You need to understand the sigmoid_cross_entropy_with_logits and multi-class problems.
Multi-class cross-entropy
Your formula is correct and it directly corresponds to tf.nn.softmax_cross_entropy_with_logits.
For example:
-tf.reduce_sum(p * tf.log(q), axis=1)
p and q are expected for probability distributions over N classes. In particular, N can be 2, as in the following example:
p = tf.placeholder(tf.float32, shape=[None, 2])
logit_q = tf.placeholder(tf.float32, shape=[None, 2])
q = tf.nn.softmax(logit_q)
feed_dict = {
p: [[0, 1],
[1, 0],
[1, 0]],
logit_q: [[0.2, 0.8],
[0.7, 0.3],
[0.5, 0.5]]
}
prob1 = -tf.reduce_sum(p * tf.log(q), axis=1)
prob2 = tf.nn.softmax_cross_entropy_with_logits(labels=p, logits=logit_q)
print(prob1.eval(feed_dict)) # [ 0.43748799 0.51301527 0.69314718]
print(prob2.eval(feed_dict)) # [ 0.43748799 0.51301527 0.69314718]
Note that q is computing tf.nn.softmax, i.e. outputs a probability distribution. So it's still a multi-class cross-entropy formula, only for N = 2.
Binary cross-entropy
This correct formula for this problem is
p * -tf.log(q) + (1 - p) * -tf.log(1 - q)
But it's a multi-class case, the meaning of p and q is different here. Each p and q is a number, corresponding to a probability of class A.
The common part is the p * -tf.log(q) part and the sum. p was a one-hot vector, it can be a number, zero or one. Same for q - it was a probability distribution, now it's a number (probability).
If p is a vector, then each individual component is an independent binary classification. This answer outlines the difference between softmax and sigmoid functions in tensorflow. So the definition p = [0, 0, 0, 1, 0] doesn't mean a one-hot vector, but 5 different features, 4 of which are off and 1 is on. The definition q = [0.2, 0.2, 0.2, 0.2, 0.2] means that each of 5 features is on with 20% probability.
The goal of the sigmoid function is to squash the logit to [0, 1] interval.
This formula can also be used for multiple independent features, and that's what tf.nn.sigmoid_cross_entropy_with_logits computes:
p = tf.placeholder(tf.float32, shape=[None, 5])
logit_q = tf.placeholder(tf.float32, shape=[None, 5])
q = tf.nn.sigmoid(logit_q)
feed_dict = {
p: [[0, 0, 0, 1, 0],
[1, 0, 0, 0, 0]],
logit_q: [[0.2, 0.2, 0.2, 0.2, 0.2],
[0.3, 0.3, 0.2, 0.1, 0.1]]
}
prob1 = -p * tf.log(q)
prob2 = p * -tf.log(q) + (1 - p) * -tf.log(1 - q)
prob3 = p * -tf.log(tf.sigmoid(logit_q)) + (1-p) * -tf.log(1-tf.sigmoid(logit_q))
prob4 = tf.nn.sigmoid_cross_entropy_with_logits(labels=p, logits=logit_q)
print(prob1.eval(feed_dict))
print(prob2.eval(feed_dict))
print(prob3.eval(feed_dict))
print(prob4.eval(feed_dict))
You can notice that the last three tensors are equal, while the prob1 is only a part of cross-entropy, so it contains the correct value only when p is 1:
[[ 0. 0. 0. 0.59813893 0. ]
[ 0.55435514 0. 0. 0. 0. ]]
[[ 0.79813886 0.79813886 0.79813886 0.59813887 0.79813886]
[ 0.5543552 0.85435522 0.79813886 0.74439669 0.74439669]]
[[ 0.7981388 0.7981388 0.7981388 0.59813893 0.7981388 ]
[ 0.55435514 0.85435534 0.7981388 0.74439663 0.74439663]]
[[ 0.7981388 0.7981388 0.7981388 0.59813893 0.7981388 ]
[ 0.55435514 0.85435534 0.7981388 0.74439663 0.74439663]]
If you take the sum of -p * tf.log(q) along axis=1, that doesn't make sense in this syntax, so it'd be a valid formula in multi-class case.
Hope this answer helps
