ListNet Loss Function

[speeding-motor]

how to compute the loss function....we only use one sequence if enough or need to use any different sequence to get the loss

[speeding-motor]

def topkprob(self, vec, k=5): vec_sort = np.sort(vec)[-1::-1] topk = vec_sort[:k] ary = np.arange(k) return np.prod([np.exp(topk[i]) / np.sum(np.exp(topk[i:])) for i in ary])

def listwise_cost(self, list_ans, list_pred): return - np.sum(self.topkprob(list_ans) * np.log(self.topkprob(list_pred)))

the loss function in there, it's make me counfused, it's not use Probability of Different Combinatorial Sequences to compute the loss right?

[shiba24]

sorry for your confusion. as written in the readme, ListNet implementation is not the exact paper. https://github.com/shiba24/learning2rank#listnet If you want to use ListNet, maybe you can look at https://github.com/fullflu/learning-to-rank as well.

[speeding-motor]

sorry for your confusion. as written in the readme, ListNet implementation is not the exact paper. https://github.com/shiba24/learning2rank#listnet If you want to use ListNet, maybe you can look at https://github.com/fullflu/learning-to-rank as well.

thank you very very very...much to answer me，thank you.... now I understand, you use the author algorithm right， but I stil have one Doubt， why you use the sort function before to compute the loss,
for example: if there have sequence [A, B, C], and the y_true_score is [10, 5, 0], and the y_pred is [5, 10, 0], if you use the sort function here, the y_true sequence is still [A,B,C], but y_pred will be [B,A,C], so finally, when we use the: self.topkprob(list_ans) * np.log(self.topkprob(list_pred))

we are calculate the y_true[A,B,C] sequence y_pred[B, A, C] cross entry here.....so is that you want finally?