【ML-SVM案例】会有十种SVM案例,供大家用来学习。本文只是实现梯度下降,求解最优解。后面一章将会实现003梯度下降:拉格乘子法。
提示:以下是本篇文章正文内容,下面案例可供参考
代码如下(示例):
import numpy as np
import matplotlib as mpl
import matplotlib.pyplot as plt
import math
from mpl_toolkits.mplot3d import Axes3D
代码如下(示例):
# 解决中文显示问题
mpl.rcParams['font.sans-serif'] = [u'SimHei']
mpl.rcParams['axes.unicode_minus'] = False
"""一维原始图像"""
def f1(x):
return 0.5 * (x - 0.25) ** 2
# 导函数
def h1(x):
return 0.5 * 2 * (x - 0.25)
GD_X = []
GD_Y = []
x = 4
alpha = 0.5
f_change = f1(x)
f_current = f_change
GD_X.append(x)
GD_Y.append(f_current)
iter_num = 0
while f_change > 1e-10 and iter_num < 100:
iter_num += 1
x = x - alpha * h1(x)
tmp = f1(x)
f_change = np.abs(f_current - tmp)
f_current = tmp
GD_X.append(x)
GD_Y.append(f_current)
print(u"最终结果为:(%.5f, %.5f)" % (x, f_current))
print(u"迭代过程中X的取值,迭代次数:%d" % iter_num)
print(GD_X)
X = np.arange(-4, 4.5, 0.05)
Y = np.array(list(map(lambda t: f1(t), X)))
plt.figure(facecolor='w')
plt.plot(X, Y, 'r-', linewidth=2)
plt.plot(GD_X, GD_Y, 'ko--', linewidth=2)
plt.title(u'函数$y=0.5 * (θ - 0.25)^2$; \n学习率:%.3f; 最终解:(%.3f, %.3f);迭代次数:%d' % (alpha, x, f_current, iter_num))
plt.show()
"""二维原始图像"""
def f2(x, y):
return 0.6 * (x + y) ** 2 - x * y
# 导函数
def hx2(x, y):
return 0.6 * 2 * (x + y) - y
def hy2(x, y):
return 0.6 * 2 * (x + y) - x
GD_X1 = []
GD_X2 = []
GD_Y = []
x1 = 4
x2 = 4
alpha = 0.5
f_change = f2(x1, x2)
f_current = f_change
GD_X1.append(x1)
GD_X2.append(x2)
GD_Y.append(f_current)
iter_num = 0
while f_change > 1e-10 and iter_num < 100:
iter_num += 1
prex1 = x1
prex2 = x2
x1 = x1 - alpha * hx2(prex1, prex2)
x2 = x2 - alpha * hy2(prex1, prex2)
tmp = f2(x1, x2)
f_change = np.abs(f_current - tmp)
f_current = tmp
GD_X1.append(x1)
GD_X2.append(x2)
GD_Y.append(f_current)
print(u"最终结果为:(%.5f, %.5f, %.5f)" % (x1, x2, f_current))
print(u"迭代过程中X的取值,迭代次数:%d" % iter_num)
print(GD_X1)
# 构建数据
X1 = np.arange(-4, 4.5, 0.2)
X2 = np.arange(-4, 4.5, 0.2)
X1, X2 = np.meshgrid(X1, X2)
Y = np.array(list(map(lambda t: f2(t[0], t[1]), zip(X1.flatten(), X2.flatten()))))
Y.shape = X1.shape
# 画图
fig = plt.figure(facecolor='w')
ax = Axes3D(fig)
ax.plot_surface(X1, X2, Y, rstride=1, cstride=1, cmap=plt.cm.jet)
ax.plot(GD_X1, GD_X2, GD_Y, 'ko--')
ax.set_title(u'函数$y=0.6 * (θ1 + θ2)^2 - θ1 * θ2$;\n学习率:%.3f; 最终解:(%.3f, %.3f, %.3f);迭代次数:%d' % (alpha, x1, x2, f_current, iter_num))
plt.show()
import numpy as np
import matplotlib as mpl
import matplotlib.pyplot as plt
import math
from mpl_toolkits.mplot3d import Axes3D
# 解决中文显示问题
mpl.rcParams['font.sans-serif'] = [u'SimHei']
mpl.rcParams['axes.unicode_minus'] = False
"""一维原始图像"""
def f1(x):
return 0.5 * (x - 0.25) ** 2
# 导函数
def h1(x):
return 0.5 * 2 * (x - 0.25)
# 使用梯度下降法求解
GD_X = []
GD_Y = []
x = 4
alpha = 0.5
f_change = f1(x)
f_current = f_change
GD_X.append(x)
GD_Y.append(f_current)
iter_num = 0
while f_change > 1e-10 and iter_num < 100:
iter_num += 1
x = x - alpha * h1(x)
tmp = f1(x)
f_change = np.abs(f_current - tmp)
f_current = tmp
GD_X.append(x)
GD_Y.append(f_current)
print(u"最终结果为:(%.5f, %.5f)" % (x, f_current))
