时尚 mnist 的 Softmax 回归

问题描述

我正在尝试使用 softmax 来解决时尚 mnist。但是，成本函数总是在增加，并且运行训练需要花费大量时间。请帮助我改进这些代码行。我不知道我写的向量维度或函数是否正确。 x.shape 是 (784,60000)，y.shape 是 (60000,)。非常感谢。

import numpy as np

num_px = x.shape[1]
x = np.array(x)
x_test = np.array(x_test)


x = (x.reshape(x.shape[0],-1).T) / 255
x_test = (x_test.reshape(x_test.shape[0],-1).T) / 255

print (y.shape)
print ("x's shape: " + str(x.shape))
print ("x_test's shape: " + str(x_test.shape))

def softmax(z):
    z -= np.max(z)
    sm = (np.exp(z).T / np.sum(np.exp(z),axis=1))
    return sm

def initialize(dim1):
    w = np.zeros((dim1,1))
    b = 0
    return w,b

def propagate(w,b,X,Y):
    
    m = X.shape[1]  

    A = softmax((np.dot(w.T,X) + b))
    cost = (-1 / m) * np.sum(Y * np.log(A)) 
 
    # backwar prop
    dw = (1 / m) * np.dot(X,(A - Y))
    db = (1 / m) * np.sum(A - Y)

    cost = np.squeeze(cost)
    grads = {"dw": dw,"db": db}
    return grads,cost

def optimize(w,Y,num_iters,alpha,print_cost=False):
    costs = []
    for i in range(num_iters):
        grads,cost = propagate(w,Y)
        dw = grads["dw"]
        db = grads["db"]
        w = w - alpha * dw
        b = b - alpha * db

        if i % 50 == 0:
            costs.append(cost)

        # Print the cost every 100 training examples
        if print_cost and i % 50 == 0:
            print("Cost after iteration %i: %f" % (i,cost))

    params = {"w": w,"b": b}

    grads = {"dw": dw,"db": db}

    return params,grads,costs

def predict(w,X):
    y_pred = np.argmax(softmax((np.dot(w.T,X) + b)),axis=1)
    return y_pred

def model_LR(X_train,Y_train,test_x,test_y,print_cost):
    
    w,b = initialize(X_train.shape[0])
    parameters,costs = optimize(w,X_train,print_cost)

    w = parameters["w"]
    b = parameters["b"]

    y_prediction_train = predict(w,X_train)
    y_prediction_test = predict(w,test_x)
    print("Train accuracy: {} %",sum(y_prediction_train == Y_train[0]) / (float(len(Y_train))) * 100)
    print("Test accuracy: {} %",sum(y_prediction_test == test_y[0]) / (float(len(test_y))) * 100)

    d = {"costs": costs,"w": w,"b": b,"learning_rate": alpha,"num_iterations": num_iters}

    # Plot learning curve (with costs)
    costs = np.squeeze(d['costs'])
    plt.plot(costs)
    plt.ylabel('cost')
    plt.xlabel('iterations (per hundreds)')
    plt.title("Learning rate =" + str(d["learning_rate"]))
    plt.plot()
    plt.show()
    plt.close()

    return d

解决方法

暂无找到可以解决该程序问题的有效方法，小编努力寻找整理中！

如果你已经找到好的解决方法，欢迎将解决方案带上本链接一起发送给小编。

小编邮箱:dio#foxmail.com (将#修改为@）

numpy python softmax