我在矩形(小的)内有很多手写数字的扫描图像。 [![喜欢这个] [1]] [1]
请帮我裁剪包含数字的每张图片,并通过为每行指定相同的名称来保存它们。
修改
import cv2
img = cv2.imread('Data\Scan_20170612_4.jpg')
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
gray = cv2.bilateralFilter(gray, 11, 17, 17)
edged = cv2.Canny(gray, 30, 200)
_, contours, hierarchy = cv2.findContours(edged, cv2.RETR_LIST, cv2.CHAIN_APPROX_SIMPLE)
i = 0
for c in contours:
peri = cv2.arcLength(c, True)
approx = cv2.approxPolyDP(c, 0.09 * peri, True)
if len(approx) == 4:
screenCnt = approx
cv2.drawContours(img, [screenCnt], -1, (0, 255, 0), 3)
cv2.imwrite('cropped\\' + str(i) + '_img.jpg', img)
i += 1
答案 0 :(得分:4)
这是我的版本
import cv2
import numpy as np
fileName = ['9','8','7','6','5','4','3','2','1','0']
img = cv2.imread('Data\Scan_20170612_17.jpg')
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
gray = cv2.bilateralFilter(gray, 11, 17, 17)
kernel = np.ones((5,5),np.uint8)
erosion = cv2.erode(gray,kernel,iterations = 2)
kernel = np.ones((4,4),np.uint8)
dilation = cv2.dilate(erosion,kernel,iterations = 2)
edged = cv2.Canny(dilation, 30, 200)
_, contours, hierarchy = cv2.findContours(edged, cv2.RETR_TREE, cv2.CHAIN_APPROX_SIMPLE)
rects = [cv2.boundingRect(cnt) for cnt in contours]
rects = sorted(rects,key=lambda x:x[1],reverse=True)
i = -1
j = 1
y_old = 5000
x_old = 5000
for rect in rects:
x,y,w,h = rect
area = w * h
if area > 47000 and area < 70000:
if (y_old - y) > 200:
i += 1
y_old = y
if abs(x_old - x) > 300:
x_old = x
x,y,w,h = rect
out = img[y+10:y+h-10,x+10:x+w-10]
cv2.imwrite('cropped\\' + fileName[i] + '_' + str(j) + '.jpg', out)
j+=1
答案 1 :(得分:3)
如果你尝试,这是一件容易的事。这是我的输出 - (图像及其一小部分)
我做了什么?
然后用正确的名字保存那件作品 - 我没有这样做。
也许,有一种更简单的方法,但我喜欢这个。没有把代码放在一边因为 我把它弄得一团糟。如果你仍然需要它将会放置。
以下是每次一次找到轮廓时面具的外观
代码:
import cv2;
import numpy as np;
# Run the code with the image name, keep pressing space bar
# Change the kernel, iterations, Contour Area, position accordingly
# These values work for your present image
img = cv2.imread("your_image.jpg", 0);
h, w = img.shape[:2]
kernel = np.ones((15,15),np.uint8)
e = cv2.erode(img,kernel,iterations = 2)
d = cv2.dilate(e,kernel,iterations = 1)
ret, th = cv2.threshold(d, 150, 255, cv2.THRESH_BINARY_INV)
mask = np.zeros((h+2, w+2), np.uint8)
cv2.floodFill(th, mask, (200,200), 255); # position = (200,200)
out = cv2.bitwise_not(th)
out= cv2.dilate(out,kernel,iterations = 3)
cnt, h = cv2.findContours(out,cv2.RETR_TREE,cv2.CHAIN_APPROX_SIMPLE)
for i in range(len(cnt)):
area = cv2.contourArea(cnt[i])
if(area>10000 and area<100000):
mask = np.zeros_like(img)
cv2.drawContours(mask, cnt, i, 255, -1)
x,y,w,h = cv2.boundingRect(cnt[i])
crop= img[ y:h+y,x:w+x]
cv2.imshow("snip",crop )
if(cv2.waitKey(0))==27:break
cv2.destroyAllWindows()
答案 2 :(得分:0)
_, contours, hierarchy = cv2.findContours(edged, cv2.RETR_LIST, cv2.CHAIN_APPROX_SIMPLE)
您正在使用 cv2.RETR_LIST 查找图片中的轮廓。为了让您的图像获得更好的输出,请使用 cv2.RETR_EXTERNAL 。在使用图像中的第一个删除黑色 边框行之前。
cv2.RETR_LIST为您提供图像所有轮廓的列表
cv2.RETR_EXTERNAL只为您提供外部或外部轮廓,而不是内部轮廓
将行更改为
_, contours, hierarchy = cv2.findContours(edged, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
<强> Contours Hierarchy 强>
答案 3 :(得分:0)
看起来位置定义得很好..它不应该需要opencv..只是将图像分割成瓷砖。 想到了Imagemagik。我经常用它来创建类似的布局..从单个瓷砖..(与你正在做的相反) 它可以在命令行执行此操作。