normpdf表现得很奇怪

Question

以下面的方式,

function ret = f(pIx5, dS)
    sigma = 1;    

    rho = dS(1);
    theta = dS(2);

    mu_x = rho*cos(theta);

    display(pIx5);
    display(mu_x);

    pdf = normpdf(pIx5, mu_x, sigma);

    ret = max(pdf);
end

我收到以下错误消息,

pIx5 =
       54   65   11    0    0

mu_x =
       11.9218

Error using normpdf (line 36) Non-scalar arguments must match in size.

Error in f (line 11)

        pdf = normpdf(pIx5, mu_x, sigma);

但是,它可以通过以下方式正常工作,

function ret = f(pIx5, dS)
    sigma = 1;    

    rho = dS(1);
    theta = dS(2);

    pIx5 = [54,65, 11, 0, 0];

    mu_x = 11.9218;

    display(pIx5);
    display(mu_x);

    pdf = normpdf(pIx5, mu_x, sigma);

    ret = max(pdf);
end

这里发生了什么？

Answer 1

我愿意投入大量资金来解决问题与你输入的类型有关pIx5.请注意:

>> pdf = normpdf([54 65 11 0 0], 11.9218, 1);  % Works fine
>> pdf = normpdf(uint8([54 65 11 0 0]), 11.9218, 1);
Error using normpdf (line 36)
Non-scalar arguments must match in size.

为什么它会给与该类型有关的东西一个大小错误？看看代码的normpdf答案.从第33-37行,R2016b:

...
try
    y = exp(-0.5 * ((x - mu)./sigma).^2) ./ (sqrt(2*pi) .* sigma);
catch
   error(message('stats:normpdf:InputSizeMismatch'));
end

基本上,评估该等式的任何错误都会报告为大小不匹配错误.在这种情况下,它实际上是一个exp不适用于整数数据类型的问题(它只支持single和double类型):

>> x = uint8([54 65 11 0 0]);
>> mu = 11.9218;
>> sigma = 1;
>> y = exp(-0.5 * ((x - mu)./sigma).^2) ./ (sqrt(2*pi) .* sigma);
Undefined function 'exp' for input arguments of type 'uint8'.

解决方案呢？只需将您的违规输入转换为single或double首先:

pdf = normpdf(double(pIx5), mu_x, sigma);