在交叉路口找到入口出口

Question

我正在使用以下代码查找人员的进出计数。

 # check to see if the object has been counted or not
            if not to.counted:
                # if the direction is negative (indicating the object
                # is moving up) AND the centroid is above the center
                # line, count the object
                if direction < 0 and centroid[1] < H // 2:
                    totalUp += 1
                    to.counted = True

                # if the direction is positive (indicating the object
                # is moving down) AND the centroid is below the
                # center line, count the object
                elif direction > 0 and centroid[1] > H // 2:
                    totalDown += 1
                    to.counted = True

根据此代码，如果同一个人返回并再次进入，则进入计数仍然与该人已被计数的相同。我想在每次人与线相交时找到进入和退出计数。我该如何解决？

Answer 1

一个快速的方法是完全忽略 counted 属性：

# if the direction is negative (indicating the object
# is moving up) AND the centroid is above the center
# line, count the object
if direction < 0 and centroid[1] < H // 2:
      totalUp += 1

# if the direction is positive (indicating the object
# is moving down) AND the centroid is below the
# center line, count the object
elif direction > 0 and centroid[1] > H // 2:
    totalDown += 1

这是假设您的总计数不是每人而是发生的总计。如果是这样，请忽略 if to.counted 因为你不关心它们是否已经被计算过，你只关心你设置的条件是否已经满足

在交叉路口找到入口出口

Find entry exit on intersect

python

opencv

image-processing

tracking

dlib