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cs231n作业:assignment1 - softmax
阅读量:4100 次
发布时间:2019-05-25

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title: cs231n作业:assignment1 - softmax

id: cs231n-1h-3
tags:

  • cs231n
  • homework
    categories:
  • AI
  • Deep Learning
    date: 2018-09-27 16:02:57

GitHub地址:

个人博客:
softmax是最常用的分类器之一。

softmax和svm都是常用的分类器,而softmax更为常用。

具体可以参考我这篇的最后,ng老师有讲,

前面数据集的都跟SVM的一样。

直接进入loss和grads推导环节。

L i = − l o g ( e f y i ∑ j e f j ) L_i = -log(\frac{e^{f_{y_i}}}{\sum_j e^{f_j}}) Li=log(jefjefyi)

可以看到,计算的公式也就是cross-entropy,即

H ( p , q ) = − ∑ i y i l o g ( y i h a t ) H(p,q) = - \sum_i y_i log(y_{i}^{hat}) H(p,q)=iyilog(yihat)

但是,这样有一个缺点,就是指数 e f y i e^{f_{y_i}} efyi可能会特别大,这样可能导致内存不足,计算不稳定等问题。那么可以在分子分母同乘一个常数C,一般C取为 l o g C = − m a x f j logC = -max f_j logC=maxfj

f = np.array([123, 456, 789]) # 例子中有3个分类,每个评分的数值都很大p = np.exp(f) / np.sum(np.exp(f)) # 不妙:数值问题,可能导致数值爆炸# 那么将f中的值平移到最大值为0:f -= np.max(f) # f becomes [-666, -333, 0]p = np.exp(f) / np.sum(np.exp(f)) # 现在OK了,将给出正确结果

精确地说,SVM分类器使用的是折叶损失(hinge loss),有时候又被称为最大边界损失(max-margin loss)。Softmax分类器使用的是交叉熵损失(corss-entropy loss)。Softmax分类器的命名是从softmax函数那里得来的,softmax函数将原始分类评分变成正的归一化数值,所有数值和为1,这样处理后交叉熵损失才能应用。注意从技术上说“softmax损失(softmax loss)”是没有意义的,因为softmax只是一个压缩数值的函数。但是在这个说法常常被用来做简称。

求导过程参考:

最终得到的公式是:

softmax代码实现

编辑cs231n/classifiers/softmax.py,先写一下softmax_loss_naive函数,依旧是循环:

def softmax_loss_naive(W, X, y, reg):  """  Softmax loss function, naive implementation (with loops)  Inputs have dimension D, there are C classes, and we operate on minibatches  of N examples.  Inputs:  - W: A numpy array of shape (D, C) containing weights.  - X: A numpy array of shape (N, D) containing a minibatch of data.  - y: A numpy array of shape (N,) containing training labels; y[i] = c means    that X[i] has label c, where 0 <= c < C.  - reg: (float) regularization strength  Returns a tuple of:  - loss as single float  - gradient with respect to weights W; an array of same shape as W  """  # Initialize the loss and gradient to zero.  loss = 0.0  dW = np.zeros_like(W)  #############################################################################  # TODO: Compute the softmax loss and its gradient using explicit loops.     #  # Store the loss in loss and the gradient in dW. If you are not careful     #  # here, it is easy to run into numeric instability. Don't forget the        #  # regularization!                                                           #  #############################################################################  (N, D) = X.shape  C = W.shape[1]  #遍历每个样本  for i in range(N):    f_i = X[i].dot(W)    #进行公式的指数修正    f_i -= np.max(f_i)    sum_j = np.sum(np.exp(f_i))    #得到样本中每个类别的概率    p = lambda k : np.exp(f_i[k]) / sum_j    loss += - np.log(p(y[i]))    #根据softmax求导公式    for k in range(C):      p_k = p(k)      dW[:, k] += (p_k - (k == y[i])) * X[i]    loss /= N  loss += 0.5 * reg * np.sum(W * W)  dW /= N  dW += reg*W  #############################################################################  #                          END OF YOUR CODE                                 #  #############################################################################  return loss, dW

