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我正在学习神经网络，我想cross_entropy用 python 编写一个函数。它被定义为

交叉熵

其中N是样本数，k是类数， 是log自然对数，t_i,j如果样本i在类中，则为 1 j，0否则为 1，并且是样本在类p_i,j中的预测概率。为避免对数出现数值问题，请将预测剪裁到范围。ij[10^{−12}, 1 − 10^{−12}]

根据上面的描述，我通过将预测裁剪到范围来写下代码[epsilon, 1 − epsilon]，然后根据上面的公式计算 cross_entropy。

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

下面的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，也就是说我定义函数cross_entropy的代码不正确。然后我打印cross_entropy(predictions, targets). 它给出0.178389544455了正确的结果应该是ans = 0.71355817782。有人可以帮我检查我的代码有什么问题吗？

4

2 回答 2

39

您根本就没有那么远，但请记住，您正在取 N 个总和的平均值，其中 N = 2（在这种情况下）。所以你的代码可以阅读：

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，我认为如果您坚持使用np.sum(). 此外，我在 1e-9 中添加了np.log()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，1e-9如果您的 epsilon 值大于0.

于 2017-11-20T17:59:27.757 回答
3 
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))
于 2018-10-31T00:23:21.330 回答