梯度下降求解邏輯回歸
- 主要有一下幾步
- 資料預處理（前面加一列1）
- 将數學推導式代碼實作
- 設定終止條件
- 參數更新
- 設定參數訓練
注意：1
#!/usr/bin/env python
# coding: utf-8
# # Logistic Regression
# ## The data
# 我們将建立一個邏輯回歸模型來預測一個學生是否被大學錄取。假設你是一個大學系的管理者，你想根據兩次考試的結果來決定每個申請人的錄取機會。你有以前的申請人的曆史資料，你可以用它作為邏輯回歸的訓練集。對于每一個教育訓練例子，你有兩個考試的申請人的分數和錄取決定。為了做到這一點，我們将建立一個分類模型，根據考試成績估計入學機率。
# In[10]:
#三大件
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
get_ipython().run_line_magic('matplotlib', 'inline')
# In[11]:
import os
path = 'data' + os.sep + 'LogiReg_data.txt'
pdData = pd.read_csv(path, header=None, names=['Exam 1', 'Exam 2', 'Admitted'])
pdData.head()
# In[12]:
pdData.shape
# In[13]:
positive = pdData[pdData['Admitted'] == 1] # returns the subset of rows such Admitted = 1, i.e. the set of *positive* examples
negative = pdData[pdData['Admitted'] == 0] # returns the subset of rows such Admitted = 0, i.e. the set of *negative* examples
fig, ax = plt.subplots(figsize=(10,5))
ax.scatter(positive['Exam 1'], positive['Exam 2'], s=30, c='b', marker='o', label='Admitted')
ax.scatter(negative['Exam 1'], negative['Exam 2'], s=30, c='r', marker='x', label='Not Admitted')
ax.legend()
ax.set_xlabel('Exam 1 Score')
ax.set_ylabel('Exam 2 Score')
# ## The logistic regression
# 目标：建立分類器（求解出三個參數 $\theta_0 \theta_1 \theta_2 $）
#
#
# 設定門檻值，根據門檻值判斷錄取結果
#
# ### 要完成的子產品
# - `sigmoid` : 映射到機率的函數
#
# - `model` : 傳回預測結果值
#
# - `cost` : 根據參數計算損失
#
# - `gradient` : 計算每個參數的梯度方向
#
# - `descent` : 進行參數更新
#
# - `accuracy`: 計算精度
# ### `sigmoid` 函數
#
# $$
# g(z) = \frac{1}{1+e^{-z}}
# $$
# In[14]:
def sigmoid(z):
return 1 / (1 + np.exp(-z))
# In[15]:
nums = np.arange(-10, 10, step=1) #creates a vector containing 20 equally spaced values from -10 to 10
fig, ax = plt.subplots(figsize=(12,4))
ax.plot(nums, sigmoid(nums), 'r')
# ### Sigmoid
# * $g:\mathbb{R} \to [0,1]$
# * $g(0)=0.5$
# * $g(- \infty)=0$
# * $g(+ \infty)=1$
# In[16]:
def model(X, theta):
return sigmoid(np.dot(X, theta.T))
# $$
# \begin{array}{ccc}
# \begin{pmatrix}\theta_{0} & \theta_{1} & \theta_{2}\end{pmatrix} & \times & \begin{pmatrix}1\\
# x_{1}\\
# x_{2}
# \end{pmatrix}\end{array}=\theta_{0}+\theta_{1}x_{1}+\theta_{2}x_{2}
# $$
# In[17]:
pdData.insert(0, 'Ones', 1) # in a try / except structure so as not to return an error if the block si executed several times
# set X (training data) and y (target variable)
orig_data = pdData.values # convert the Pandas representation of the data to an array useful for further computations
cols = orig_data.shape[1]
X = orig_data[:,0:cols-1]
y = orig_data[:,cols-1:cols]
# convert to numpy arrays and initalize the parameter array theta
#X = np.matrix(X.values)
#y = np.matrix(data.iloc[:,3:4].values) #np.array(y.values)
theta = np.zeros([1, 3])
# In[18]:
X[:5]
# In[19]:
y[:5]
# In[20]:
theta
# In[21]:
X.shape, y.shape, theta.shape
# ### 損失函數
# 将對數似然函數去負号
#
# $$
# D(h_\theta(x), y) = -y\log(h_\theta(x)) - (1-y)\log(1-h_\theta(x))
# $$
# 求平均損失
# $$
# J(\theta)=\frac{1}{n}\sum_{i=1}^{n} D(h_\theta(x_i), y_i)
# $$
# In[22]:
def cost(X, y, theta):
left = np.multiply(-y, np.log(model(X, theta)))
right = np.multiply(1 - y, np.log(1 - model(X, theta)))
return np.sum(left - right) / (len(X))
# In[23]:
cost(X, y, theta)
# ### 計算梯度
#
#
# $$
# \frac{\partial J}{\partial \theta_j}=-\frac{1}{m}\sum_{i=1}^n (y_i - h_\theta (x_i))x_{ij}
# $$
#
# In[24]:
def gradient(X, y, theta):
grad = np.zeros(theta.shape)
error = (model(X, theta)- y).ravel()
for j in range(len(theta.ravel())): #for each parmeter
term = np.multiply(error, X[:,j])
grad[0, j] = np.sum(term) / len(X)
return grad
# ### Gradient descent
# 比較3中不同梯度下降方法
#
# In[25]:
STOP_ITER = 0
STOP_COST = 1
STOP_GRAD = 2
def stopCriterion(type, value, threshold):
#設定三種不同的停止政策
if type == STOP_ITER: return value > threshold
elif type == STOP_COST: return abs(value[-1]-value[-2]) < threshold
