Question

我的代码是：

import numpy as np
import matplotlib as plt
polyCoeffiecients = [1,2,3,4,5]
plt.plot(PolyCoeffiecients)
plt.show()

这样的结果是描述1,2,3,4,5中的点和它们之间的直线的直线，而不是具有1,2,3,4,5的5的多项式。其系数（P（x）= 1 + 2x + 3x + 4x + 5x）

我怎么想用一个只用系数绘制多项式？

Answer 1

Eyzuky，看看这是不是你想要的：

import numpy as np
from matplotlib import pyplot as plt

def PolyCoefficients(x, coeffs):
    """ Returns a polynomial for ``x`` values for the ``coeffs`` provided.

    The coefficients must be in ascending order (``x**0`` to ``x**o``).
    """
    o = len(coeffs)
    print(f'# This is a polynomial of order {ord}.')
    y = 0
    for i in range(o):
        y += coeffs[i]*x**i
    return y

x = np.linspace(0, 9, 10)
coeffs = [1, 2, 3, 4, 5]
plt.plot(x, PolyCoefficients(x, coeffs))
plt.show()

Answer 2

一个非常Python的解决方案是使用列表推导来计算该函数的值。

import numpy as np
from matplotlib import pyplot as plt

x = np.linspace(0, 10, 11)
coeffs = [1, 2, 3, 4, 5]
y = np.array([np.sum(np.array([coeffs[i]*(j**i) for i in range(len(coeffs))])) for j in x])
plt.plot(x, y)
plt.show()

Answer 3

您可以通过获取许多x值并使用np.polyval()来在x值处获取多项式的y值来近似绘制多项式。然后，您只需绘制x值和y值即可。

import numpy as np
import matplotlib.pyplot as plt

curve = np.array([1,2,3,4,5])
x = np.linspace(0,10,100)
y = [np.polyval(curve, i) for i in x]
plt.plot(x,y)

Answer 4

通用的矢量化实现：

from typing import Sequence, Union
import numpy as np
import matplotlib.pyplot as plt


Number = Union[int, float, complex]

def polyval(coefficients: Sequence[Number], x: Sequence[Number]) -> np.ndarray:
    # expand dimensions to allow broadcasting (constant time + inexpensive)
    # axis=-1 allows for arbitrarily shaped x
    x = np.expand_dims(x, axis=-1)
    powers = x ** np.arange(len(coefficients))
    return powers @ coefficients

def polyplot(coefficients: Sequence[Number], x: Sequence[Number]) -> None:
    y = polyval(coefficients, x)
    plt.plot(x, y)

polyplot(np.array([0, 0, -1]), np.linspace(-10, 10, 210))
plt.show()

使用Matplotlib和coeffiecients绘制多项式

4 个答案: