python
复制代码
import cv2
import numpy as np
def xyxy2xywh(rect):
'''
(x1,y1,x2,y2) -> (x,y,w,h)
'''
return [rect[0],rect[1],rect[2]-rect[0],rect[3]-rect[1]]
def xywh2xyxy(rect):
'''
(x,y,w,h) -> (x1,y1,x2,y2)
'''
return [rect[0],rect[1],rect[0]+rect[2],rect[1]+rect[3]]
def is_RecA_RecB_interSect(RecA, RecB): # Rec = [xmin,ymin,xmax,ymax]
# 获取交集区域的[xmin,ymin,xmax,ymax]
x_A_and_B_min = max(RecA[0], RecB[0])
y_A_and_B_min = max(RecA[1], RecB[1])
x_A_and_B_max = min(RecA[2], RecB[2])
y_A_and_B_max = min(RecA[3], RecB[3])
# 计算交集部分面积, 当(xmax - xmin)为负时，说明A与B框无交集，直接置为0。 (ymax - ymin)同理。
interArea = max(0, x_A_and_B_max - x_A_and_B_min) * max(0, y_A_and_B_max - y_A_and_B_min)
return interArea > 0
def merge_RecA_RecB(RecA, RecB): # Rec = [xmin,ymin,xmax,ymax]
# 获取合并区域的[xmin,ymin,xmax,ymax]
xmin = min(RecA[0], RecB[0])
ymin = min(RecA[1], RecB[1])
xmax = max(RecA[2], RecB[2])
ymax = max(RecA[3], RecB[3])
return [xmin,ymin, xmax,ymax]
# def merge_rect(box,box_len):
# if box_len== 1:
# return box
# for i in range(box_len):
# RecA_xywh = box[i]
# RecA_xyxy = xywh2xyxy(RecA_xywh)
# for j in range(i+1,box_len):
# RecB_xywh = box[j]
# RecB_xyxy = xywh2xyxy(RecB_xywh)
# print(is_RecA_RecB_interSect(RecA_xyxy, RecB_xyxy))
# if is_RecA_RecB_interSect(RecA_xyxy, RecB_xyxy):
# rect_xyxy = merge_RecA_RecB(RecA_xyxy, RecB_xyxy)
# rect_xywh = xyxy2xywh(rect_xyxy)
# box.remove(RecA_xywh)
# box.remove(RecB_xywh)
# box.append(rect_xywh)
# box_len = len(box)
# merge_rect(box,box_len)
# # 此处少了return box会报错
# return box
# def merge_rect(box, box_len):
# if box_len == 1:
# return box
# for i in range(box_len):
# RecA_xywh = box[i]
# RecA_xyxy = xywh2xyxy(RecA_xywh)
# for j in range(i+1, box_len):
# RecB_xywh = box[j]
# RecB_xyxy = xywh2xyxy(RecB_xywh)
# if is_RecA_RecB_interSect(RecA_xyxy, RecB_xyxy):
# rect_xyxy = merge_RecA_RecB(RecA_xyxy, RecB_xyxy)
# rect_xywh = xyxy2xywh(rect_xyxy)
# # 使用remove(elem)来移除元素
# box.remove(RecA_xywh)
# box.remove(RecB_xywh)
# box.append(rect_xywh)
# box_len = len(box)
# merge_rect(box, box_len)
# # 返回上一级循环，避免重复处理已合并的矩形
# return box
# return box
'''
递归是一个过程或函数在其定义或说明中有直接或间接调用自身的一种方法，
它通常把一个大型复杂的问题层层转化为一个与原问题相似的规模较小的问题来求解。
因此递归过程，最重要的就是查看能不能讲原本的问题分解为更小的子问题，这是使用递归的关键。
终止条件：矩形框数为1或者为空。
返回值： 新合并的矩形框
本级任务： 每一级需要做的就是遍历从它开始的后续矩形框，寻找可以和他合并的矩形
'''
def merge_rect(box):
'''
合并重叠框
输入参数: box :[[x,y,w,h],...]
返回:
合并后的box:[[x,y,w,h],...]
'''
if len(box) == 1 or len(box) == 0 : # 矩形框数为1或者为空
return box
for i in range(len(box)):
RecA_xywh = box[i]
RecA_xyxy = xywh2xyxy(RecA_xywh)
for j in range(i+1, len(box)):
RecB_xywh = box[j]
RecB_xyxy = xywh2xyxy(RecB_xywh)
if is_RecA_RecB_interSect(RecA_xyxy, RecB_xyxy):
rect_xyxy = merge_RecA_RecB(RecA_xyxy, RecB_xyxy)
rect_xywh = xyxy2xywh(rect_xyxy)
# 使用remove(elem)来移除元素
box.remove(RecA_xywh)
box.remove(RecB_xywh)
box.append(rect_xywh)
merge_rect(box)
# 返回上一级循环，避免重复处理已合并的矩形
return box
return box
if __name__=="__main__":
# 原始
box = [[256,256,10,10],[10,10,15,15],[20,20,10,10],[100,100,150,150],
[200,200,100,100],[400,400,15,15],[410,410,15,15],[420,420,10,10]] # (x,y,w,h)
print("原始的矩形框:",box)
img = np.ones([512, 512, 3], np.uint8)
for x,y,w,h in box:
img = cv2.rectangle(img, (x,y), (x+w,y+h), (0, 255, 0), 2)
cv2.imshow('origin', img)
# 合并后
merged_box = merge_rect(box)
print("合并的矩形框:",merged_box)
img = np.ones([512, 512, 3], np.uint8)
for x,y,w,h in merged_box:
img = cv2.rectangle(img, (x,y), (x+w,y+h), (0, 0, 255), 2)
cv2.imshow('merged', img)
cv2.waitKey(0)
cv2.destroyAllWindows()
