#!/usr/bin/python
#coding=utf-8
import numpy as np
from scipy import stats
import matplotlib.pyplot as plt
 
# 構(gòu)造訓(xùn)練數(shù)據(jù)
x = np.arange(0., 10., 0.2)
m = len(x) # 訓(xùn)練數(shù)據(jù)點(diǎn)數(shù)目
print m
x0 = np.full(m, 1.0)
input_data = np.vstack([x0, x]).T # 將偏置b作為權(quán)向量的第一個(gè)分量
target_data = 2 * x + 5 + np.random.randn(m)
 
# 兩種終止條件
loop_max = 10000 # 最大迭代次數(shù)(防止死循環(huán))
epsilon = 1e-3
 
# 初始化權(quán)值
np.random.seed(0)
theta = np.random.randn(2)
 
alpha = 0.001 # 步長(zhǎng)(注意取值過(guò)大會(huì)導(dǎo)致振蕩即不收斂,過(guò)小收斂速度變慢)
diff = 0.
error = np.zeros(2)
count = 0 # 循環(huán)次數(shù)
finish = 0 # 終止標(biāo)志
 
while count < loop_max:
 count += 1
 
 # 標(biāo)準(zhǔn)梯度下降是在權(quán)值更新前對(duì)所有樣例匯總誤差，而隨機(jī)梯度下降的權(quán)值是通過(guò)考查某個(gè)訓(xùn)練樣例來(lái)更新的
 # 在標(biāo)準(zhǔn)梯度下降中，權(quán)值更新的每一步對(duì)多個(gè)樣例求和，需要更多的計(jì)算
 sum_m = np.zeros(2)
 for i in range(m):
 dif = (np.dot(theta, input_data[i]) - target_data[i]) * input_data[i]
 sum_m = sum_m + dif # 當(dāng)alpha取值過(guò)大時(shí),sum_m會(huì)在迭代過(guò)程中會(huì)溢出
 
 theta = theta - alpha * sum_m # 注意步長(zhǎng)alpha的取值,過(guò)大會(huì)導(dǎo)致振蕩
 # theta = theta - 0.005 * sum_m # alpha取0.005時(shí)產(chǎn)生振蕩,需要將alpha調(diào)小
 
 # 判斷是否已收斂
 if np.linalg.norm(theta - error) < epsilon:
 finish = 1
 break
 else:
 error = theta
 print 'loop count = %d' % count, '\tw:',theta
print 'loop count = %d' % count, '\tw:',theta
 
# check with scipy linear regression
slope, intercept, r_value, p_value, slope_std_error = stats.linregress(x, target_data)
print 'intercept = %s slope = %s' % (intercept, slope)
 
plt.plot(x, target_data, 'g*')
plt.plot(x, theta[1] * x + theta[0], 'r')
plt.show()

#!/usr/bin/python
#coding=utf-8
import numpy as np
from scipy import stats
import matplotlib.pyplot as plt
 
# 構(gòu)造訓(xùn)練數(shù)據(jù)
x = np.arange(0., 10., 0.2)
m = len(x) # 訓(xùn)練數(shù)據(jù)點(diǎn)數(shù)目
x0 = np.full(m, 1.0)
input_data = np.vstack([x0, x]).T # 將偏置b作為權(quán)向量的第一個(gè)分量
target_data = 2 * x + 5 + np.random.randn(m)
 
# 兩種終止條件
loop_max = 10000 # 最大迭代次數(shù)(防止死循環(huán))
epsilon = 1e-3
 
# 初始化權(quán)值
np.random.seed(0)
theta = np.random.randn(2)
# w = np.zeros(2)
 
alpha = 0.001 # 步長(zhǎng)(注意取值過(guò)大會(huì)導(dǎo)致振蕩,過(guò)小收斂速度變慢)
diff = 0.
error = np.zeros(2)
count = 0 # 循環(huán)次數(shù)
finish = 0 # 終止標(biāo)志
######-隨機(jī)梯度下降算法
while count < loop_max:
 count += 1
 
 # 遍歷訓(xùn)練數(shù)據(jù)集，不斷更新權(quán)值
 for i in range(m):
 diff = np.dot(theta, input_data[i]) - target_data[i] # 訓(xùn)練集代入,計(jì)算誤差值
 
