在Python方法中记住单个参数

时间:2015-05-26 15:20:50

标签: python numpy recursion memoization

我编写了一个Numpy实现,它使用Cox-de Boor递归算法来计算B样条基函数。我想memoize给定order的对象实例,但保留相对于xi可调用的函数。

换句话说,在实例化对象之后,递归函数应该是" set",但仍然可以在xi调用。我真的需要这个速度,因为我将多次调用该函数,并且不想重复重建递归函数。

以下是当前的实施:

import numpy as np

#Turn off divide by zero warning because we explicitly check for it
np.seterr(divide='ignore')

class Bspline():

    def __init__(self, knot_vector, order):

        self.knot_vector = knot_vector
        self.p = order


    def __basis0(self, xi):

        return np.where(np.all([self.knot_vector[:-1] <=  xi, 
                            xi < self.knot_vector[1:]],axis=0), 1.0, 0.0)

    def __basis(self, xi, p):

        if p == 0:
            return self.__basis0(xi)
        else:
            basis_p_minus_1 = self.__basis(xi, p - 1)

            first_term_numerator = xi - self.knot_vector[:-p] 
            first_term_denominator = self.knot_vector[p:] - self.knot_vector[:-p]

            second_term_numerator = self.knot_vector[(p + 1):] - xi
            second_term_denominator = self.knot_vector[(p + 1):] - self.knot_vector[1:-p]

            first_term = np.where(first_term_denominator > 1.0e-12, 
                              first_term_numerator / first_term_denominator, 0)
            second_term = np.where(second_term_denominator > 1.0e-12,
                               second_term_numerator / second_term_denominator, 0)

            return  first_term[:-1] * basis_p_minus_1[:-1] + second_term * basis_p_minus_1[1:]


    def __call__(self, xi):

        return self.__basis(xi, self.p)

并用作

knot_vector = np.array([0,0,0,0,0,1,2,2,3,3,3,4,4,4,4,5,5,5,5,5])
basis = Bspline(knot_vector,4)
basis(1.2)

返回在1.2评估的基函数。但是,我需要多次调用此函数,并且正如现在编写的那样,每次调用都会重构递归函数,这不是必需的,因为递归级别在实例化时设置为4

2 个答案:

答案 0 :(得分:3)

在Python3中使用functools.lru_cache或使用this之类的内容在Python2.7中记忆任何内容都非常容易:

class Bspline(object):
    ...

    # Python2.7
    @memoize
    # or, Python3*
    @functools.lru_cache()
    def op(self, args):
        return self._internal_op(xi)

答案 1 :(得分:0)

创建一个保存函数结果的字典,然后在再次调用该函数之前检查该值是否在dict中。根据您的设置,类似的东西应该有效。

results = {}
for value in values:
    if results.get(value):
        answer = results[value]
    else:
        answer = basis(value)
        results[value] = answer
        print(answer)  # or just display the results dict once you are done