处理连续属性的朴素贝叶斯实现

import numpy as np class naiveBayes(): def __init__(self,): """ cols: 标识样本属性是离散的还是连续的 1: 离散属性 0: 连续 example: cols = [1,1,0,1] """ pass def fit(self,x,y): """ x: ndarray, (n_samples, n_features) y: ndarray, (n_samples, ) """ self.params = self.getMeanAndStd(x,y) def predict(self,x): """ x: ndarray, (n_samples, n_features) """ flag,probs = False,None for pc,mu_c,std_c in self.params: pxc = self.normal_distribution(x,mu_c,std_c) if not flag: flag = True probs = (pxc.prod(axis=1) * pc).reshape(-1,1) else: pxc = (pxc.prod(axis=1) * pc).reshape(-1,1) probs = np.hstack([probs, pxc]) return np.argmax(probs, axis=1) def getMeanAndStd(self,x,y): """ 计算连续属性的均值和标准差,即计算正态分布的参数 """ params = [] for c in np.unique(y): tx = x[y==c] params.append([np.sum(y == c) / y.shape[0], tx.mean(axis=0),tx.std(axis=0)]) return params def normal_distribution(self,x,mu,sigma): """ 功能: 计算P(x|c) x: 样本属性,(n_samples, n_features) mu: 均值列表,(N,) sigma: 标准差,(N, ) """ return 1 / (np.sqrt(2 * np.pi) * sigma) * np.exp(-1 * np.power(x - mu, 2) / (2 * np.power(sigma,2))) if __name__ == "__main__": pass