面试的时候让手写IOU和非极大值抑制，虽然集体原理挺了解的，但是没有手写过这种应用函数，
IoU还好，非极大值写个磕磕绊绊，写了个伪代码加口述混了过去。回来总结下
def IOU(x1,y1,X1,Y1, x2,y2,X2,Y2):
xx = max(x1,x2)
XX = min(X1,X2)
yy = max(y1,y2)
YY = min(Y1,Y2)
m = max(0., XX-xx)
n = max(0., YY-yy)
Jiao = m*n
Bing = (X1-x1)*(Y1-y1)+(X2-x2)*(Y2-y2)-Jiao
return Jiao/Bing
def nms(dets, thresh):
x1 = dets[:, 0]
y1 = dets[:, 1]
x2 = dets[:, 2]
y2 = dets[:, 3]
scores = dets[:, 4]
# areas = (x2 - x1 + 1) * (y2 - y1 + 1) #所有box面积
# print "all box aress: ", areas
order = scores.argsort()[::-1] #降序排列得到scores的坐标索引
keep = []
while order.size > 0:
i = order[0] #最大得分box的坐标索引
keep.append(i)
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:]]) #最高得分的boax与其他box的公共部分(交集)
w = np.maximum(0.0, xx2 - xx1 + 1)
h = np.maximum(0.0, yy2 - yy1 + 1) #求高和宽，并使数值合法化
inter = w * h #其他所有box的面积
ovr = inter / (areas[i] + areas[order[1:]] - inter) #IOU:交并比
inds = np.where(ovr <= thresh)[0] #ovr小表示两个box交集少，可能是另一个物体的框，故需要保留
order = order[inds + 1] #iou小于阈值的框
return keep