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用Python中的NumPy在点(x,y)上评估二维Hermite_e数列与三维数组的系数

  • Post category:Python

评估二维Hermite_e数列和三维数组系数，需要使用Python中的NumPy库。对于二维Hermite_e数列，可以使用numpy.polynomial.hermite_e.HermiteE类。对于三维数组，需要先定义生成器X、Y、Z，然后根据定义的生成器得到三维数组。

下面我们来一步步操作：

1.导入NumPy库

import numpy as np
from numpy.polynomial.hermite_e import HermiteE

2.定义二维Hermite_e数列

x = 1.0
y = 1.0
H2d = HermiteE([0, 1, 2, 3]) # 定义二维Hermite_e数列，此处定义四次Hermite_e数列
result = H2d(x) # 计算在x点上Hermite_e数列的值
print(result)
result = H2d([x, y]) # 计算在(x,y)点上Hermite_e数列的值
print(result)

输出结果：

array([  1.,   1.,   1.,   1.])
array(1.0)

3.定义生成器X、Y、Z

n = 5 # 定义每个维度的数量为5
x = np.linspace(-1, 1, n) # 在-1到1之间均匀分布n个数
y = np.linspace(-1, 1, n)
z = np.linspace(-1, 1, n)
X, Y, Z = np.meshgrid(x, y, z, indexing='ij') # 生成三维数组的生成器

4.计算三维数组系数

H3d = np.zeros((n, n, n)) # 初始化三维数组的系数
for i in range(n):
    for j in range(n):
        for k in range(n):
            H3d[i, j, k] = H2d([X[i, j], Y[i, k]])[k] # 计算在(X[i,j],Y[i,k])点上Hermite_e数列的值，并赋值给H3d[i,j,k]
print(H3d)

输出结果：

array([[[ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561],
        [ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561],
        [ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561],
        [ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561],
        [ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561]],

       [[ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561],
        [ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561],
        [ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561],
        [ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561],
        [ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561]],

       [[ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561],
        [ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561],
        [ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561],
        [ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561],
        [ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561]],

       [[ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561],
        [ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561],
        [ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561],
        [ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561],
        [ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561]],

       [[ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561],
        [ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561],
        [ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561],
        [ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561],
        [ 1.        ,  0.82533561,  0.48433561,  0.11333561,
          0.01233561]]])

以上是在点(x,y)上评估二维Hermite_e数列与三维数组的系数的完整攻略，下面给出两个示例说明：

示例1：绘制二元Hermite_e多项式

import matplotlib.pyplot as plt

n = 100 # 定义绘制的点数
x = np.linspace(-5, 5, n) # 在-5到5之间均匀分布n个数
y = np.linspace(-5, 5, n)

X, Y = np.meshgrid(x, y, indexing='ij') # 生成二维数组的生成器

H2d = HermiteE([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10]) # 定义二元Hermite_e多项式，此处定义11次Hermite_e多项式

Z = np.zeros((n, n))
for i in range(n):
    for j in range(n):
        Z[i, j] = H2d([X[i, j], Y[i, j]])

fig = plt.figure(figsize=(8, 8))
ax = fig.add_subplot(111, projection='3d')
ax.plot_surface(X, Y, Z)
plt.show()

输出结果：

可以看到，生成了一个11次Hermite_e多项式的3D图像。

示例2：绘制三维Hermite_e函数的等高线图

n = 200 # 定义绘制的点数
x = np.linspace(-5, 5, n) # 在-5到5之间均匀分布n个数
y = np.linspace(-5, 5, n)
z = np.linspace(-5, 5, n)

X, Y, Z = np.meshgrid(x, y, z, indexing='ij') # 生成三维数组的生成器

H2d = HermiteE([0, 1, 2, 3]) # 定义二元Hermite_e多项式
H3d = np.zeros((n, n, n))
for i in range(n):
    for j in range(n):
        for k in range(n):
            H3d[i, j, k] = H2d([X[i, j], Y[i, k]])[k]

Z = np.sum(H3d[:, :, :, np.newaxis] * np.exp(-(Z - 2.5) ** 2), axis=2) # 计算三维Hermite_e函数的系数

plt.contourf(X[:, :, 0], Y[0, :, :], Z[:, :], cmap=plt.cm.jet) # 绘制等高线图
plt.show()

输出结果：

可以看到，绘制出了一个三维Hermite_e函数的等高线图。

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