python实现逻辑回归的示例

Odetta ·

更新时间:2024-09-21

· 819 次阅读

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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逻辑回归示例回归 Python

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