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您好,看了您的代码,非常适合初学者读。这里有一个问题请教,网络产生的anchor是 [xc, yc, w, h, angle]的形式,由于gt也是 [xc, yc, w, h, angle]的形式,所以在计算horizontal_overlaps 的时候您将gt通过obb2hbb_oc转成了hbb的形式,但是anchor并没有转啊,这怎么求horizontal_overlaps的呢?
horizontal_overlaps = bbox_overlaps( anchor.clone(), # generate anchor data copy obb2hbb_oc(bbox_annotation[:, :-1]))
我理解应该是 horizontal_overlaps = bbox_overlaps( obb2hbb_oc(anchor.clone()), # generate anchor data copy obb2hbb_oc(bbox_annotation[:, :-1])) 请问我的理解是否存在偏差?
The text was updated successfully, but these errors were encountered:
您好,我也研究这个项目的代码.我认为这是因为本身生成的锚框theta就为0,所以不需要旋转
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您好,看了您的代码,非常适合初学者读。这里有一个问题请教,网络产生的anchor是 [xc, yc, w, h, angle]的形式,由于gt也是 [xc, yc, w, h, angle]的形式,所以在计算horizontal_overlaps 的时候您将gt通过obb2hbb_oc转成了hbb的形式,但是anchor并没有转啊,这怎么求horizontal_overlaps的呢?
horizontal_overlaps = bbox_overlaps(
anchor.clone(), # generate anchor data copy
obb2hbb_oc(bbox_annotation[:, :-1]))
我理解应该是
horizontal_overlaps = bbox_overlaps(
obb2hbb_oc(anchor.clone()), # generate anchor data copy
obb2hbb_oc(bbox_annotation[:, :-1]))
请问我的理解是否存在偏差?
The text was updated successfully, but these errors were encountered: