nms演算法思路**於：
把置信度最高的乙個boundingbox(bbox)作為目標，然後對比剩下bbox與目標bbox之間的交叉區域
如果交叉區域大於設定的閾值，那麼在剩下的bbox中去除該bbox（即使該bbox的置信度與目標bbox的置信度一樣）—-這個操作就是抑制最大重疊區域
把第二置信度高的bbox作為目標，重複1、2
import numpy as np
def nms(dets, thresh):
x1 = dets[:, 0]
y1 = dets[:, 1]
x2 = dets[:, 2]
y2 = dets[:, 3]
score = dets[:, 4]
sort_id_list = score.argsort()[::-1].tolist()
res =
while len(sort_id_list) >= 1:
i = sort_id_list.pop(0)
#intersect area left top point(xx1, yy1): xx1 >= x1, yy1 >= y1
#intersect area right down point(xx2, yy2): xx2 <= x2, yy2 <= y2
xx1 = np.maximum(x1[i], x1[sort_id_list[:]])
yy1 = np.maximum(y1[i], y1[sort_id_list[:]])
xx2 = np.minimum(x2[i], x2[sort_id_list[:]])
yy2 = np.minimum(y2[i], y2[sort_id_list[:]])
inter_w = np.maximum(0, (xx2 - xx1))
inter_h = np.maximum(0, (yy2 - yy1))
intersect = inter_w * inter_h
#iou = intersect area / union; union = box1 + box2 - intersect
iou = intersect / ((x2[i] - x1[i]) * (y2[i] - y1[i]) +\
(x2[sort_id_list[:]] - x1[sort_id_list[:]]) * (y2[sort_id_list[:]] - y1[sort_id_list[:]]) -\
intersect)
for i in reversed(range(len(sort_id_list))):
if iou[i] > thresh:
sort_id_list.pop(i)
return res
if __name__ == "__main__":
dets = np.array([
[204, 102, 358, 250, 0.5],
[257, 118, 380, 250, 0.7],
[280, 135, 400, 250, 0.6],
[255, 118, 360, 235, 0.7]
])thresh = 0.3
res = nms(dets, thresh)
print(res)
import numpy as np
def nms(dets, thresh):
x1 = dets[:, 0]
y1 = dets[:, 1]
x2 = dets[:, 2]
y2 = dets[:, 3]
score = dets[:, 4]
order = score.argsort()[::-1]
area = (x2 - x1) * (y2 - y1)
res =
while order.size >= 1:
i = order[0]
#intersect area left top point(xx1, yy1): xx1 >= x1, yy1 >= y1
#intersect area right down point(xx2, yy2): xx2 <= x2, yy2 <= y2
xx1 = np.maximum(x1[i], x1[order[1:]])
yy1 = np.maximum(y1[i], y1[order[1:]])
xx2 = np.minimum(x2[i], x2[order[1:]])
yy2 = np.minimum(y2[i], y2[order[1:]])
w = np.maximum(0.0, (xx2 - xx1))
h = np.maximum(0.0, (yy2 - yy1))
intersect = w * h
#iou = intersect area / union; union = box1 + box2 - intersect
iou = intersect / (area[i] + area[order[1:]] - intersect)
#update order index;ind +1:because ind is obtain by index [1:]
ind = np.where(iou <= thresh)[0]
order = order[ind +1]
return res
if __name__ == "__main__":
dets = np.array([
[204, 102, 358, 250, 0.5],
[257, 118, 380, 250, 0.7],
[280, 135, 400, 250, 0.6],
[255, 118, 360, 235, 0.7]
])thresh = 0.7
res = nms(dets, thresh)
print(res)
2、使用python高階庫有助於快速實現。
3、np.array 陣列本身可以作為另乙個np.array陣列的索引下標進行陣列訪問，如order[ind + 1](ind也為np.array陣列)
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