Question

有一个MATLAB函数quadgk可以计算复数积分，或至少有极点和奇点的函数。在Python中，有一个通用的scipy.integrate.quad，它可以方便地沿着实轴进行集成。是否有Python等同于MATLAB＆＃39; quadgk？我在另一个SO问题中找到的只是dr jimbob's code，它似乎不再适用于Python 3.4。

Answer 1

我不认为SciPy确实提供了相当于MATLAB的output = subprocess.check_output(['python', 'script.py']).decode('utf-8')，但是对于你在this question中链接到的代码是值得的只需稍作修改即可在Python 3中工作：

quadgk

例如，

import scipy
from scipy import array

def quad_routine(func, a, b, x_list, w_list):
    c_1 = (b-a)/2.0
    c_2 = (b+a)/2.0
    eval_points = map(lambda x: c_1*x+c_2, x_list)
    func_evals = list(map(func, eval_points))    # Python 3: make a list here
    return c_1 * sum(array(func_evals) * array(w_list))

def quad_gauss_7(func, a, b):
    x_gauss = [-0.949107912342759, -0.741531185599394, -0.405845151377397, 0, 0.405845151377397, 0.741531185599394, 0.949107912342759]
    w_gauss = array([0.129484966168870, 0.279705391489277, 0.381830050505119, 0.417959183673469, 0.381830050505119, 0.279705391489277,0.129484966168870])
    return quad_routine(func,a,b,x_gauss, w_gauss)

def quad_kronrod_15(func, a, b):
    x_kr = [-0.991455371120813,-0.949107912342759, -0.864864423359769, -0.741531185599394, -0.586087235467691,-0.405845151377397, -0.207784955007898, 0.0, 0.207784955007898,0.405845151377397, 0.586087235467691, 0.741531185599394, 0.864864423359769, 0.949107912342759, 0.991455371120813]
    w_kr = [0.022935322010529, 0.063092092629979, 0.104790010322250, 0.140653259715525, 0.169004726639267, 0.190350578064785, 0.204432940075298, 0.209482141084728, 0.204432940075298, 0.190350578064785, 0.169004726639267, 0.140653259715525,  0.104790010322250, 0.063092092629979, 0.022935322010529]
    return quad_routine(func,a,b,x_kr, w_kr)

class Memoize:                     # Python 3: no need to inherit from object
    def __init__(self, func):
        self.func = func
        self.eval_points = {}
    def __call__(self, *args):
        if args not in self.eval_points:
            self.eval_points[args] = self.func(*args)
        return self.eval_points[args]

def quad(func,a,b):
    ''' Output is the 15 point estimate; and the estimated error '''
    func = Memoize(func) #  Memoize function to skip repeated function calls.
    g7 = quad_gauss_7(func,a,b)
    k15 = quad_kronrod_15(func,a,b)
    # I don't have much faith in this error estimate taken from wikipedia
    # without incorporating how it should scale with changing limits
    return [k15, (200*scipy.absolute(g7-k15))**1.5]

Python中的复杂集成

1 个答案: