Question

我有一个理性（这里：双线性）表达，并希望我同情收集系数。但是如何？

from sympy import symbols, Wild, pretty_print

a, b, c, d, x, s = symbols("a b c d x s")

def coeffs(expr):
    n0 = Wild("n0", exclude=[x])
    n1 = Wild("n1", exclude=[x])
    d0 = Wild("d0", exclude=[x])
    d1 = Wild("d1", exclude=[x])
    match = expr.match((n0 + n1*x) / (d0 + d1*x))
    n0 = n0.xreplace(match)
    n1 = n1.xreplace(match)
    d0 = d0.xreplace(match)
    d1 = d1.xreplace(match)
    return [n0, n1, d0, d1]

if __name__ == '__main__':
    pretty_print(coeffs((a + b*x) / (c + d*x)))
    pretty_print(coeffs(2 * (a + b*x) / (c + d*x)))
    pretty_print(coeffs(s * (a + b*x) / (c + d*x)))

我使用match尝试了这一点，但它几乎总是失败，例如在最后一行（带有符号前因子的那个＆＃34; s＆＃34;）我得到了

Traceback (most recent call last):
  File "...", line 20, in <module>
    pretty_print(coeffs(s * (a + b*x) / (c + d*x)))
  File "...", line 11, in coeffs
    n0 = n0.xreplace(match)
  File "/usr/lib/python3/dist-packages/sympy/core/basic.py", line 1626, in xreplace
    return rule.get(self, self)
AttributeError: 'NoneType' object has no attribute 'get'

所以比赛没有用。

Answer 1

如果您知道您将考虑的表达式是合理的，您可以提取它们的分子和分母并独立找到它们的系数。

这是一种方法：

import sympy as sp

def get_rational_coeffs(expr):
    num, denom = expr.as_numer_denom()

    return [sp.Poly(num, x).all_coeffs(), sp.Poly(denom, x).all_coeffs()]

a, b, c, d, x, s = sp.symbols("a b c d x s") 
expr = (a + b*x) / (c + d*x)

# note the order of returned coefficients
((n1, n0), (d1, d0)) = get_rational_coeffs(s*expr)
print(((n0, n1), (d0, d1)))

((a*s, b*s), (c, d))

上述方法也比coeffs快。对于需要%timeit系数的情况（expr成功的情况），我得到以下时间（基于Jupyter的coeffs魔法）：

%timeit get_rational_coeffs(expr)
%timeit coeffs(expr)

1000 loops, best of 3: 1.33 ms per loop

1000 loops, best of 3: 1.99 ms per loop

在同情中，我如何得到理性表达式的系数？

1 个答案: