我正在阅读有关 opencv 和 python 模板匹配的文档,在关于与多个对象进行模板匹配的最后一部分中,代码检测到 mario 图像上的 19 个硬币,但是,是否可以计算使用某些函数检测到的对象数量在 python 上,比如 len() 或任何 opencv 方法?
这是教程中显示的代码: http://docs.opencv.org/3.1.0/d4/dc6/tutorial_py_template_matching.html
模板匹配代码:
import cv2
import numpy as np
from matplotlib import pyplot as plt
img_rgb = cv2.imread('mario.png')
img_gray = cv2.cvtColor(img_rgb, cv2.COLOR_BGR2GRAY)
template = cv2.imread('mario_coin.png',0)
w, h = template.shape[::-1]
res = cv2.matchTemplate(img_gray,template,cv2.TM_CCOEFF_NORMED)
threshold = 0.8
loc = np.where( res >= threshold)
for pt in zip(*loc[::-1]):
cv2.rectangle(img_rgb, pt, (pt[0] + w, pt[1] + h), (0,0,255), 2)
cv2.imwrite('res.png',img_rgb)
那么,有没有什么方法可以统计图像上检测到的硬币并在终端上打印数字呢? 比如:
The Template Matching code showed before...
print "Function that detect number of coins with template matching"
>>> 19
我找到了一个合适的解决方案(适合我的应用程序),按照乌尔里希的建议计算唯一匹配项。这并不理想,但对于我的应用程序来说,使用“灵敏度”通常会产生 +/- 2% 以内的结果。
import cv2
import numpy as np
from matplotlib import pyplot as plt
img_rgb = cv2.imread('mario.png')
img_gray = cv2.cvtColor(img_rgb, cv2.COLOR_BGR2GRAY)
template = cv2.imread('mario_coin.png',0)
w, h = template.shape[::-1]
res = cv2.matchTemplate(img_gray,template,cv2.TM_CCOEFF_NORMED)
threshold = 0.8
loc = np.where( res >= threshold)
f = set()
for pt in zip(*loc[::-1]):
cv2.rectangle(img_rgb, pt, (pt[0] + w, pt[1] + h), (0,0,255), 2)
sensitivity = 100
f.add((round(pt[0]/sensitivity), round(pt[1]/sensitivity)))
cv2.imwrite('res.png',img_rgb)
found_count = len(f)
我列出了所有匹配项,对于每个新匹配项,我检查列表中的任何匹配项是否与边界框有交集:
res = cv.matchTemplate(image,template,cv.TM_CCOEFF_NORMED)
threshold = 0.5
loc = np.where(res >= threshold)
matches = []
for pt in zip(*loc[::-1]):
intersection = 0
for match in matches:
if intersected(match, (match[0] + w, match[1] + h), pt, (pt[0] + w, pt[1] + h)):
intersection = 1
break
if intersection == 0:
matches.append(pt)
rect = cv.rectangle(image, pt, (pt[0] + w, pt[1] + h), (0, 0, 255), 2)
这是检查交叉点的代码:
def intersected(bottom_left1, top_right1, bottom_left2, top_right2):
if top_right1[0] < bottom_left2[0] or bottom_left1[0] > top_right2[0]:
return 0
if top_right1[1] < bottom_left2[1] or bottom_left1[1] > top_right2[1]:
return 0
return 1
我使用一个列表来存储许多相同对象检测的第一个 (x,y)。然后,对于找到的检测中的每个 (x,y)(同一对象上必须有许多检测),我计算新的 (x,y) 与列表中每个点之间的距离。如果距离足够大,那么一定是新的检测最先发现的。然后我将新的 (x,y) 放入列表中。虽然很蠢,但确实有效。
目的是删除第一次检测到物体的 (x,y) 附近的点并仅保留该“组”中的一个点,然后迭代 loc 中的所有点以定位更多“组”并找到一个和每组中唯一的一分。
import cv2
import numpy as np
import matplotlib.pyplot as plt
import math
def notInList(newObject):
for detectedObject in detectedObjects:
if math.hypot(newObject[0]-detectedObject[0],newObject[1]-detectedObject[1]) < thresholdDist:
return False
return True
img_rgb = cv2.imread("7.jpg")
img_gray = cv2.cvtColor(img_rgb, cv2.COLOR_BGR2GRAY)
template = cv2.imread("face.jpg",0)
w, h = template.shape[::-1]
res = cv2.matchTemplate(img_gray,template,cv2.TM_CCOEFF_NORMED)
