2D函数梯度优化
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首先我们构建一个2D函数方程
$$
f(x,y)=(x^2+y-11)^2+(x+y^2-7)^2
$$
该方程为Himmelblau方程，是科学家们研究出来专门用于检测一个优化器效果的方程。该方程所绘制的图像如下
由图可见，四个蓝色圆圈即为该方程的极小值点，其平面上的图像如右图所示。该方程虽然有四个极小值点，但四个点对应的值均为0- f(3.0,2.0)=0.0f(3.0,2.0)=0.0f(3.0,2.0)=0.0
- f(−2.805118,3.131312)=0.0f(-2.805118,3.131312)=0.0f(−2.805118,3.131312)=0.0
- f(−3.779310,−3.283186)=0.0f(-3.779310,-3.283186)=0.0f(−3.779310,−3.283186)=0.0
- f(3.584428,−1.848126)=0.0f(3.584428,-1.848126)=0.0f(3.584428,−1.848126)=0.0
我们可以通过测试优化器能否找到四个极小值点，来判断其优劣
import torch import torch.nn.functional as F import numpy as np import matplotlib.pyplot as plt from mpl_toolkits.mplot3d import Axes3D def himmelblau(x, y): return (x**2 + y - 11)**2 + (x + y**2 - 7)**2 x = np.arange(-6, 6, 0.1) # x轴的范围 y = np.arange(-6, 6, 0.1) # y轴的范围 X, Y = np.meshgrid(x, y) Z = himmelblau(X, Y) fig = plt.figure('himmelblau') ax = fig.gca(projection='3d') ax.plot_surface(X, Y, Z) ax.view_init(60, -30) ax.set_xlabel('x') ax.set_ylabel('y') plt.show()输出图像为
关于
meshgrid()函数，这篇博客解释的很好，如果不理解的可以看看下面以梯度下降法为例子来试着找到Himmelblau方程的极小值。这里是以优化预测值pred为目标，而不是误差Error
x = torch.tensor([0., 0.], requires_grad=True) # 设定初始值(0, 0) optimizer = torch.optim.Adam([x], lr=1e-3) # 优化器对x进行优化，设定学习率为0.001 for step in range(20000): pred = himmelblau(x[0], x[1]) optimizer.zero_grad() # 梯度信息清零 pred.backward() optimizer.step() # 进行一次优化器优化，根据梯度信息更新x[0]和x[1] if step % 2000 == 0: print("step{}: x={}, f(x) = {}".format(step, x.tolist(), pred.item()))输出为
step0: x=[0.0009999999310821295, 0.0009999999310821295], f(x) = 170.0 step2000: x=[2.3331806659698486, 1.9540694952011108], f(x) = 13.730916023254395 step4000: x=[2.9820079803466797, 2.0270984172821045], f(x) = 0.014858869835734367 step6000: x=[2.999983549118042, 2.0000221729278564], f(x) = 1.1074007488787174e-08 step8000: x=[2.9999938011169434, 2.0000083446502686], f(x) = 1.5572823031106964e-09 step10000: x=[2.999997854232788, 2.000002861022949], f(x) = 1.8189894035458565e-10 step12000: x=[2.9999992847442627, 2.0000009536743164], f(x) = 1.6370904631912708e-11 step14000: x=[2.999999761581421, 2.000000238418579], f(x) = 1.8189894035458565e-12 step16000: x=[3.0, 2.0], f(x) = 0.0 step18000: x=[3.0, 2.0], f(x) = 0.0由结果可见，运行到16000次后，找到极小值点。若改变初始值点，则可能会改变输出结果，比方说把初始值点由(0,0)改为(4,0)
