【发布时间】:2015-04-25 06:13:42
【问题描述】:
我想使用 Python 来实现 Simpson 集成。如果我不需要它自动收敛(这里我需要abs(result - my_expect) < 0.001)并不难。
但我想要一个用 Python 编写的自动融合程序。所以我尝试使用我从SICP中学到的方法——递归实现。
# imports
from __future__ import division
import numpy as np
# constants
MU = 50
SIGMA = 15
X0 = 0
XN = 100
# interface functions
def f(x):
"""Integrand.
It will be used for Lagrangian interpolation and
calculating corresponding function value in function
`improve_precision`.
"""
y = (1 / (SIGMA * np.sqrt(2 * np.pi))
* np.exp(-1/2 * (x-MU)**2 / SIGMA**2))
return y
# classes
class Integration:
def __init__(self, start, end):
self.start = start
self.end = end
self.a = (self.end - self.start) / 2
self.h = 2 * self.a
self.rp = 0
self.s = 0
self.i = 1
def integrate(self):
return self.integrate_iter(1, 0, 0.001)
def integrate_iter(self, expectation, summation, accuracy):
if accuracy > abs(summation - expectation):
return summation
else:
self.integrate_iter(expectation,
self.improve_precision(self.i+1),
accuracy)
def improve_precision(self, ii):
rc = np.sum([f(-self.a - self.h/2 + k*self.h)
for k in range(1, 2**ii + 1)])
self.s = self.h/6 * (self.rp + 4*rc)
self.h /= 2
self.rp = self.rp + 2*rc
return self.s
calc = Integration(X0, XN)
print calc.integrate()
当我运行它时,出现了很多错误:
File "/Users/*/integrate.py", line 72, in <module>
print calc.integrate()
File "/Users/*/integrate.py", line 56, in integrate
return self.integrate_iter(1, 0, 0.001)
File "/Users/*/integrate.py", line 62, in integrate_iter
self.integrate_iter(expectation, self.improve_precision(self.i+1), accuracy)
File "/Users/*/integrate.py", line 62, in integrate_iter
self.integrate_iter(expectation, self.improve_precision(self.i+1), accuracy)
...
File "/Users/*/integrate.py", line 65, in improve_precision
rc = np.sum([f(-self.a - self.h/2 + k*self.h) for k in range(1, 2**ii + 1)])
File "/usr/local/lib/python2.7/site-packages/numpy/core/fromnumeric.py", line 1708, in sum
if isinstance(a, _gentype):
RuntimeError: maximum recursion depth exceeded while calling a Python object
Process finished with exit code 1
那我该怎么办?有什么办法可以改正吗?
编辑:
好吧,我对函数integrate_iter 进行了更改,并添加了一个函数is_rough。我的最终代码是这样的:
...
def integrate(self):
"""A function encapsulate calculating process."""
integrate_value = self.integrate_iter(0)
interval_count = self.i
return integrate_value, interval_count
def integrate_iter(self, summation):
"""Main part of recursive process."""
last_summation = summation
summation = self.improve_precision(self.i)
return self.is_good(summation, last_summation)
def is_good(self, summ, last_summ):
# if precision < 0.001, over; else, call `integrate_iter` again.
if abs(summ - last_summ) <= self.accuracy:
return summ
else:
self.i += 1
return self.integrate_iter(summ)
...
calc = Integration(X0, XN, 0.001)
val = calc.integrate()
scipy_result = scipy.integrate.quad(f, X0, XN, epsabs=0.001)[0]
print 'The integration is: {0:.8f}'.format(val[0])
print 'The error is: {0:.8f}'.format(abs(val[0] - scipy_result))
print 'The number of recursions is: {0:d}'.format(val[1])
print 'The number of intervals is: {0:d}'.format(2**val[1])
我对递归的次数不是很满意,不知道是不是应该收敛这么慢(根据我的经验)。
【问题讨论】:
-
向我们展示完整的回溯。这不是“很多错误”。这是一个错误的回溯。
-
@dbliss 我愿意。但是回溯太长了。但是这个问题的正文限制是 30000 个字符。我该怎么办?
-
my_expect的值是多少?了解您希望输出的内容将帮助我们找到适合您的答案。 -
@dbliss 是 1。对不起,我没有澄清它,但我在这个声明中使用了它:
return self.integrate_iter(1, 0, 0.001) -
^ 啊,我的错,错过了。
标签: python numpy numerical-methods