拼图
作者:互联网
import cv2
import numpy as np
def stackImages(scale,imgArray):
rows = len(imgArray)
cols = len(imgArray[0])
# & 输出一个 rows * cols 的矩阵(imgArray)
print(rows,cols)
# & 判断imgArray[0] 是不是一个list
rowsAvailable = isinstance(imgArray[0], list)
# & imgArray[][] 是什么意思呢?
# & imgArray[0][0]就是指[0,0]的那个图片(我们把图片集分为二维矩阵,第一行、第一列的那个就是第一个图片)
# & 而shape[1]就是width,shape[0]是height,shape[2]是
width = imgArray[0][0].shape[1]
height = imgArray[0][0].shape[0]
# & 例如,我们可以展示一下是什么含义
# cv2.imshow("img", imgArray[0][1])
if rowsAvailable:
for x in range (0, rows):
for y in range(0, cols):
# & 判断图像与后面那个图像的形状是否一致,若一致则进行等比例放缩;否则,先resize为一致,后进行放缩
if imgArray[x][y].shape[:2] == imgArray[0][0].shape [:2]:
imgArray[x][y] = cv2.resize(imgArray[x][y], (0, 0), None, scale, scale)
else:
imgArray[x][y] = cv2.resize(imgArray[x][y], (imgArray[0][0].shape[1], imgArray[0][0].shape[0]), None, scale, scale)
# & 如果是灰度图,则变成RGB图像(为了弄成一样的图像)
if len(imgArray[x][y].shape) == 2: imgArray[x][y]= cv2.cvtColor( imgArray[x][y], cv2.COLOR_GRAY2BGR)
# & 设置零矩阵
imageBlank = np.zeros((height, width, 3), np.uint8)
hor = [imageBlank]*rows
hor_con = [imageBlank]*rows
for x in range(0, rows):
hor[x] = np.hstack(imgArray[x])
ver = np.vstack(hor)
# & 如果不是一组照片,则仅仅进行放缩 or 灰度转化为RGB
else:
for x in range(0, rows):
if imgArray[x].shape[:2] == imgArray[0].shape[:2]:
imgArray[x] = cv2.resize(imgArray[x], (0, 0), None, scale, scale)
else:
imgArray[x] = cv2.resize(imgArray[x], (imgArray[0].shape[1], imgArray[0].shape[0]), None,scale, scale)
if len(imgArray[x].shape) == 2: imgArray[x] = cv2.cvtColor(imgArray[x], cv2.COLOR_GRAY2BGR)
hor= np.hstack(imgArray)
ver = hor
return ver
img = cv2.imread(‘default.png’)
grayimg = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
blurimg = cv2.GaussianBlur(grayimg,(3,3),0)
canny = cv2.Canny(blurimg,50,100)
stackimg = stackImages(0.6,([img,grayimg,blurimg],
[canny,grayimg,blurimg]))
contours,hierarchy = cv2.findContours(img,cv2.RETR_EXTERNAL,cv2.CHAIN_APPROX_NONE)
cv2.imshow(‘stack’,stackimg)
cv2.waitKey(0)
标签:scale,拼图,cv2,shape,rows,hor,imgArray 来源: https://blog.csdn.net/tel_1392/article/details/113785339