python - 在 scipy.optimize 中分配 fmin 的输出

Question

我有一个单一变量的函数，我想找到它的最小值，以及达到最小值的变量的值。目前，我通过以下 Python 脚本实现了这一点:

import numpy as np
from scipy.optimize import fmin
import math

x1=0.
y1=800.
x2=1100.
y2=-800.

v1=2000.
v2=4000.

def T(xi):
    time=sqrt((x1-xi)**2+y1**2)/v1+sqrt((x2-xi)**2+y2**2)/v2
    return time

fmin(T,0)

运行此脚本会产生以下回显:

import numpy as np
from scipy.optimize import fmin
import math

x1=0.
y1=800.
x2=1100.
y2=-800.

v1=2000.
v2=4000.

def T(xi):
    time=sqrt((x1-xi)**2+y1**2)/v1+sqrt((x2-xi)**2+y2**2)/v2
    return time

fmin(T,0)
Optimization terminated successfully.
         Current function value: 0.710042
         Iterations: 41
         Function evaluations: 82
Out[24]: array([ 301.9498125])

因此函数的最小值是 ~0.71，并且是通过 ~302 的参数值获得的。但是，我想按如下方式分配这些值:

(Tmin,xmin)=fmin(T,0)
Optimization terminated successfully.
         Current function value: 0.710042
         Iterations: 41
         Function evaluations: 82
---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
C:\Users\Kurt.Peek\<ipython-input-25-aec613726d59> in <module>()
----> 1 (Tmin,xmin)=fmin(T,0)

ValueError: need more than 1 value to unpack

所以我得到一个错误“ValueError:需要超过 1 个值来解包”。有人知道如何防止此错误并分配这两个输出吗？

Answer 1

fmin has a full_output=True parameter :

xopt, fopt, iter, funcalls, warnflag = fmin(T,0, full_output=True, disp=False)
print(xopt, fopt)
# (array([ 301.9498125]), 0.71004171552448025)