问题描述
给定一个符号多元多项式P
,我需要将其系数和相应的单项式提取为列表:
Given a symbolic multivariate polynomial P
, I need to extract both its coefficients and corresponding monomials as lists:
def poly_decomp(P):
....
return coeffs, monoms
使得 P
是系数和单项式的点积,例如,如果 P(x,y) = ax**2 + bxy + cy**2
那么我们应该得到 coeffs = [a, b, c]
和 monoms = [x**2, x*y, y**2]
.
such that P
is the dot product of coefficients and monomials, e.g., if P(x,y) = ax**2 + bxy + cy**2
then we should get coeffs = [a, b, c]
and monoms = [x**2, x*y, y**2]
.
获得系数很容易,因为函数内置于coeffs = P.coeffs()
.但是,我无法获得单项式.这里内置函数返回一个指数列表,例如,在上面的例子中,我们将得到 P.monoms() = [(2,0),(1,1),(0,2)]代码>.
Getting the coefficients is easy since the function is built in coeffs = P.coeffs()
. However, I'm having trouble getting the monomials. Here the build in function returns a list of exponents, e.g., in the example above we would get P.monoms() = [(2,0),(1,1),(0,2)]
.
显然的想法是,提供一个变量列表var=[x,y]
,来做类似的事情
Obviously the idea would be, provided a list of the variables var=[x,y]
, to do something like
powers = P.monoms()
monoms = [sympy.prod(x**k for x,k in zip(var, mon)) for mon in powers ]
然而,多项式类似乎没有提供返回变量列表的函数.我能找到的只有 free_symbols
和 free_symbols_in_domain
方法,它们返回集合 {a, b, c, x, y}
和 {a, b, c}
.因此,通过取它们的差异,可以得到 set {x, y}
.
However the polynomial class doesn't seem to offer a function that returns a list of variables. All I could find were the methods free_symbols
and free_symbols_in_domain
which return the sets {a, b, c, x, y}
and {a, b, c}
. So by taking their difference one could get the set {x, y}
.
然而,我们面临的问题是集合是无序的,因此将其转换为列表可能会根据变量的数量以不同的方式弄乱顺序.
However then we are faced with the issue that the sets are unordered, hence converting it into a list might mess up the order in different ways depending on the number of variables.
我有点不知所措.有什么提示吗?
I am kind of at a loss here. Any tips?
推荐答案
属性 gens
(generators 的缩写)包含一个 tuple
符号或其他合适的对象,定义了多项式.
The property gens
(short for generators) holds a tuple
of symbols or other suitable objects that the polynomial is defined over.
from sympy import symbols, Poly
x, y = symbols('x y')
p = Poly(x**3 + 2*x**2 + 3*x*y + 4*y**2 + 5*y**3, x, y)
q = Poly(x**3 + 2*x**2 + 3*x*y + 4*y**2 + 5*y**3, y, x)
print(p.gens) # (x, y)
print(q.gens) # (y, x)
所以,
[prod(x**k for x, k in zip(p.gens, mon)) for mon in p.monoms()]
返回[x**3, x**2, x*y, y**3, y**2]
.
另请注意,生成器可以是符号以外的类型,例如:
Note also that the generators can be types other than symbols, for example:
import sympy
x = sympy.symbols('x')
poly = sympy.poly(sympy.sqrt(2) * x**2)
print('generators: {g}'.format(g=poly.gens))
print('monomials: {m}'.format(m=poly.monoms()))
print('coefficients: {c}'.format(c=poly.coeffs()))
打印:
generators: (x, sqrt(2))
monomials: [(2, 1)]
coefficients: [1]
哪里:
type(poly.gens[0])
是,和
type(poly.gens[1])
是.
type(poly.gens[0])
is<class 'sympy.core.symbol.Symbol'>
, andtype(poly.gens[1])
is<class 'sympy.core.power.Pow'>
.
一个相关的方法是sympy.polys.polytools.Poly.as_dict
,它返回一个dict
,键是单项式,值是相应的系数.
A relevant method is sympy.polys.polytools.Poly.as_dict
, which returns a dict
with keys that are monomials, and values that are the corresponding coefficients.
这篇关于从 SymPy 中的给定多项式中提取系数和相应的单项式的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!