What is the problem with my implementation of the cross-entropy function?

问题

I am learning the neural network and I want to write a function cross_entropy in python. Where it is defined as

where N is the number of samples, k is the number of classes, log is the natural logarithm, t_i,j is 1 if sample i is in class j and 0 otherwise, and p_i,j is the predicted probability that sample i is in class j. To avoid numerical issues with logarithm, clip the predictions to [10^{−12}, 1 − 10^{−12}] range.

According to the above description, I wrote down the codes by clipping the predictions to [epsilon, 1 − epsilon] range, then computing the cross_entropy based on the above formula.

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

The following code will be used to check if the function cross_entropy are correct.

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))

The output of the above codes is False, that to say my codes for defining the function cross_entropy is not correct. Then I print the result of cross_entropy(predictions, targets). It gave 0.178389544455 and the correct result should be ans = 0.71355817782. Could anybody help me to check what is the problem with my codes?

回答1:

You're not that far off at all, but remember you are taking the average value of N sums, where N = 2 (in this case). So your code could read:

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))

Here, I think it's a little clearer if you stick with np.sum(). Also, I added 1e-9 into the np.log() to avoid the possibility of having a log(0) in your computation. Hope this helps!

NOTE: As per @Peter's comment, the offset of 1e-9 is indeed redundant if your epsilon value is greater than 0.

回答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))

来源：https://stackoverflow.com/questions/47377222/what-is-the-problem-with-my-implementation-of-the-cross-entropy-function