python实现逻辑回归的示例

代码

import numpy as np

import matplotlib.pyplot as plt

from sklearn.datasets.samples_generator import make_classification

def initialize_params(dims):

w = np.zeros((dims, 1))

b = 0

return w, b

def sigmoid(x):

z = 1 / (1 + np.exp(-x))

return z

def logistic(X, y, w, b):

num_train = X.shape[0]

y_hat = sigmoid(np.dot(X, w) + b)

loss = -1 / num_train * np.sum(y * np.log(y_hat) + (1-y) * np.log(1-y_hat))

cost = -1 / num_train * np.sum(y * np.log(y_hat) + (1 - y) * np.log(1 - y_hat))

dw = np.dot(X.T, (y_hat - y)) / num_train

db = np.sum(y_hat - y) / num_train

return y_hat, cost, dw, db

def linear_train(X, y, learning_rate, epochs):

# 参数初始化

w, b = initialize_params(X.shape[1])

loss_list = []

for i in range(epochs):

# 计算当前的预测值、损失和梯度

y_hat, loss, dw, db = logistic(X, y, w, b)

loss_list.append(loss)

# 基于梯度下降的参数更新

w += -learning_rate * dw

b += -learning_rate * db

# 打印迭代次数和损失

if i % 10000 == 0:

print("epoch %d loss %f" % (i, loss))

# 保存参数

params = {

'w': w,

'b': b

}

# 保存梯度

grads = {

'dw': dw,

'db': db

}

return loss_list, loss, params, grads

def predict(X, params):

w = params['w']

b = params['b']

y_pred = sigmoid(np.dot(X, w) + b)

return y_pred

if __name__ == "__main__":

# 生成数据

X, labels = make_classification(n_samples=100,

n_features=2,

n_informative=2,

n_redundant=0,

random_state=1,

n_clusters_per_class=2)

print(X.shape)

print(labels.shape)

# 生成伪随机数

rng = np.random.RandomState(2)

X += 2 * rng.uniform(size=X.shape)

# 划分训练集和测试集

offset = int(X.shape[0] * 0.9)

X_train, y_train = X[:offset], labels[:offset]

X_test, y_test = X[offset:], labels[offset:]

y_train = y_train.reshape((-1, 1))

y_test = y_test.reshape((-1, 1))

print('X_train=', X_train.shape)

print('y_train=', y_train.shape)

print('X_test=', X_test.shape)

print('y_test=', y_test.shape)

# 训练

loss_list, loss, params, grads = linear_train(X_train, y_train, 0.01, 100000)

print(params)

# 预测

y_pred = predict(X_test, params)

print(y_pred[:10])

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