Stanford-CS231n-assignment1-Softmax梯度推导及代码

下图是softmax的第一题，也就是循环法求梯度公式推导，推导中跟随一个特定的元素的前世今生，来展现整个softmax梯度的推导过程，其中字母有些混乱，i/j傻傻分不清楚，有多个地方都用到j，但代表的含义不尽相同：

Softmax梯度推导

1. softmax.py

from builtins import range
import numpy as np
from random import shuffle
from past.builtins import xrange

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!                                                           #
    #############################################################################
    # *****START OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****

    num_train = X.shape[0]
    num_classes = W.shape[1]
    
    for i in xrange(num_train):
        scores = X[i].dot(W) # X的第i行，也就是第i个图像点乘参数矩阵，得到的矩阵维度为(1,C)
        scores -= max(scores) # 减去最大值防止计算时数值错误
        loss += -np.log(np.exp(scores[y[i]]) / np.sum(np.exp(scores))) # 跟下面的公式是一样的，下面的使用了对数化简公式log(a/b)=log(a)-log(b)
        # loss += -score[y[i]] + np.log(np.sum(np.exp(scores)))
        # 对W的更新要放在i的循环中，每轮都要对所有W进行更新，也即每求出一个X的预测矩阵后都要反向更新W
        for j in range(num_classes):
            if j == y[i]:
                dW[:, j] += (np.exp(scores[j]) / np.sum(np.exp(scores))) * X[i] - X[i] # 代入公式，图中有详细推导
            else:
                dW[:, j] += (np.exp(scores[j]) / np.sum(np.exp(scores))) * X[i]

    loss /= num_train
    loss += 0.5 * reg * np.sum(W*W)
    dW /= num_train
    dW += reg * W
    pass

    # *****END OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****

    return loss, dW


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!                                                           #
    #############################################################################
    # *****START OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****

    num_train = X.shape[0]
    # scores是N*C矩阵，N对应N个实例，C代表每一项的得分
    scores = X.dot(W)
    # 减去最大值防止出现数值错误
    scores -= np.max(scores, axis=1, keepdims=True)
    # 求概率，用指数矩阵除以每一行的指数行和
    prob = np.exp(scores) / np.sum(np.exp(scores), axis=1, keepdims=True)
    # loss
    loss = -np.sum(np.log(prob[range(num_train), y]))
    loss = loss / num_train + 0.5 * reg * np.sum(W * W)
    
    # 根据循环法的公式，先给出一个掩膜矩阵，表示当j=y[i]时，-X[i]那一项
    mask_minus_ones = np.zeros_like(scores)
    mask_minus_ones[range(num_train), y] = -1
    # 另外一个矩阵，用来乘以概率，表示循环法中两种情况下的prob*X[i]
    mask_ones = np.ones_like(scores)
    # 两个矩阵相加得到最终的更新矩阵，该更新方法与SVM类似
    update = mask_ones * prob + mask_minus_ones
    dW += np.dot(X.T, update) / num_train + reg * W
    pass

    # *****END OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****

    return loss, dW

 

2. softmax.ipynb

# Use the validation set to tune hyperparameters (regularization strength and
# learning rate). You should experiment with different ranges for the learning
# 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.                          #
################################################################################
# *****START OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****

for learning_rate in learning_rates:
    for regularization_strength in regularization_strengths:
        softmax = Softmax()
        softmax.train(X_train, y_train, learning_rate, regularization_strength, num_iters = 500)
        acc_train = np.mean(y_train == softmax.predict(X_train))
        acc_val = np.mean(y_val == softmax.predict(X_val))
        
        results[(learning_rate,regularization_strength)] = (acc_train,acc_val)
        
        if acc_val > best_val :
            best_val = acc_val
            best_softmax = softmax
pass

# *****END OF YOUR CODE (DO NOT DELETE/MODIFY THIS LINE)*****
    
# 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)

个人感觉Softmax和SVM非常相似，都需要在求梯度的过程中分情况讨论，而且向量化梯度的更新几乎完全相同，都比较抽象，但是只要理解了其中一个另外一个就没有太大的问题。