python - 将 Matlab @ 翻译成 Python 代码(RuntimeWarning : invalid value encountered in true_divide )

Question

我正在将 Matlab 代码翻译成 python，但我被困在这里了。我认为使用@它会创建一个局部变量，对吗？我尝试用嵌套函数来引用代码，但在处理其中一个变量时似乎存在问题。

#Code's variable:

#LL0 = first tentative value
#Sad and f = returns of psd estimation via welch's method
#Sad = is the psd of discrete time-series
#f = is an array of the sampled frequencies
#u_mean = is a scalar

% Matlab Code
f = @(LL, Sad, n, U_media)sum((Sad - 4 * n * LL / U_media .* (1 + 70.8 * (n * LL / U_media).^2).^(-5/6)).^2);
fun = @(LL)f(LL, Sad, n, U_media);
LL = fminsearch(fun, LL0);

def f1(LL, Sad, n, u_mean):
    a = sum((Sad - 4 * n * LL / u_mean * (1 + 70.8 * (n * LL / u_mean)**2)**(-5/6))**2)
    return(a)
f2 = lambda LL, Sad, u_mean, f: f1(LL, Sad, n, u_mean)
fun = lambda LL: f2(LL, Sad, n, u_mean)
LL = scipy.optimize.fmin(func=fun, x0=LL0, maxfun=100000, xtol=1e-6, maxiter=10000, disp=True)

代码正在运行，但迭代给了我这个输出

RuntimeWarning: invalid value encountered in true_divide
  sum((S_adim - 4 * f * LL / u_mean * (1 + 70.8 * (f * LL / u_mean**2)**(-5/6)))**2)
RuntimeWarning: invalid value encountered in reduce
  return ufunc.reduce(obj, axis, dtype, out, **passkwargs)
Warning: Maximum number of iterations has been exceeded.

我得到的返回值与我用作暂定值的值相同谁能帮我？提前致谢

Answer 1

终于找到解决办法了；我不太确定为什么现在它起作用了，我认为我分配了同一个变量两次。无论如何，这是我找到的解决方案:

    def f(LL, Sad, n, u_mean):
        f = sum((Sad - (4 * n * LL / u_mean) *
                 ((1 + 70.8 * ((n * LL / u_mean) ** (2)))**(-5 / 6)))**2)
        return(f)

    fun = lambda LL: f(LL, Sad, n, u_mean)
    res = scipy.optimize.minimize(fun=fun, x0=LL0)
    Lux = res.x

顺便说一句，spyder ID 一直说我不应该使用 lambda 函数，而应该定义一个函数。谁能告诉我为什么？我能够将“乐趣”转化为一种功能；我尝试过:

def fun(LL):
   f(LL, Sad, n, u_mean)
   return(f)

res = scipy.optimize.minimize(fun=fun, x0=LL0)

但它不起作用。如果有人能告诉我原因，那就太好了。

感谢您的帮助