Question

我想集成一个3x3矩阵，并让它以3x3矩阵的形式返回解决方案，但是我不确定在python中是否可行。有没有我可以使用的命令？如果没有，我是否应该设置某种for循环（如果是这种情况，是否可以在此方面提供一些帮助）？

我尝试使用其他建议，例如vectorize（integrate.quad）（integrandtemp，0、10）和vectorize（quad）（integrandtemp，0、10），但我遇到了与我相同的错误：“仅size-1数组可以转换为Python标量”。

def integrandtemp(s):
    K = 17.5
    r = 0.7
    x0 = 0.1

    t = 5
    x = K/(1+((K/x0)-1)*np.exp(-r*s))

    dxdK = (x0*x0-x0*x0*np.exp(-r*s))/((x0+K*np.exp(-r*s)-x0*np.exp(-r*s))**2)
    dxdr = (K*K*x0*s*np.exp(-r*s)-K*x0*x0*s*np.exp(-r*s))/((x0+K*np.exp(-r*s)-x0*np.exp(-r*s))**2)
    dxdx0 = (K*x0+K*K*np.exp(-r*s)-K*x0*np.exp(-r*s)-K*x0+K*x0*np.exp(-r*s))/((x0+K*np.exp(-r*s)-x0*np.exp(-r*s))**2)

    M = [dxdK, dxdr, dxdx0]
    M = np.array([M])
    print(M)

    transpose = M.T
    print(transpose)
    var = 0.16
    print(transpose@M)
    return (1/var)*transpose@M

F = vectorize(integrate.quad)(integrandtemp, 0, 10)
F_inv = inv(F)

我收到此错误消息：

TypeError                                 Traceback (most recent call last)
<ipython-input-106-a2757b67abc5> in <module>
     34     return (1/var)*transpose@M
     35 
---> 36 F = vectorize(integrate.quad)(integrandtemp, 0, 10)
     37 F_inv = inv(F)
     38 

~/anaconda3/lib/python3.7/site-packages/numpy/lib/function_base.py in __call__(self, *args, **kwargs)
   1970             vargs.extend([kwargs[_n] for _n in names])
   1971 
-> 1972         return self._vectorize_call(func=func, args=vargs)
   1973 
   1974     def _get_ufunc_and_otypes(self, func, args):

~/anaconda3/lib/python3.7/site-packages/numpy/lib/function_base.py in _vectorize_call(self, func, args)
   2040             res = func()
   2041         else:
-> 2042             ufunc, otypes = self._get_ufunc_and_otypes(func=func, args=args)
   2043 
   2044             # Convert args to object arrays first

~/anaconda3/lib/python3.7/site-packages/numpy/lib/function_base.py in _get_ufunc_and_otypes(self, func, args)
   2000 
   2001             inputs = [arg.flat[0] for arg in args]
-> 2002             outputs = func(*inputs)
   2003 
   2004             # Performance note: profiling indicates that -- for simple

~/anaconda3/lib/python3.7/site-packages/scipy/integrate/quadpack.py in quad(func, a, b, args, full_output, epsabs, epsrel, limit, points, weight, wvar, wopts, maxp1, limlst)
    339     if weight is None:
    340         retval = _quad(func, a, b, args, full_output, epsabs, epsrel, limit,
--> 341                        points)
    342     else:
    343         retval = _quad_weight(func, a, b, args, full_output, epsabs, epsrel,

~/anaconda3/lib/python3.7/site-packages/scipy/integrate/quadpack.py in _quad(func, a, b, args, full_output, epsabs, epsrel, limit, points)
    446     if points is None:
    447         if infbounds == 0:
--> 448             return _quadpack._qagse(func,a,b,args,full_output,epsabs,epsrel,limit)
    449         else:
    450             return _quadpack._qagie(func,bound,infbounds,args,full_output,epsabs,epsrel,limit)

TypeError: only size-1 arrays can be converted to Python scalars

我需要它返回另一个3x3矩阵，以便我可以求逆并将其用于实际尝试求解的事情。

Answer 1

quadpy（属于我的项目）对数组值函数进行正交运算：

<div class="imgs">
  <img src="https://example.com/upload/zxcvbn.jpg">
  <img src="https://example.com/upload/qwerty.jpg">
</div>

import numpy as np
import quadpy


def integrandtemp(S):
    out = []
    for s in S:
        K = 17.5
        r = 0.7
        x0 = 0.1

        dxdK = (x0 * x0 - x0 * x0 * np.exp(-r * s)) / (
            (x0 + K * np.exp(-r * s) - x0 * np.exp(-r * s)) ** 2
        )
        dxdr = (K * K * x0 * s * np.exp(-r * s) - K * x0 * x0 * s * np.exp(-r * s)) / (
            (x0 + K * np.exp(-r * s) - x0 * np.exp(-r * s)) ** 2
        )
        dxdx0 = (
            K * x0
            + K * K * np.exp(-r * s)
            - K * x0 * np.exp(-r * s)
            - K * x0
            + K * x0 * np.exp(-r * s)
        ) / ((x0 + K * np.exp(-r * s) - x0 * np.exp(-r * s)) ** 2)

        M = [dxdK, dxdr, dxdx0]
        M = np.array([M])

        transpose = M.T
        var = 0.16

        out.append((1 / var) * transpose @ M)

    out = np.array(out)
    out = np.moveaxis(out, 0, -1)
    return out


scheme = quadpy.line_segment.gauss_legendre(10)
val = scheme.integrate(integrandtemp, [0, 10])
print(val)

除了循环遍历[[4.78541989e+00 2.77802115e+02 3.35326738e+02] [2.77802115e+02 2.32166322e+04 3.12478945e+04] [3.35326738e+02 3.12478945e+04 4.37373431e+04]]之外，您还可以向量化函数以提高速度。请注意，输出的形状必须为S。

有没有一种方法可以整合矩阵的每个项并将解决方案作为另一个矩阵返回？

1 个答案: