整个代码主要分为以下几步：（1）对图像预处理
（2）抠出每个停车位的图，以及其地址
（3）使用keras训练一个2分类的模型，卷积网络选择vgg，为了节省时间，采用keras提供的api，并冻结前10层。
（4）从视频中取出每一帧，挨个训练，实现实时
def imggg():
#读入一张原始图
img=cv2.imread("../test_images/scene1410.jpg")
#实现2值化操作，将在（120，255）以内的变成变成白色，将以外的变成黑色
low=np.uint8([120,120,120])
upper=np.uint8([255,255,255])
white_img=cv2.inRange(img,low,upper)
print(white_img.shape)
cv2.imshow("111",white_img)
cv2.waitKeyEx(0)
cv2.destroyAllWindows()
#与原图进行与运算，求出原图
img=cv2.bitwise_and(img,img,mask=white_img)
print(img.shape)
cv2.imshow("111",img)
cv2.waitKeyEx(0)
cv2.destroyAllWindows()
img=cv2.cvtColor(img,cv2.COLOR_RGB2GRAY)
#边缘检测
img=cv2.Canny(img,100,200)
cv2.imshow("111",img)
cv2.waitKeyEx(0)
cv2.destroyAllWindows()
cv2.imwrite("canny.jpg",img)以上主要是将图像2值化并边缘检测，为了后面的调用霍夫曼直线做准备
（2）制作一个mask，将停车场从图中抠出来
（1）先认为的选取图中的6个点，该6个点组成的一个多边形将停车场包围。
（2）生成一张和原始图同样大小的mask图，使用fillpoly（）函数将上述6个点内填充为白色
（3）使用bitwith_and ()将原图与mask做mask运算，即可提取出停车场
import cv2
import numpy as np
img1=cv2.imread("../test_images/scene1380.jpg")
img=cv2.imread("canny.jpg",0)
img=img.astype("uint8")
rows, cols = img.shape[:2]
pt_1 = [cols*0.05, rows*0.90]
pt_2 = [cols*0.05, rows*0.70]
pt_3 = [cols*0.30, rows*0.55]
pt_4 = [cols*0.6, rows*0.15]
pt_5 = [cols*0.90, rows*0.15]
pt_6 = [cols*0.90, rows*0.90]
#注意这里采用的是int32类型，一定要指明，否则会报错
point=np.array([pt_1,pt_2,pt_3,pt_4,pt_5,pt_6],dtype=np.int32)
print(point)
for i in point:
img1=cv2.circle(img1,(i[0],i[1]),10,(0,0,255))
cv2.imshow("111",img1)
cv2.waitKey(0)
cv2.destroyAllWindows()
#填充图形
zreos=np.zeros((rows,cols))
cv2.imshow("111",zreos)
cv2.waitKeyEx(0)
print(zreos.shape)
# cv2.destroyAllWindows()
mask=cv2.fillPoly(zreos,[point],255)
cv2.imshow("111",mask)
cv2.waitKey(0)
cv2.destroyAllWindows()
print(img.shape)
print(mask.shape)
#一定要指明是uint8，不说又报错。
mask=mask.astype("uint8")
img4=cv2.bitwise_and(img,img,mask=mask)
cv2.imshow("111",img4)
cv2.waitKey(0)
cv2.destroyAllWindows()
cv2.imwrite("mask_aprk.jpg",img4)（3）提取出停车场后下一步就是要标识每一个停车位的位置，采用霍夫直线检测，这里需要反复的调整大小，同时也要注意图中除了停车线，还有许多其他的线，要将其他的线剔除（本段代码较长，逻辑比较复杂，就不细讲了）
（1）lines = cv2.HoughLinesP(edges,0.1,np.pi/180,15,minLineLength=9,maxLineGap=4)
1.霍夫变换是将坐标空间转换为霍夫空间，在霍夫空间中，笛卡尔坐标系中的每个点都是一条曲线，同一条直线上的点会过一个定点
2.cv中的霍夫空间是有极坐标系构成
3.各个参数分别代表 图片，像素精度，角度精度，最少多少个焦点可被判断为一条直线，相隔几个点能被判断为一条直线
def hough_lines(self,image):
#输入的图像需要是边缘检测后的结果
#minLineLengh(线的最短长度，比这个短的都被忽略)和MaxLineCap（两条直线之间的最大间隔，小于此值，认为是一条直线）
#rho距离精度,theta角度精度,threshod超过设定阈值才被检测出线段
return cv2.HoughLinesP(image, rho=0.1, theta=np.pi/10, threshold=15, minLineLength=9, maxLineGap=4)
def draw_lines(self,image, lines, color=[255, 0, 0], thickness=2, make_copy=True):
# 过滤霍夫变换检测到直线
if make_copy:
image = np.copy(image)
cleaned = []
for line in lines:
for x1,y1,x2,y2 in line:
if abs(y2-y1) <=1 and abs(x2-x1) >=25 and abs(x2-x1) <= 55:
cleaned.append((x1,y1,x2,y2))
cv2.line(image, (x1, y1), (x2, y2), color, thickness)
print(" No lines detected: ", len(cleaned))
return image
可以看到，识别并不是很精准，这是由于汽车的存在导致的干扰。
def identify_blocks(self,image, lines, make_copy=True):
if make_copy:
new_image = np.copy(image)
#Step 1: 过滤部分直线
cleaned = []
for line in lines:
