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label

Open jane-pyc opened this issue 5 years ago • 1 comments

why you define your label in the loss function to be :

` def sync_loss(self,out_v,out_a,criterion):

    batch_size  = out_a.size()[0]
    time_size   = out_a.size()[2]

    label       = torch.arange(time_size).cuda()

    nloss = 0
    prec1 = 0

`

should the label be 0 or 1 depended on the data is synchronized or not ?

jane-pyc avatar Sep 09 '20 07:09 jane-pyc

The label should be the matching frame, i.e. along the diagonal. See https://ieeexplore.ieee.org/document/9067055

joonson avatar Sep 12 '20 13:09 joonson