帮我写用朴素贝叶斯分类器对手写数字进行识别的python代码不使用库函数

当然可以帮您写这个代码。以下是一个简单的手写数字识别的朴素贝叶斯分类器Python代码，不使用库函数：

import numpy as np

# 加载数据
def load_data(path):
    with open(path, 'r') as f:
        data = f.readlines()
    X = []
    y = []
    for line in data:
        line = line.strip().split(',')
        y.append(int(line[0]))
        X.append([int(i) for i in line[1:]])
    return np.array(X), np.array(y)

# 训练朴素贝叶斯分类器
def train(X_train, y_train):
    num_samples, num_features = X_train.shape
    num_classes = len(set(y_train))
    # 计算每个类别中每个特征的均值和标准差
    mean = np.zeros((num_classes, num_features))
    std = np.zeros((num_classes, num_features))
    for i in range(num_classes):
        X_i = X_train[y_train == i]
        mean[i, :] = X_i.mean(axis=0)
        std[i, :] = X_i.std(axis=0)
    # 计算每个类别的先验概率
    prior = np.zeros(num_classes)
    for i in range(num_classes):
        prior[i] = len(X_train[y_train == i]) / float(num_samples)
    return mean, std, prior

# 预测
def predict(X, mean, std, prior):
    num_samples, num_features = X.shape
    num_classes = len(prior)
    # 计算每个样本在每个类别下的概率
    likelihood = np.zeros((num_samples, num_classes))
    for i in range(num_classes):
        likelihood[:, i] = np.sum(-0.5 * ((X - mean[i, :]) / std[i, :]) ** 2 - 0.5 * np.log(2 * np.pi * std[i, :]), axis=1)
    # 计算后验概率
    posterior = np.zeros((num_samples, num_classes))
    for i in range(num_classes):
        posterior[:, i] = likelihood[:, i] + np.log(prior[i])
    # 返回每个样本最有可能属于哪个类别
    return np.argmax(posterior, axis=1)

# 测试
def test(X_test, y_test, mean, std, prior):
    y_pred = predict(X_test, mean, std, prior)
    acc = np.mean(y_pred == y_test)
    return acc

# 加载数据集
X_train, y_train = load_data('train.csv')
X_test, y_test = load_data('test.csv')

# 训练模型
mean, std, prior = train(X_train, y_train)

# 测试模型
acc = test(X_test, y_test, mean, std, prior)
print('Accuracy:', acc)

希望能够帮到您！如果您还有其他问题，可以继续问我。