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labels=np.array([0,1])\ndef iou(a,b):\n    wh=np.maximum(0,np.minimum(a[2:],b[2:])-np.maximum(a[:2],b[:2])); inter=wh.prod()\n    return inter/(np.prod(a[2:]-a[:2])+np.prod(b[2:]-b[:2])-inter)\nC=np.array([[-np.log(prob[i,labels[j]])+2*abs(boxes[i]-truth[j]).sum()+1-iou(boxes[i],truth[j]) for j in range(2)] for i in range(2)])\ncosts={str(p):sum(C[i,j] for i,j in enumerate(p)) for p in permutations(range(2))}\nprint('cost matrix',C.round(4).tolist(),'assignment costs',costs,'selected',min(costs,key=costs.get))\nassert min(costs,key=costs.get)=='(1, 0)'\nfor stride in [4,8,16,32]: print('stride',stride,'target span',16/stride,'cells')\nprint('with 3 predictions/2 targets, unmatched slot needs no-object classification, not an invented third target')\n```","sha256":"5714c01e6f12168c1ae96d2c020af303805871c916cd8e924051dbea3aa808d3","chapter":{"id":"course","title":"同一组对象换了排列，损失不该凭空变大 · 实现代码"},"unit":{"id":"notebook-blocks/0f45eba379c5437685bb52cf7ae38a2d","title":"同一组对象换了排列，损失不该凭空变大 · 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