About matcher code for top-k links (M)

[jinyoung0121]

Hi! Thank you for your wonderful work!

The code that determines the top-k links(M) seems to be 'build_matcher'.
However, it seems like that only the code for top-k=1 has been uploaded.
Is there a code uploaded for top-k≠1? (e.g., top-k=3)

Thanks.

def build_matcher(name, cost_class, cost_bbox, cost_giou, topk=1):
    if topk == 1:
        return MATCHER_REGISTRY.get(name)(cost_class=cost_class, cost_bbox=cost_bbox, cost_giou=cost_giou)
    else:
        return MATCHER_REGISTRY.get(name)(cost_class=cost_class, cost_bbox=cost_bbox, cost_giou=cost_giou, topk=topk)

@MATCHER_REGISTRY.register()
class IterativeHungarianMatcher(nn.Module):
    """This class computes an assignment between the targets and the predictions of the network
    For efficiency reasons, the targets don't include the no_object. Because of this, in general,
    there are more predictions than targets. In this case, we do a 1-to-1 matching of the best predictions,
    while the others are un-matched (and thus treated as non-objects).
    """

    def __init__(self, cost_class: float = 1, cost_bbox: float = 1, cost_giou: float = 1):
        """Creates the matcher
        Params:
            cost_class: This is the relative weight of the classification error in the matching cost
            cost_bbox: This is the relative weight of the L1 error of the bounding box coordinates in the matching cost
            cost_giou: This is the relative weight of the giou loss of the bounding box in the matching cost
        """
        super().__init__()
        self.cost_class = cost_class
        self.cost_bbox = cost_bbox
        self.cost_giou = cost_giou
        assert cost_class != 0 or cost_bbox != 0 or cost_giou != 0, "all costs cant be 0"

[Lil-Shake]

Hi! I have the same question as you. Has the question been solved?
Thanks:)

[Scarecrow0]

Same question; I'm wondering about the implementation of how the author achieves M=3.