吴恩达机器学习 || 实验一作业 || python || 梯度下降
导入pandas包用于处理数据 导入seborn和pyplot包用于画图 tensorflow包提供placeholder占位符
import pandas as pd
import seaborn as sns
sns.set(context="notebook", style="whitegrid", palette="dark")
import matplotlib.pyplot as plt
import tensorflow as tf
import numpy as np
这里定义为人口和利润
df = pd.read_csv('ex1data1.txt', names=['population','profit'])df.head()
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population
profit
0
6.1101
17.5920
1
5.5277
9.1302
2
8.5186
13.6620
3
7.0032
11.8540
4
5.8598
6.8233
</div>
df.info()
<class 'pandas.core.frame.DataFrame'>
RangeIndex: 97 entries, 0 to 96
Data columns (total 2 columns):
# Column Non-Null Count Dtype
--- ------ -------------- -----
0 population 97 non-null float64
1 profit 97 non-null float64
dtypes: float64(2)
memory usage: 1.6 KB
lmplot绘制出散点图
sns.lmplot(x='population', y='profit',data=df, height=6, fit_reg=False)
plt.show()
get_X函数 创建ones（dataframe类型）m行1列的1元素，并将ones这一列与传入的df合并，传出 类似 1,a 1,b 1,c get_Y函数获取最后一列数据[a,b,c]
normalize_feature函数将传入的df全部按列归一化方便后续梯度下降
def get_X(df):#读取特征
ones = pd.DataFrame({'ones':np.ones(len(df))})#ones是和pf等高的1元素
data = pd.concat([ones,df], axis=1)#拼接ones和pd为data
return data.iloc[:,:-1].values
def get_Y(df):#读取标签
return np.array(df.iloc[:, -1])#获取df最后一列
def normalize_feature(df):#特征缩放，与平均数的差值/方差
return df.apply(lambda column:(column - column.mean()) / column.std())
linear_regression函数进行梯度下降
def linear_regression(X_data,y_data, alpha,epoch,optimizer=tf.compat.v1.train.GradientDescentOptimizer):
#放数据的graph
X = tf.placeholder(tf.float32, shape = X_data.shape)
y = tf.placeholder(tf.float32, shape = y_data.shape)
with tf.varialbe_scope('liner-regression'):
W = tf.get_variable("weights",
shape=(X_data.shape[1],1),#2行1列
initializer=tf.constant_initializer()) #n*1
y_pred = tf.matmul(X, W) #m*n的矩阵乘 n*1 ->m*1矩阵
# (m*1)T*(m*1) = 1*1
#transpose_a/b说明矩阵a/b的转置
#loss就是梯度下降中的代价函数J(θ)
loss = 1/ (2 * len(X_data)) *tf.matmul((y_pred -y), (y_pred - y), transpose_a=True)
opt = optimizer(learning_rate=alpha)
opt_operation = opt.minimize(loss)
#run session
with tf.Session() as sess:
sess.run(tf.global_variables_initializer())
sess_data = []
for i in range(epoch):
_, loss_val,W_val = sess.run([opt_operation, loss, W], feed_dict = {X:X_data, y:y_data})
loss_data.append(loss_val[0,0]) #损失函数都是1*1矩阵
if len(loss_data)>1 and np.abs(loss_data[-1] - loss_data[-2])<10 ** -9:
print('Converged at epoch {}'.format(i))
break
#清除 graph
tf.reset_default_gragh()
return {'loss':loss_data, 'parameters': W_val}
data = pd.read_csv('ex1data1.txt', names=['population', 'profit'])
data.head()
.dataframe tbody tr th {
vertical-align: top;
}
.dataframe thead th {
text-align: right;
}</style>
population
profit
0
6.1101
17.5920
1
5.5277
9.1302
2
8.5186
13.6620
3
7.0032
11.8540
4
5.8598
6.8233
</div>
X = get_X(data)
print(X.shape,type(X))
y = get_Y(data)
print(y.shape, type(y))
(97, 2) <class 'numpy.ndarray'>
(97,) <class 'numpy.ndarray'>
theta = np.zeros(X.shape[1])#X有两列，shape[0]表行数，shpae[1]=2是两列
lr_cost函数传入θ，样本里的x和y（需要拟合的值） 定义m为样本数 线性拟合函数最初始参数为θ0=0，θ1=0，h(x) = θ0+θ1x 用x计算出的值h(x)-y(x)为差值，函数中的inner即为差值数组 因为损失函数为差值平方的和/2m 直接对差值矩阵inner乘其转置矩阵得差值平方的和
def lr_cost(theta, X, y):
# X:R(m*n) ,m样本数，n 特征数
# y：R(m)
# theta:R(n) ,线性回归参数
m = X.shape[0] #m样本数
inner = X @ theta - y #R(m*1) ,X @ theta 等价于X.dot(theta)也即是矩阵乘法
square_sum = inner.T @ inner
cost = square_sum / (2 * m)
return cost
lr_cost(theta, X, y)#返回损失函数的值
32.072733877455676
批量梯度下降函数（每一次梯度下降的过程中，都用到了所有样本，也就是所有的x，y） 对于θ0θ1求偏导，式子为 
def gradient(theta, X, y):
m = X.shape[0]
inner = X.T @ (X @ theta - y)
i
return inner / m
[ -5.83913505 -65.32884975]
拟合线性回归，返回最后的两个参数θ0和θ1 _theta拷贝原来的theta 在epoch范围内循环执行梯度下降，并将损失函数添加到cost_data数组中
def batch_gradient_decent(theta, X, y, epoch, alpha=0.01):
# 拟合线性回归，返回参数和代价
cost_data = [lr_cost(theta, X, y)]
_theta = theta.copy() #备份的theta
for _ in range(epoch):
_theta = _theta - alpha * gradient(_theta, X, y)
cost_data.append(lr_cost(_theta, X, y))
return _theta, cost_data
epoch = 500
final_theta, cost_data = batch_gradient_decent(theta, X, y, epoch)
final_theta
array([-2.28286727, 1.03099898])
cost_data
[32.072733877455676,
6.737190464870011,
5.931593568604956,
5.901154707081388,
5.895228586444221,
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lr_cost(final_theta, X, y)
4.713809531116866
c = {'epoch':list(range(epoch+1)),'cost':cost_data}
cost_df = pd.DataFrame(c)
ax = sns.lineplot(x='epoch',y='cost',data = cost_df)
plt.show()
b = final_theta[0] #intercept,Y轴上的截距
m = final_theta[1] # slope ，斜率
print('fx = ')
plt.scatter(data.population,data.profit, label="Training data")
plt.plot(data.population, data.population*m + b,label="prediction")
plt.legend(loc=2)
plt.show()
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