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SGD/BGD/MBGD使用python简单实现

时间:2018-08-20 20:13:55      阅读:476      评论:0      收藏:0      [点我收藏+]

标签:ack   while   int   梯度   pytho   psi   算法   python   utf-8   

算法具体可以参照其他的博客:

随机梯度下降:

# coding=utf-8
‘‘‘
随机梯度下降
‘‘‘
import numpy as np

# 构造训练数据
x = np.arange(0., 10., 0.2)
m = len(x)
x0 = np.full(m, 1.0)
input_data = np.vstack([x0, x]).T  # 将偏置b作为权向量的第一个分量
target_data = 3 * x + 8 + np.random.randn(m)

max_iter = 10000  # 最大迭代次数
epsilon = 1e-5

# 初始化权值
w = np.random.randn(2)
# w = np.zeros(2)

alpha = 0.001  # 步长
diff = 0.
error = np.zeros(2)
count = 0  # 循环次数

print 随机梯度下降算法.center(60, =)

while count < max_iter:
    count += 1
    for j in range(m):
        diff = np.dot(w, input_data[j]) - target_data[j]  # 训练集代入,计算误差值
        # 最大的特点就是迭代一下就更新一下模型的参数
        w = w - alpha * diff * input_data[j]

    if np.linalg.norm(w - error) < epsilon:  # 直接通过np.linalg包求两个向量的范数
        break
    else:
        error = w
print loop count = %d % count, \tw:[%f, %f] % (w[0], w[1])
# coding=utf-8
"""
批量梯度下降
"""
import numpy as np

# 构造训练数据
x = np.arange(0., 10., 0.2)
m = len(x)
x0 = np.full(m, 1.0)
input_data = np.vstack([x0, x]).T  # 将偏置b作为权向量的第一个分量
target_data = 3 * x + 8 + np.random.randn(m)

# 停止条件
max_iter = 10000
epsilon = 1e-5

# 初始化权值
w = np.random.randn(2)
# w = np.zeros(2)

alpha = 0.001  # 步长
diff = 0.
error = np.zeros(2)
count = 0  # 循环次数

while count < max_iter:
    count += 1

    sum_m = np.zeros(2)

    for i in range(m):
        dif = (np.dot(w, input_data[i]) - target_data[i]) * input_data[i]
        sum_m = sum_m + dif
    ‘‘‘
    for j in range(m):
        diff = np.dot(w, input_data[j]) - target_data[j]  # 训练集代入,计算误差值
        w = w - alpha * diff * input_data[j]
    ‘‘‘
    w = w - alpha * sum_m

    if np.linalg.norm(w - error) < epsilon:
        break
    else:
        error = w
print loop count = %d % count, \tw:[%f, %f] % (w[0], w[1])

小批量梯度下降:

# coding=utf-8
"""
小批量梯度下降
"""
import numpy as np
import random

# 构造训练数据
x = np.arange(0., 10., 0.2)
m = len(x)
x0 = np.full(m, 1.0)
input_data = np.vstack([x0, x]).T  # 将偏置b作为权向量的第一个分量
target_data = 3 * x + 8 + np.random.randn(m)

# 两种终止条件
max_iter = 10000
epsilon = 1e-5

# 初始化权值
np.random.seed(0)
w = np.random.randn(2)
# w = np.zeros(2)

alpha = 0.001  # 步长
diff = 0.
error = np.zeros(2)
count = 0  # 循环次数

while count < max_iter:
    count += 1

    sum_m = np.zeros(2)
    index = random.sample(range(m), int(np.ceil(m * 0.2)))

    for i in range(len(input_data)):
        dif = (np.dot(w, input_data[i]) - target_data[i]) * input_data[i]
        sum_m = sum_m + dif

    w = w - alpha * sum_m

    if np.linalg.norm(w - error) < epsilon:
        break
    else:
        error = w
print loop count = %d % count, \tw:[%f, %f] % (w[0], w[1])

通过迭代,结果会收敛到8和3

loop count = 704 	w:[8.025972, 2.982300]

SGD/BGD/MBGD使用python简单实现

标签:ack   while   int   梯度   pytho   psi   算法   python   utf-8   

原文地址:https://www.cnblogs.com/leixingzhi7/p/9507718.html

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