cs231n作业：assignment1 - softmax


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GitHub地址：https://github.com/ZJUFangzh/cs231n

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，softmax

前面数据集的都跟SVM的一样。

直接进入loss和grads推导环节。

Li=−log(efyi∑jefj)

可以看到，计算的公式也就是cross-entropy，即

H(p,q)=−∑iyilog(yhati)

但是，这样有一个缺点，就是指数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logC=−maxfj

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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：cs231n softmax求导

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：

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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：

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

向量化softmax

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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：

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naive loss: 2.357905e+00 computed in 0.091724s
vectorized loss: 2.357905e+00 computed in 0.002995s
Loss difference: 0.000000
Gradient 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：

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# 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 Softmax
results = {}
best_val = -1
best_softmax = None
learning_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)
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lr 1.000000e-07 reg 2.500000e+04 train accuracy: 0.350592 val accuracy: 0.354000
lr 1.000000e-07 reg 5.000000e+04 train accuracy: 0.329551 val accuracy: 0.342000
lr 5.000000e-07 reg 2.500000e+04 train accuracy: 0.347286 val accuracy: 0.359000
lr 5.000000e-07 reg 5.000000e+04 train accuracy: 0.328551 val accuracy: 0.337000
best validation accuracy achieved during cross-validation: 0.359000
赏

谢谢你请我吃糖果

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