for x1,y1,x2,y2 in line:
if abs(y2-y1) <=1 and abs(x2-x1) >=25 and abs(x2-x1) <= 55:
cleaned.append((x1,y1,x2,y2))
#Step 2: 对直线按照x1进行排序
import operator
list1 = sorted(cleaned, key=operator.itemgetter(0, 1))
#Step 3: 找到多个列，相当于每列是一排车
clusters = {}
dIndex = 0
clus_dist = 5
for i in range(len(list1) - 1):
distance = abs(list1[i+1][0] - list1[i][0])
if distance <clus_dist:
if not dIndex in clusters.keys(): clusters[dIndex] = []
clusters[dIndex].append(list1[i])
clusters[dIndex].append(list1[i + 1])
else:
dIndex += 1
#Step 4: 得到坐标
rects = {}
i = 0
for key in clusters:
all_list = clusters[key]
cleaned = list(set(all_list))
if len(cleaned) > 5:
cleaned = sorted(cleaned, key=lambda tup: tup[1])
avg_y1 = cleaned[0][1]
avg_y2 = cleaned[-1][1]
avg_x1 = 0
avg_x2 = 0
for tup in cleaned:
avg_x1 += tup[0]
avg_x2 += tup[2]
avg_x1 = avg_x1/len(cleaned)
avg_x2 = avg_x2/len(cleaned)
rects[i] = (avg_x1, avg_y1, avg_x2, avg_y2)
i += 1
print("Num Parking Lanes: ", len(rects))
#Step 5: 把列矩形画出来
buff = 7
for key in rects:
tup_topLeft = (int(rects[key][0] - buff), int(rects[key][1]))
tup_botRight = (int(rects[key][2] + buff), int(rects[key][3]))
cv2.rectangle(new_image, tup_topLeft,tup_botRight,(0,255,0),3)
return new_image, rects
def draw_parking(self,image, rects, make_copy = True, color=[255, 0, 0], thickness=2, save = True):
if make_copy:
new_image = np.copy(image)
gap = 15.5
spot_dict = {} # 字典：一个车位对应一个位置
tot_spots = 0
#微调
adj_y1 = {0: 20, 1:-10, 2:0, 3:-11, 4:28, 5:5, 6:-15, 7:-15, 8:-10, 9:-30, 10:9, 11:-32}
adj_y2 = {0: 30, 1: 50, 2:15, 3:10, 4:-15, 5:15, 6:15, 7:-20, 8:15, 9:15, 10:0, 11:30}
adj_x1 = {0: -8, 1:-15, 2:-15, 3:-15, 4:-15, 5:-15, 6:-15, 7:-15, 8:-10, 9:-10, 10:-10, 11:0}
adj_x2 = {0: 0, 1: 15, 2:15, 3:15, 4:15, 5:15, 6:15, 7:15, 8:10, 9:10, 10:10, 11:0}
for key in rects:
tup = rects[key]
x1 = int(tup[0]+ adj_x1[key])
x2 = int(tup[2]+ adj_x2[key])
y1 = int(tup[1] + adj_y1[key])
y2 = int(tup[3] + adj_y2[key])
cv2.rectangle(new_image, (x1, y1),(x2,y2),(0,255,0),2)
num_splits = int(abs(y2-y1)//gap)
for i in range(0, num_splits+1):
y = int(y1 + i*gap)
cv2.line(new_image, (x1, y), (x2, y), color, thickness)
if key > 0 and key < len(rects) -1 :
#竖直线
x = int((x1 + x2)/2)
cv2.line(new_image, (x, y1), (x, y2), color, thickness)
# 计算数量
if key == 0 or key == (len(rects) -1):
tot_spots += num_splits +1
else:
tot_spots += 2*(num_splits +1)
# 字典对应好
if key == 0 or key == (len(rects) -1):
for i in range(0, num_splits+1):
cur_len = len(spot_dict)
y = int(y1 + i*gap)
spot_dict[(x1, y, x2, y+gap)] = cur_len +1
else:
for i in range(0, num_splits+1):
cur_len = len(spot_dict)
y = int(y1 + i*gap)
x = int((x1 + x2)/2)
spot_dict[(x1, y, x, y+gap)] = cur_len +1
spot_dict[(x, y, x2, y+gap)] = cur_len +2
print("total parking spaces: ", tot_spots, cur_len)
if save:
filename = 'with_parking.jpg'
cv2.imwrite(filename, new_image)
return new_image, spot_dict
又懒了，大家自己看，直接上结果
每个停车位的位置已经被获得，接下来，只要将图片抠出来即可。到这图像处理工作算是完成了，接下来就是训练模型的事了，有空在写。
本文地址：https://blog.csdn.net/qq_38669789/article/details/111110281
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