import torch # Original author: Francisco Massa: # https://github.com/fmassa/object-detection.torch # Ported to PyTorch by Max deGroot (02/01/2017) def nms(boxes, scores, overlap=0.5, top_k=200): """Apply non-maximum suppression at test time to avoid detecting too many overlapping bounding boxes for a given object. Args: boxes: (tensor) The location preds for the img, Shape: [num_priors,4]. scores: (tensor) The class predscores for the img, Shape:[num_priors]. overlap: (float) The overlap thresh for suppressing unnecessary boxes. top_k: (int) The Maximum number of box preds to consider. Return: The indices of the kept boxes with respect to num_priors. """ keep = scores.new(scores.size(0)).zero_().long() if boxes.numel() == 0: return keep x1 = boxes[:, 0] y1 = boxes[:, 1] x2 = boxes[:, 2] y2 = boxes[:, 3] area = torch.mul(x2 - x1, y2 - y1) v, idx = scores.sort(0) # sort in ascending order # I = I[v >= 0.01] idx = idx[-top_k:] # indices of the top-k largest vals xx1 = boxes.new() yy1 = boxes.new() xx2 = boxes.new() yy2 = boxes.new() w = boxes.new() h = boxes.new() # keep = torch.Tensor() count = 0 while idx.numel() > 0: i = idx[-1] # index of current largest val # keep.append(i) keep[count] = i count += 1 if idx.size(0) == 1: break idx = idx[:-1] # remove kept element from view # load bboxes of next highest vals torch.index_select(x1, 0, idx, out=xx1) torch.index_select(y1, 0, idx, out=yy1) torch.index_select(x2, 0, idx, out=xx2) torch.index_select(y2, 0, idx, out=yy2) # store element-wise max with next highest score xx1 = torch.clamp(xx1, min=x1[i]) yy1 = torch.clamp(yy1, min=y1[i]) xx2 = torch.clamp(xx2, max=x2[i]) yy2 = torch.clamp(yy2, max=y2[i]) w.resize_as_(xx2) h.resize_as_(yy2) w = xx2 - xx1 h = yy2 - yy1 # check sizes of xx1 and xx2.. after each iteration w = torch.clamp(w, min=0.0) h = torch.clamp(h, min=0.0) inter = w*h # IoU = i / (area(a) + area(b) - i) rem_areas = torch.index_select(area, 0, idx) # load remaining areas) union = (rem_areas - inter) + area[i] IoU = inter/union # store result in iou # keep only elements with an IoU <= overlap idx = idx[IoU.le(overlap)] return keep, count