交叉熵函数（python）

Question

我正在学习神经网络，我想在python中编写一个函数cross_entropy。在哪里定义为

其中N是样本数，k是类的数量，log是自然对数，t_i,j是1，如果样本i在类j和0否则，p_i,j是样本i在类j的预测概率。要避免对数的数值问题，请将预测剪辑到[10^{−12}, 1 − 10^{−12}]范围。

根据上面的描述，我通过clippint对[epsilon, 1 − epsilon]范围的预测写下代码，然后根据上面的公式计算交叉熵。

def cross_entropy(predictions, targets, epsilon=1e-12):
    """
    Computes cross entropy between targets (encoded as one-hot vectors)
    and predictions. 
    Input: predictions (N, k) ndarray
           targets (N, k) ndarray        
    Returns: scalar
    """
    predictions = np.clip(predictions, epsilon, 1. - epsilon)
    ce = - np.mean(np.log(predictions) * targets) 
    return ce

以下代码将用于检查函数cross_entropy是否正确。

predictions = np.array([[0.25,0.25,0.25,0.25],
                        [0.01,0.01,0.01,0.96]])
targets = np.array([[0,0,0,1],
                  [0,0,0,1]])
ans = 0.71355817782  #Correct answer
x = cross_entropy(predictions, targets)
print(np.isclose(x,ans))

上述代码的输出为False，即我的用于定义函数cross_entropy的代码不正确。然后我打印cross_entropy(predictions, targets)的结果。它给了0.178389544455，正确的结果应该是ans = 0.71355817782。有人可以帮我查一下我的代码有什么问题吗？

Answer 1

你根本不是那么遥远，但记住你正在取N个和的平均值，其中N = 2（在这种情况下）。所以你的代码可以读取：

def cross_entropy(predictions, targets, epsilon=1e-12):
    """
    Computes cross entropy between targets (encoded as one-hot vectors)
    and predictions. 
    Input: predictions (N, k) ndarray
           targets (N, k) ndarray        
    Returns: scalar
    """
    predictions = np.clip(predictions, epsilon, 1. - epsilon)
    N = predictions.shape[0]
    ce = -np.sum(targets*np.log(predictions+1e-9))/N
    return ce

predictions = np.array([[0.25,0.25,0.25,0.25],
                        [0.01,0.01,0.01,0.96]])
targets = np.array([[0,0,0,1],
                   [0,0,0,1]])
ans = 0.71355817782  #Correct answer
x = cross_entropy(predictions, targets)
print(np.isclose(x,ans))

在这里，我认为如果你坚持使用np.sum()会更清楚一些。另外，我在np.log()中添加了1e-9，以避免在计算中出现log（0）的可能性。希望这可以帮助！

注意：根据@ Peter的评论，如果你的epsilon值大于1e-9，那么0的偏移确实是多余的。

Answer 2

def cross_entropy(x, y):
    """ Computes cross entropy between two distributions.
    Input: x: iterabale of N non-negative values
           y: iterabale of N non-negative values
    Returns: scalar
    """

    if np.any(x < 0) or np.any(y < 0):
        raise ValueError('Negative values exist.')

    # Force to proper probability mass function.
    x = np.array(x, dtype=np.float)
    y = np.array(y, dtype=np.float)
    x /= np.sum(x)
    y /= np.sum(y)

    # Ignore zero 'y' elements.
    mask = y > 0
    x = x[mask]
    y = y[mask]    
    ce = -np.sum(x * np.log(y)) 
    return ce

def cross_entropy_via_scipy(x, y):
        ''' SEE: https://en.wikipedia.org/wiki/Cross_entropy'''
        return  entropy(x) + entropy(x, y)

from scipy.stats import entropy, truncnorm

x = truncnorm.rvs(0.1, 2, size=100)
y = truncnorm.rvs(0.1, 2, size=100)
print np.isclose(cross_entropy(x, y), cross_entropy_via_scipy(x, y))