验证一下loss和grad得到:

numerical: -0.621593 analytic: -0.621593, relative error: 7.693773e-09numerical: -2.576505 analytic: -2.576505, relative error: 4.492083e-09numerical: -1.527801 analytic: -1.527801, relative error: 4.264914e-08numerical: 1.101379 analytic: 1.101379, relative error: 9.735173e-09numerical: 2.375620 analytic: 2.375620, relative error: 3.791861e-08numerical: 3.166961 analytic: 3.166960, relative error: 8.526285e-09numerical: -1.440997 analytic: -1.440998, relative error: 4.728898e-08numerical: 0.563304 analytic: 0.563304, relative error: 2.409996e-08numerical: -2.057292 analytic: -2.057292, relative error: 1.820335e-08numerical: -0.450338 analytic: -0.450338, relative error: 8.075985e-08numerical: -0.233090 analytic: -0.233090, relative error: 4.136546e-08numerical: 0.251391 analytic: 0.251391, relative error: 4.552523e-08numerical: 0.787031 analytic: 0.787031, relative error: 5.036469e-08numerical: -1.801593 analytic: -1.801594, relative error: 3.159903e-08numerical: -0.294108 analytic: -0.294109, relative error: 1.792497e-07numerical: -1.974307 analytic: -1.974307, relative error: 1.160708e-08numerical: 2.986921 analytic: 2.986921, relative error: 2.788065e-08numerical: -0.247281 analytic: -0.247281, relative error: 8.957573e-08numerical: 0.569337 analytic: 0.569337, relative error: 2.384912e-08numerical: -1.579298 analytic: -1.579298, relative error: 1.728733e-08

向量化softmax

def softmax_loss_vectorized(W, X, y, reg):  """  Softmax loss function, vectorized version.  Inputs and outputs are the same as softmax_loss_naive.  """  # Initialize the loss and gradient to zero.  loss = 0.0  dW = np.zeros_like(W)  #############################################################################  # TODO: Compute the softmax loss and its gradient using no explicit loops.  #  # Store the loss in loss and the gradient in dW. If you are not careful     #  # here, it is easy to run into numeric instability. Don't forget the        #  # regularization!                                                           #  #############################################################################  (N, D) = X.shape  C = W.shape[1]  f = X.dot(W)  #在列方向进行指数修正  f -= np.max(f,axis=1,keepdims=True)  #求得softmax各个类的概率  p = np.exp(f) / np.sum(np.exp(f),axis=1,keepdims=True)  y_lable = np.zeros((N,C))  #y_lable就是(N,C)维的矩阵,每一行中只有对应的那个正确类别 = 1,其他都是0  y_lable[np.arange(N),y] = 1  #cross entropy  loss = -1 * np.sum(np.multiply(np.log(p),y_lable)) / N  loss += 0.5 * reg * np.sum( W * W)  #求导公式,很清晰  dW = X.T.dot(p-y_lable)  dW /= N  dW += reg*W  #############################################################################  #                          END OF YOUR CODE                                 #  #############################################################################  return loss, dW

检验一下向量化和非向量化的时间:

naive loss: 2.357905e+00 computed in 0.091724svectorized loss: 2.357905e+00 computed in 0.002995sLoss difference: 0.000000Gradient difference: 0.000000

softmax的函数已经编写完成了,接下来调一下学习率和正则化两个超参数:

# rates and regularization strengths; if you are careful you should be able to# get a classification accuracy of over 0.35 on the validation set.from cs231n.classifiers import Softmaxresults = {
}best_val = -1best_softmax = Nonelearning_rates = [1e-7, 5e-7]regularization_strengths = [2.5e4, 5e4]################################################################################# TODO: ## Use the validation set to set the learning rate and regularization strength. ## This should be identical to the validation that you did for the SVM; save ## the best trained softmax classifer in best_softmax. #################################################################################for lr in learning_rates: for reg in regularization_strengths: softmax = Softmax() loss_hist = softmax.train(X_train, y_train, learning_rate=lr, reg=reg, num_iters=1500, verbose=True) y_train_pred = softmax.predict(X_train) y_val_pred = softmax.predict(X_val) y_train_acc = np.mean(y_train_pred==y_train) y_val_acc = np.mean(y_val_pred==y_val) results[(lr,reg)] = [y_train_acc, y_val_acc] if y_val_acc > best_val: best_val = y_val_acc best_softmax = softmax################################################################################# END OF YOUR CODE ################################################################################# # Print out results.for lr, reg in sorted(results): train_accuracy, val_accuracy = results[(lr, reg)] print('lr %e reg %e train accuracy: %f val accuracy: %f' % ( lr, reg, train_accuracy, val_accuracy)) print('best validation accuracy achieved during cross-validation: %f' % best_val)
lr 1.000000e-07 reg 2.500000e+04 train accuracy: 0.350592 val accuracy: 0.354000lr 1.000000e-07 reg 5.000000e+04 train accuracy: 0.329551 val accuracy: 0.342000lr 5.000000e-07 reg 2.500000e+04 train accuracy: 0.347286 val accuracy: 0.359000lr 5.000000e-07 reg 5.000000e+04 train accuracy: 0.328551 val accuracy: 0.337000best validation accuracy achieved during cross-validation: 0.359000

转载地址:http://eerii.baihongyu.com/

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