拥有列表并尝试在Python中创建多项式（符号计算）

Question

假设我有以下列表：

f=[1, 3, 4, 5, 3, 4, 5, 6, 6, 3, 3, 1, 1, 4, 2]

我可以使用两种方法计算列表中每个整数的数量。

第一：

from collections import Counter
Counter(f)
#Counter({1: 3, 2: 1, 3: 4, 4: 3, 5: 2, 6: 2})

或第二：

[[x,f.count(x)] for x in set(f)]
#[[1, 3], [2, 1], [3, 4], [4, 3], [5, 2], [6, 2]]

第二种方法更可取，因为它会向我抛出一个列表。 现在，我想将输出转换为多项式，其中子列表的第一个元素是x的幂，子列表的第二个元素是系数，最后将它们求和以形成多项式，这样我得到：

3 x + x^2 + 4 x^3 + 3 x^4 + 2 x^5 + 2 x^6

为了使这个多项式，我使用Sympy如下：

from sympy import Array, tensorproduct
from sympy.abc import x
from sympy import transpose
A = Array([x])
B = Array([x[1] for x in [[x,f.count(x)] for x in set(f)]])
tensorproduct(A, B)
#[[3*x, x, 4*x, 3*x, 2*x, 2*x]]

现在我不确定如何提高x的正确幂？ 也有更好的解决方案吗？

Answer 1

这应该为您工作：

from sympy import poly, var
from collections import Counter

x = var("x")
f = [1, 3, 4, 5, 3, 4, 5, 6, 6, 3, 3, 1, 1, 4, 2]
count = Counter(f)
p = sum(coeff * x ** exp for exp, coeff in count.items())
# 2*x**6 + 2*x**5 + 3*x**4 + 4*x**3 + x**2 + 3*x

其中**用于求幂 。

Answer 2

您是要实际评估多项式，还是只是获得漂亮的字符串？

from collections import Counter
superscript_digits = "⁰¹²³⁴⁵⁶⁷⁸⁹"

f=[1, 3, 4, 5, 3, 4, 5, 6, 6, 3, 3, 1, 1, 4, 2]
coeffs_by_exp = Counter(f)
print(' + '.join("{}x{}".format(coeff, superscript_digits[exp])
                     for exp, coeff in coeffs_by_exp.items()))

得到：

3x¹ + 1x² + 4x³ + 3x⁴ + 2x⁵ + 2x⁶

省略1个系数以及0和1个指数的特殊处理作为OP的练习。

Answer 3

您也可以看看这个它运作良好，但与你想实现（你可以计算出导数，...）相比，有什么可能过于复杂。

更具体地说，这段代码可以帮助您：

import numpy as np
import matplotlib.pyplot as plt
class Polynomial:

def __init__(self, *coefficients):
    """ input: coefficients are in the form a_n, ...a_1, a_0 
    """
    # for reasons of efficiency we save the coefficients in reverse order,
    # i.e. a_0, a_1, ... a_n
    self.coefficients = coefficients[::-1] # tuple is also turned into list

def __repr__(self):
    """
    method to return the canonical string representation 
    of a polynomial.

    """
    # The internal representation is in reverse order,
    # so we have to reverse the list
    return "Polynomial" + str(self.coefficients[::-1])

def __call__(self, x):    
    res = 0
    for index, coeff in enumerate(self.coefficients):
        res += coeff * x** index
    return res 

def degree(self):
    return len(self.coefficients)

@staticmethod
def zip_longest(iter1, iter2, fillchar=None):    
    for i in range(max(len(iter1), len(iter2))):
        if i >= len(iter1):
            yield (fillchar, iter2[i])
        elif i >= len(iter2):
            yield (iter1[i], fillchar)
        else:
            yield (iter1[i], iter2[i])
        i += 1   

def __add__(self, other):
    c1 = self.coefficients
    c2 = other.coefficients
    res = [sum(t) for t in Polynomial.zip_longest(c1, c2, 0)]
    return Polynomial(*res[::-1])

def __sub__(self, other):
    c1 = self.coefficients
    c2 = other.coefficients
    res = [t1-t2 for t1, t2 in Polynomial.zip_longest(c1, c2, 0)]
    return Polynomial(*res[::-1])

def derivative(self):
    derived_coeffs = []
    exponent = 1
    for i in range(1, len(self.coefficients)):
        derived_coeffs.append(self.coefficients[i] * exponent)
        exponent += 1
    return Polynomial(*derived_coeffs[::-1])

def __str__(self):
    res = ""
    for i in range(len(self.coefficients)-1, -1, -1):
        res +=  str(self.coefficients[i]) + "x^" + str(i) + " + "
    if res.endswith(" + "):
        res = res[:-3]
    return res

然后，您可以使用：

 p = Polynomial(-0.8, 2.3, 0.5, 1, 0.2)

要得到：

希望这可以帮到你！

Answer 4

Poly可以从系数列表中实例化。 defaultdict可用于填写缺少的权力：

>>> d = defaultdict(int, multiset(f))
>>> Poly.from_list([d[i] for i in range(max(d), -1, -1)], x)
Poly(2*x**6 + 2*x**5 + 3*x**4 + 4*x**3 + x**2 + 3*x, x, domain='ZZ')
>>> _.as_expr()
2*x**6 + 2*x**5 + 3*x**4 + 4*x**3 + x**2 + 3*x