elif type == STOP_GRAD: return np.linalg.norm(value) < threshold
# In[26]:
import numpy.random
#洗牌
def shuffleData(data):
np.random.shuffle(data)
cols = data.shape[1]
X = data[:, 0:cols-1]
y = data[:, cols-1:]
return X, y
# In[27]:
import time
def descent(data, theta, batchSize, stopType, thresh, alpha):
#梯度下降求解
init_time = time.time()
i = 0 # 疊代次數
k = 0 # batch
X, y = shuffleData(data)
grad = np.zeros(theta.shape) # 計算的梯度
costs = [cost(X, y, theta)] # 損失值
while True:
grad = gradient(X[k:k+batchSize], y[k:k+batchSize], theta)
k += batchSize #取batch數量個資料
if k >= n:
k = 0
X, y = shuffleData(data) #重新洗牌
theta = theta - alpha*grad # 參數更新
costs.append(cost(X, y, theta)) # 計算新的損失
i += 1
if stopType == STOP_ITER: value = i
elif stopType == STOP_COST: value = costs
elif stopType == STOP_GRAD: value = grad
if stopCriterion(stopType, value, thresh): break
return theta, i-1, costs, grad, time.time() - init_time
# In[28]:
def runExpe(data, theta, batchSize, stopType, thresh, alpha):
#import pdb; pdb.set_trace();
theta, iter, costs, grad, dur = descent(data, theta, batchSize, stopType, thresh, alpha)
name = "Original" if (data[:,1]>2).sum() > 1 else "Scaled"
name += " data - learning rate: {} - ".format(alpha)
if batchSize==n: strDescType = "Gradient"
elif batchSize==1: strDescType = "Stochastic"
else: strDescType = "Mini-batch ({})".format(batchSize)
name += strDescType + " descent - Stop: "
if stopType == STOP_ITER: strStop = "{} iterations".format(thresh)
elif stopType == STOP_COST: strStop = "costs change < {}".format(thresh)
else: strStop = "gradient norm < {}".format(thresh)
name += strStop
print ("***{}\nTheta: {} - Iter: {} - Last cost: {:03.2f} - Duration: {:03.2f}s".format(
name, theta, iter, costs[-1], dur))
fig, ax = plt.subplots(figsize=(12,4))
ax.plot(np.arange(len(costs)), costs, 'r')
ax.set_xlabel('Iterations')
ax.set_ylabel('Cost')
ax.set_title(name.upper() + ' - Error vs. Iteration')
return theta
# ### 不同的停止政策
# #### 設定疊代次數
# In[29]:
#選擇的梯度下降方法是基于所有樣本的
n=100
runExpe(orig_data, theta, n, STOP_ITER, thresh=5000, alpha=0.000001)
# #### 根據損失值停止
# 設定門檻值 1E-6, 差不多需要110 000次疊代
# In[30]:
runExpe(orig_data, theta, n, STOP_COST, thresh=0.000001, alpha=0.001)
# #### 根據梯度變化停止
# 設定門檻值 0.05,差不多需要40 000次疊代
# In[31]:
runExpe(orig_data, theta, n, STOP_GRAD, thresh=0.05, alpha=0.001)
# ### 對比不同的梯度下降方法
# #### Stochastic descent
# In[32]:
runExpe(orig_data, theta, 1, STOP_ITER, thresh=5000, alpha=0.001)
# 有點爆炸。。。很不穩定,再來試試把學習率調小一些
# In[33]:
runExpe(orig_data, theta, 1, STOP_ITER, thresh=15000, alpha=0.000002)
# 速度快，但穩定性差，需要很小的學習率
# #### Mini-batch descent
# In[34]:
runExpe(orig_data, theta, 16, STOP_ITER, thresh=15000, alpha=0.001)
# 浮動仍然比較大，我們來嘗試下對資料進行标準化
# 将資料按其屬性(按列進行)減去其均值，然後除以其方差。最後得到的結果是，對每個屬性/每列來說所有資料都聚集在0附近，方內插補點為1
# In[35]:
from sklearn import preprocessing as pp
scaled_data = orig_data.copy()
scaled_data[:, 1:3] = pp.scale(orig_data[:, 1:3])
runExpe(scaled_data, theta, n, STOP_ITER, thresh=5000, alpha=0.001)
# 它好多了！原始資料，隻能達到達到0.61，而我們得到了0.38個在這裡！
# 是以對資料做預處理是非常重要的
# In[ ]:
runExpe(scaled_data, theta, n, STOP_GRAD, thresh=0.02, alpha=0.001)
# 更多的疊代次數會使得損失下降的更多！
# In[ ]:
theta = runExpe(scaled_data, theta, 1, STOP_GRAD, thresh=0.002/5, alpha=0.001)
# 随機梯度下降更快，但是我們需要疊代的次數也需要更多，是以還是用batch的比較合适！！！
# In[ ]:
runExpe(scaled_data, theta, 16, STOP_GRAD, thresh=0.002*2, alpha=0.001)
# ## 精度
# In[ ]:
#設定門檻值
def predict(X, theta):
return [1 if x >= 0.5 else 0 for x in model(X, theta)]
# In[ ]:
scaled_X = scaled_data[:, :3]
y = scaled_data[:, 3]
predictions = predict(scaled_X, theta)
correct = [1 if ((a == 1 and b == 1) or (a == 0 and b == 0)) else 0 for (a, b) in zip(predictions, y)]
accuracy = (sum(map(int, correct)) % len(correct))
print ('accuracy = {0}%'.format(accuracy))
- 運作時注意環境版本 ↩︎
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