Python 使用Matplotlib和系数系数绘制多项式

我的代码是：

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

其结果是描述1,2,3,4,5中点及其之间直线的直线，而不是以1,2,3,4,5为系数的5次多项式（p（x）=1+2x+3x+4x+5x）

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我该如何绘制一个只有系数的多项式呢？

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()

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一个非常类似Python的解决方案是使用列表理解来计算函数的值

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()

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您可以通过获得大量的x值并使用来获得多项式在x值处的y值来近似绘制多项式。然后你可以画出x-vals和y-vals

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)

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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()

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定义x的值，并使用感谢的答案。我现在明白了我的问题：我有矩阵乘法：（np.linalg.pinv（XXt[I]）*X[I]*Yt）现在这段代码返回一个数组：[[X1]，[X2]，…[Xm]]我想把它转换成[X1，X2，…，Xm]我怎么做？如果你喜欢这个答案，请投赞成票；-）要展开列表，请参见。