为什么80%的码农都做不了架构师？>>> 
图像处理基础，等大小图片相似度匹配
基于RGB通道的像素点相似度匹配算法思路
def compare_by_RGB(image_1,image_2):
"""
基于通道和的差
:param image_1:
:param image_2:
:return:
"""
G_1 = 0
B_1 = 0
R_1 = 0
G_2 = 0
B_2 = 0
R_2 = 0
#第一个图像矩阵通道和
for x in image_1:
for y in x:
G_1 += y[0]
B_1 += y[1]
R_1 += y[2]
#第二个图像矩阵通道和
for x in image_2:
for y in x:
G_2 += y[0]
B_2 += y[1]
R_2 += y[2]
#图像矩阵各通道相似度
inc_G = 1 - math.fabs(G_1 - G_2) / G_2
inc_B = 1 - math.fabs(B_1 - B_2) / B_2
inc_R = 1 - math.fabs(R_1 - R_2) / R_2
dec = (inc_G + inc_B + inc_R) / 3
return dec
根据不相似的像素点统计
def compare_by_pixe(pic1, pic2):
"""
方法二，基于像素点相似数量统计
:param pic1:
:param pic2:
:return:totlepix 像素点个数,diffcount 相似度 ,nptg,ptg
"""
res={}
#获取行数列数和通道数
sp1 = pic1.shape
sp2 = pic2.shape
#获取矩阵大小
res['totlepix'] = pic1.size # pix*3 矩阵大小等于像素点数乘以3
#判断两张图片的行列数是否相同
if sp1 != sp2:
ret = "The two picture is in Different range"
return ret
DiffCount = 0
#遍历矩阵行和列
for index1 in range(sp1[0]):
for index2 in range(sp1[1]):
(b1, g1, r1) = pic1[index1, index2]
(b2, g2, r2) = pic2[index1, index2]
if (b1, g1, r1) != (b2, g2, r2):
DiffCount = DiffCount + 1
res['totlepix'] = res['totlepix'] / 3
#相似度=不同的像素点数/像素点总数
res['diffcount'] =1- DiffCount/ res['totlepix']
return res
主函数测试
if __name__=="__main__":
#相似度阈值
threshold_value=0.75
path="./lab1_data/lab1_video.mp4"
cap=cv2.VideoCapture(path)
count=0
if cap.isOpened():
#锁，用来判断是否为第一帧
flag=False
while 1:
ret,frame=cap.read()
#缩小图片
image=cv2.resize(frame,(32,32),interpolation=cv2.INTER_CUBIC)
if flag==True:
res=compare_by_pixe(image, temp)
if res["diffcount"]