The softmax exp(x)/sum(exp(x)) is actually numerically well-behaved. It has only positive terms, so we needn’t worry about loss of significance, and the denominator is at least as large as the numerator, so the result is guaranteed to fall between 0 and 1. The only accident that might happen is over- or under-flow in the exponentials. … Read more
Your problem is caused by the NaN or Inf elements in your out_vec array. You could use the following code to avoid this problem: if np.isnan(np.sum(out_vec)): out_vec = out_vec[~numpy.isnan(out_vec)] # just remove nan elements from vector out_vec[out_vec > 709] = 709 … or you could use the following code to leave the NaN values in … Read more
As stated in the torch.nn.CrossEntropyLoss() doc: This criterion combines nn.LogSoftmax() and nn.NLLLoss() in one single class. Therefore, you should not use softmax before.
One reason to use the temperature function is to change the output distribution computed by your neural net. It is added to the logits vector according to this equation : 𝑞𝑖 =exp(𝑧𝑖/𝑇)/ ∑𝑗exp(𝑧𝑗/𝑇) where 𝑇 is the temperature parameter. You see, what this will do is change the final probabilities. You can choose T to … Read more
Having two different functions is a convenience, as they produce the same result. The difference is simple: For sparse_softmax_cross_entropy_with_logits, labels must have the shape [batch_size] and the dtype int32 or int64. Each label is an int in range [0, num_classes-1]. For softmax_cross_entropy_with_logits, labels must have the shape [batch_size, num_classes] and dtype float32 or float64. Labels … Read more
There is one nice attribute of Softmax as compared with standard normalisation. It react to low stimulation (think blurry image) of your neural net with rather uniform distribution and to high stimulation (ie. large numbers, think crisp image) with probabilities close to 0 and 1. While standard normalisation does not care as long as the … Read more
They’re both correct, but yours is preferred from the point of view of numerical stability. You start with e ^ (x – max(x)) / sum(e^(x – max(x)) By using the fact that a^(b – c) = (a^b)/(a^c) we have = e ^ x / (e ^ max(x) * sum(e ^ x / e ^ max(x))) … Read more