print(u"迭代过程中X的取值,迭代次数:%d" % iter_num)
print(GD_X)
# 构建数据
X = np.arange(-4, 4.5, 0.05)
Y = np.array(list(map(lambda t: f1(t), X)))
# 画图
plt.figure(facecolor='w')
plt.plot(X, Y, 'r-', linewidth=2)
plt.plot(GD_X, GD_Y, 'ko--', linewidth=2)
plt.title(u'函数$y=0.5 * (θ - 0.25)^2$; \n学习率:%.3f; 最终解:(%.3f, %.3f);迭代次数:%d' % (alpha, x, f_current, iter_num))
plt.show()
"""二维原始图像"""
def f2(x, y):
return 0.6 * (x + y) ** 2 - x * y
# 导函数
def hx2(x, y):
return 0.6 * 2 * (x + y) - y
def hy2(x, y):
return 0.6 * 2 * (x + y) - x
# 使用梯度下降法求解
GD_X1 = []
GD_X2 = []
GD_Y = []
x1 = 4
x2 = 4
alpha = 0.5
f_change = f2(x1, x2)
f_current = f_change
GD_X1.append(x1)
GD_X2.append(x2)
GD_Y.append(f_current)
iter_num = 0
while f_change > 1e-10 and iter_num < 100:
iter_num += 1
prex1 = x1
prex2 = x2
x1 = x1 - alpha * hx2(prex1, prex2)
x2 = x2 - alpha * hy2(prex1, prex2)
tmp = f2(x1, x2)
f_change = np.abs(f_current - tmp)
f_current = tmp
GD_X1.append(x1)
GD_X2.append(x2)
GD_Y.append(f_current)
print(u"最终结果为:(%.5f, %.5f, %.5f)" % (x1, x2, f_current))
print(u"迭代过程中X的取值,迭代次数:%d" % iter_num)
print(GD_X1)
# 构建数据
X1 = np.arange(-4, 4.5, 0.2)
X2 = np.arange(-4, 4.5, 0.2)
X1, X2 = np.meshgrid(X1, X2)
Y = np.array(list(map(lambda t: f2(t[0], t[1]), zip(X1.flatten(), X2.flatten()))))
Y.shape = X1.shape
# 画图
fig = plt.figure(facecolor='w')
ax = Axes3D(fig)
ax.plot_surface(X1, X2, Y, rstride=1, cstride=1, cmap=plt.cm.jet)
ax.plot(GD_X1, GD_X2, GD_Y, 'ko--')
ax.set_title(u'函数$y=0.6 * (θ1 + θ2)^2 - θ1 * θ2$;\n学习率:%.3f; 最终解:(%.3f, %.3f, %.3f);迭代次数:%d' % (alpha, x1, x2, f_current, iter_num))
plt.show()
"""二维原始图像"""
def f2(x, y):
return 0.15 * (x + 0.5) ** 2 + 0.25 * (y - 0.25) ** 2 + 0.35 * (1.5 * x - 0.2 * y + 0.35 ) ** 2
## 偏函数
def hx2(x, y):
return 0.15 * 2 * (x + 0.5) + 0.25 * 2 * (1.5 * x - 0.2 * y + 0.35 ) * 1.5
def hy2(x, y):
return 0.25 * 2 * (y - 0.25) - 0.25 * 2 * (1.5 * x - 0.2 * y + 0.35 ) * 0.2
# 使用梯度下降法求解
GD_X1 = []
GD_X2 = []
GD_Y = []
x1 = 4
x2 = 4
alpha = 0.5
f_change = f2(x1, x2)
f_current = f_change
GD_X1.append(x1)
GD_X2.append(x2)
GD_Y.append(f_current)
iter_num = 0
while f_change > 1e-10 and iter_num < 100:
iter_num += 1
prex1 = x1
prex2 = x2
x1 = x1 - alpha * hx2(prex1, prex2)
x2 = x2 - alpha * hy2(prex1, prex2)
tmp = f2(x1, x2)
f_change = np.abs(f_current - tmp)
f_current = tmp
GD_X1.append(x1)
GD_X2.append(x2)
GD_Y.append(f_current)
print(u"最终结果为:(%.5f, %.5f, %.5f)" % (x1, x2, f_current))
print(u"迭代过程中X的取值,迭代次数:%d" % iter_num)
print(GD_X1)
# 构建数据
X1 = np.arange(-4, 4.5, 0.2)
X2 = np.arange(-4, 4.5, 0.2)
X1, X2 = np.meshgrid(X1, X2)
Y = np.array(list(map(lambda t: f2(t[0], t[1]), zip(X1.flatten(), X2.flatten()))))
Y.shape = X1.shape
# 画图
fig = plt.figure(facecolor='w')
ax = Axes3D(fig)
ax.plot_surface(X1, X2, Y, rstride=1, cstride=1, cmap=plt.cm.jet)
ax.plot(GD_X1, GD_X2, GD_Y, 'ko--')
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('z')
ax.set_title(u'函数;\n学习率:%.3f; 最终解:(%.3f, %.3f, %.3f);迭代次数:%d' % (alpha, x1, x2, f_current, iter_num))
plt.show()
001梯度下降:一维和二维图像
003梯度下降:拉格乘子法