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Pytorch NMS implementation
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
@SamvitJ

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@SamvitJ SamvitJ commented Sep 17, 2019

There is a major bug in this implementation:
(1) Line 67: should be IoU = inter.float() / union.float(). Currently, IoU is always 0 (as integer division is performed), so the overlap argument to nms() has no effect.

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@usholanb usholanb commented Mar 31, 2020

great, thank you so much!

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