 # 采用隨機(jī)梯度下降算法,更新一次權(quán)值只使用一組訓(xùn)練數(shù)據(jù)
 theta = theta - alpha * diff * input_data[i]
 
 # ------------------------------終止條件判斷-----------------------------------------
 # 若沒(méi)終止，則繼續(xù)讀取樣本進(jìn)行處理，如果所有樣本都讀取完畢了,則循環(huán)重新從頭開(kāi)始讀取樣本進(jìn)行處理。
 
 # ----------------------------------終止條件判斷-----------------------------------------
 # 注意：有多種迭代終止條件，和判斷語(yǔ)句的位置。終止判斷可以放在權(quán)值向量更新一次后,也可以放在更新m次后。
 if np.linalg.norm(theta - error) < epsilon: # 終止條件：前后兩次計(jì)算出的權(quán)向量的絕對(duì)誤差充分小
 finish = 1
 break
 else:
 error = theta
print 'loop count = %d' % count, '\tw:',theta
 
 
# check with scipy linear regression
slope, intercept, r_value, p_value, slope_std_error = stats.linregress(x, target_data)
print 'intercept = %s slope = %s' % (intercept, slope)
 
plt.plot(x, target_data, 'g*')
plt.plot(x, theta[1] * x + theta[0], 'r')
plt.show()

#!/usr/bin/python
#coding=utf-8
import numpy as np
from scipy importstats
import matplotlib.pyplot as plt
 
# 構(gòu)造訓(xùn)練數(shù)據(jù)
x = np.arange(0.,10.,0.2)
m = len(x) # 訓(xùn)練數(shù)據(jù)點(diǎn)數(shù)目
print m
x0 = np.full(m, 1.0)
input_data = np.vstack([x0, x]).T # 將偏置b作為權(quán)向量的第一個(gè)分量
target_data = 2 *x + 5 +np.random.randn(m)
 
# 兩種終止條件
loop_max = 10000 #最大迭代次數(shù)(防止死循環(huán))
epsilon = 1e-3
 
# 初始化權(quán)值
np.random.seed(0)
theta = np.random.randn(2)
 
alpha = 0.001 #步長(zhǎng)(注意取值過(guò)大會(huì)導(dǎo)致振蕩即不收斂,過(guò)小收斂速度變慢)
diff = 0.
error = np.zeros(2)
count = 0 #循環(huán)次數(shù)
finish = 0 #終止標(biāo)志
minibatch_size = 5 #每次更新的樣本數(shù)
while count < loop_max:
 count += 1
 
 # minibatch梯度下降是在權(quán)值更新前對(duì)所有樣例匯總誤差，而隨機(jī)梯度下降的權(quán)值是通過(guò)考查某個(gè)訓(xùn)練樣例來(lái)更新的
 # 在minibatch梯度下降中，權(quán)值更新的每一步對(duì)多個(gè)樣例求和，需要更多的計(jì)算
 
 for i inrange(1,m,minibatch_size):
 sum_m = np.zeros(2)
 for k inrange(i-1,i+minibatch_size-1,1):
  dif = (np.dot(theta, input_data[k]) - target_data[k]) *input_data[k]
  sum_m = sum_m + dif #當(dāng)alpha取值過(guò)大時(shí),sum_m會(huì)在迭代過(guò)程中會(huì)溢出
 
 theta = theta- alpha * (1.0/minibatch_size) * sum_m #注意步長(zhǎng)alpha的取值,過(guò)大會(huì)導(dǎo)致振蕩
 
 # 判斷是否已收斂
 if np.linalg.norm(theta- error) < epsilon:
 finish = 1
 break
 else:
 error = theta
 print 'loopcount = %d'% count, '\tw:',theta
print 'loop count = %d'% count, '\tw:',theta
 
# check with scipy linear regression
slope, intercept, r_value, p_value,slope_std_error = stats.linregress(x, target_data)
print 'intercept = %s slope = %s'% (intercept, slope)
 
plt.plot(x, target_data, 'g*')
plt.plot(x, theta[1]* x +theta[0],'r')
plt.show()