Many thanks.
Edit. As pointed out by Cris Luengo, I did negligently not use wise-element .* in my question. Note that, even if using .*, my approach does not work.
y = @(x) (x>=0) .* x^(1/2) - NaN.* (x<0);
Test = y([-1,1,4,9])
答案1
得分: 2
函数的目的是什么？如果在内联函数中实现if语句很困难，那么使用如下的关系运算符如何？
x = [-1, 1, 4, 9];
x(x>=0) = x(x>=0).^(1/2);
x(x<0) = NaN;
>> x
x =
NaN 1 2 3
英文:
What is the purpose of having a function? If statement is hard to implement in an inline function. What about using relational operators like below?
x = [-1, 1, 4, 9];
x(x>=0) = x(x>=0).^(1/2);
x(x<0) = NaN;
>> x
x =
NaN 1 2 3
答案2
得分: 2
代码中的问题在于给定的代码中，NaN 乘以任何数都是 NaN。NaN 会通过设计在所有计算中传播；它是一个错误条件。
（代码的另一个问题是它计算了矩阵乘法和矩阵幂。应该使用元素级运算符 .* 和 .^：(x>=0) .* x.^(1/2)。）
通常，你可以在两行代码中实现你的目标：
y = x.^(1/2);
y(x<0) = NaN;
但对于这种特殊情况，我们可以通过将 x>=0 除以自身来创建一个在 x<0 时为 NaN 的值：(x>=0)./(x>=0) .* x.^(1/2)。除法的结果要么是 NaN，要么是 1，乘以 x 的平方根将得到 NaN 或 x 的平方根。
在这里，我发现另一个问题：如果 x 的任何元素为负数，x.^(1/2) 就是复数。因此，我们需要取其结果的实部：
y = @(x) (x>=0)./(x>=0) .* real(x.^(1/2));
计算这个的最快方式需要使用一个常规函数（而不是在 OP 中使用的匿名函数）。它会像这样：
function y = sqrt_or_nan(x)
y = nan(size(x));
index = x < 0;
y(index) = sqrt(x(index));
这避免了重复计算和不必要的计算，并使用 sqrt 而不是 .^(1/2)（我假设 sqrt 更快，尽管我没有计时验证）。
英文:
The problem with the code as given is that NaN times anything is NaN. NaN propagates through all calculations, by design; it's an error condition.
(Another problem with the code is that it computes the matrix multiplication and the matrix exponent. Use the element-wise operators .* and .^ instead: (x>=0) .* x.^(1/2).)
Typically you'd accomplish your goal in two lines:
y = x.^(1/2);
y(x<0) = NaN;
But for this particular case we can create a NaN where x<0 by dividing x>=0 by itself: (x>=0)./(x>=0) .* x.^(1/2). The result of the division is either NaN or 1, which multiplied by the square root of x yields either NaN or the square root of x.
Here I found yet another issue: x.^(1/2) is complex if any element of x is negative. So we need to take the real component of its result:
y = @(x) (x>=0)./(x>=0) .* real(x.^(1/2));
The fastest way to compute this would need to be in a regular function (as opposed to an anonymous function as you have in the OP. It would look like this:
function y = sqrt_or_nan(x)
y = nan(size(x));
index = x < 0;
y(index) = sqrt(x(index));
This avoids duplicate computations and unnecessary computations, and uses sqrt instead of .^(1/2) (which I presume is faster, though I haven't timed it).
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