python - 如何在特定间隔下绘制符号函数

Question

我打算绘制以下函数：

X = [x, y]
f = c'X + norm(c)* (sum(cos^2(X)) - sum(log(b-aX))
c = [7,2]
a = [2,3;4,6]
b = [10, 50]

所以我想得到这个

7x+2y+7.2*(cos^2(pi*x)+cos^2(pi*y))-(log(10-2x-3y)+ log(50-4x-6y)+log(x)+log(y))

如何绘制上述函数，比方说，-5<=x<=5-3<=y<=2？

我尝试了以下方法：

import numpy as np
import matplotlib.pyplot as plt
from sympy import symbols
x,y = symbols(['x','y'])

c = np.array([7,2])                
A = np.array([[2,3],[4,6]])               
b = np.array([10,50])

那么接下来该怎么办？？不知道要不要用linspace？？

plt.plot(A*x+norm*(np.sum(np.cos^2(pi*x))-np.sum(log10(b-A*x))

标签： pythonnumpymatplotlibsympy

Answer 1

您可以使用 sympy 函数sympy.Function('cos')(x)和('log')(x)，然后使用 lambdify。我没有写下确切的功能太长，但简化版本是这样的

import sympy
from sympy import *
from sympy.utilities.lambdify import lambdify, implemented_function
import numpy as np
import matplotlib.pyplot as plt
from matplotlib import cm

if __name__=="__main__":
    
    x, y = symbols('x y')
    
    X = np.array([x,y])
    
    a = np.array([[2,3],[4,6]])
    b = np.array([10, 50])  # np.tile(b, (2,1))
    c = np.array([7, 2])
    
    fPart1 = sum(c*np.array( X ) )
    fPart2 = sum(np.array( c*np.array([sympy.Function('cos')(x) for x in X])))
    fPart3a = sum(c*np.array( sympy.Function('log')(sum(b - a[0,:]*X) )))
    fPart3b = sum(c*np.array( sympy.Function('log')(sum(b - a[1,:]*X) )))

    fPart3 = fPart3a + fPart3b

    
    zFunction = lambdify([x, y], fPart1 + fPart2*fPart2 + fPart3)

    xValues = np.linspace(-5, 5, 100)
    yValues = np.linspace(-3, 2, 100)
   
    X, Y = np.meshgrid(xValues, yValues)

    Z = zFunction(X, Y)

    fig, ax = plt.subplots(subplot_kw={"projection": "3d"})

    surf = ax.plot_surface(X, Y, Z, cmap=cm.coolwarm, linewidth=0, antialiased=False)
    ax.set_xlabel('X', fontsize=20, rotation=150)
    ax.set_ylabel('Y', fontsize=20)
    ax.set_zlabel('Z', fontsize=20)

    plt.show()

打印功能以了解您想要的功能

In [4]: fPart1
Out[4]: 7*x + 2*y

In [5]: fPart2
Out[5]: 7*cos(x) + 2*cos(y)

In [6]: fPart3
Out[6]: 9*log(-4*x - 6*y + 60) + 9*log(-2*x - 3*y + 60)