手写数字识别流程
MNIST手写数字集7000*10张图片
60k张图片训练,10k张图片测试
每张图片是28*28,如果是彩色图片是28*28*3
0-255表示图片的灰度值,0表示纯白,255表示纯黑
打平28*28的矩阵,得到28*28=784的向量
对于b张图片得到[b,784];然后对于b张图片可以给定编码
把上述的普通编码给定成独热编码,但是独热编码都是概率值,并且概率值相加为1,类似于softmax回归
套用线性回归公式
X[b,784] W[784,10] b[10] 得到 [b,10]
高维图片实现非常复杂,一个线性模型无法完成,因此可以添加非线性因子
f(X@W+b),使用激活函数让其非线性化,引出relu函数
1 =relu(X@W1+b1)
H2 = relu(h1@W2+b2)
Out = relu(h2@W3+b3)
第一步,把[1,784]变成[1,512]变成[1,256]变成[1,10]
得到[1,10]后将结果进行独热编码
使用欧氏距离或者使用mse进行误差度量
[1,784]通过三层网络输出一个[1,10]
# [b,784] ==> [b,256] ==> [b,128] ==> [b,10]
# [dim_in,dim_out],[dim_out]
w1 = tf.Variable(tf.random.truncated_normal([784, 256], stddev=0.1))
b1 = tf.Variable(tf.zeros([256]))
w2 = tf.Variable(tf.random.truncated_normal([256, 128], stddev=0.1))
b2 = tf.Variable(tf.zeros([128]))
w3 = tf.Variable(tf.random.truncated_normal([128, 10], stddev=0.1))
b3 = tf.Variable(tf.zeros([10]))
# learning rate
lr = 1e-3
for epoch in range(10):  # iterate db for 10
    # tranin every train_db
    for step, (x, y) in enumerate(train_db):
        # x: [128,28,28]
        # y: [128]
        # [b,28,28] ==> [b,28*28]
        x = tf.reshape(x, [-1, 28*28])
        with tf.GradientTape() as tape:  # only data types of tf.variable are logged
            # x: [b,28*28]
            # h1 = x@w1 + b1
            # [b,784]@[784,256]+[256] ==> [b,256] + [256] ==> [b,256] + [b,256]
            h1 = x @ w1 + tf.broadcast_to(b1, [x.shape[0], 256])
            h1 = tf.nn.relu(h1)
            # [b,256] ==> [b,128]
            # h2 = x@w2 + b2  # b2 can broadcast automatic
            h2 = h1 @ w2 + b2
            h2 = tf.nn.relu(h2)
            # [b,128] ==> [b,10]
            out = h2 @ w3 + b3
            # compute loss
            # out: [b,10]
            # y:[b] ==> [b,10]
            y_onehot = tf.one_hot(y, depth=10)
            # mse = mean(sum(y-out)^2)
            # [b,10]
            loss = tf.square(y_onehot - out)
            # mean:scalar
            loss = tf.reduce_mean(loss)
        # compute gradients
        grads = tape.gradient(loss, [w1, b1, w2, b2, w3, b3])
        # w1 = w1 - lr * w1_grad
        # w1 = w1 - lr * grads[0]  # not in situ update
        # in situ update
        w1.assign_sub(lr * grads[0])
        b1.assign_sub(lr * grads[1])
        w2.assign_sub(lr * grads[2])
        b2.assign_sub(lr * grads[3])
        w3.assign_sub(lr * grads[4])
        b3.assign_sub(lr * grads[5])
        if(step % 100 == 0):
            print(f'epoch:{epoch}, step: {step}, loss:{float(loss)}')