threshold = 0.85
loc = np.where( res >= threshold)
detectedObjects=[]
thresholdDist=30
for pt in zip(*loc[::-1]):
if len(detectedObjects) == 0 or notInList(pt):
detectedObjects.append(pt)
cellImage=img_rgb[pt[1]:pt[1]+h, pt[0]:pt[0]+w]
cv2.imwrite("results/"+str(pt[1])+"_"+str(pt[0])+".jpg",cellImage,
[int(cv2.IMWRITE_JPEG_QUALITY), 50])
对于仍然想知道的人:首先对列表“zip(*loc[::-1})”进行排序会更容易。
例如,我的脚本返回了这样的结果:
(580, 822)
(871, 822)
(1017, 822)
(434, 823)
(726, 823)
(871, 823)
(1017, 823)
7
您会注意到有多个重复项,但不仅仅是按顺序重复。只需简单地使用“sorted(zip(*loc[::-1]))”进行排序,只需计算相邻的 2 个点并检查每个循环的距离,即可轻松获取距离。
通过在循环中添加条件并检查当前点的距离是否小于所需的距离,可以很好地完成工作。我从来没有正确地学习过Python,所以我不确定这是否有效。至少这对我的用例有用。您可以在下面查看。
示例代码:
img = np.array( *YOUR_SCREENSHOT_HERE* )
img_gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
template = cv2.imread( *TARGET_IMAGE_HERE* , 0)
w, h = template.shape[::-1]
res = cv2.matchTemplate(img_gray, template, cv2.TM_CCOEFF_NORMED)
loc = np.where(res >= precision)
count = 0
last_pt = [0, 0] # Set this negative if target image is at top-left corner.
for pt in sorted(zip(*loc[::-1])):
if sqrt(abs(last_pt[0]-pt[0])**2 + abs(last_pt[0]-pt[0])**2) < threshold*min([h, w]):
continue
else:
last_pt = pt
print(pt)
count = count + 1
cv2.rectangle(img, pt, (pt[0] + w, pt[1] + h), (0, 0, 255), 2)
cv2.imwrite('res.png', img)
return count
我也遇到过类似的问题。 这是我的解决方案:
import cv2
import numpy as np
import os
from os import listdir
from os.path import isfile, join
from matplotlib import pyplot as plt
from imutils.object_detection import non_max_suppression
def count_similar_objects(main_image_path, template_image_path, times_of_run):
# Load the main image and the template image
main_image = cv2.imread(main_image_path)
template_image = cv2.imread(template_image_path)
W, H = template_image.shape[:2]
# Convert the images to grayscale
main_gray = cv2.cvtColor(main_image, cv2.COLOR_BGR2GRAY)
template_gray = cv2.cvtColor(template_image, cv2.COLOR_BGR2GRAY)
# Perform template matching
result = cv2.matchTemplate(main_gray, template_gray, cv2.TM_CCOEFF_NORMED)
threshold = 0.6
count = 0
(y_points, x_points) = np.where(result >= threshold)
# Draw rectangles around the matched objects with a thicker outline
thickness = 3 # Adjust the thickness as desired
# initialize our list of rectangles
boxes = list()
# loop over the starting (x, y)-coordinates again
for (x, y) in zip(x_points, y_points):
# update our list of rectangles
boxes.append((x, y, x + W, y + H))
# apply non-maxima suppression to the rectangles
# this will create a single bounding box
boxes = non_max_suppression(np.array(boxes))
# loop over the final bounding boxes
for (x1, y1, x2, y2) in boxes:
# draw the bounding box on the image
cv2.rectangle(main_image, (x1, y1), (x2, y2),
(0, 255, 0), 3)
count += 1
cv2.imwrite('result'+str(times_of_run)+'.png',main_image)
return count
我不明白它是如何工作的。我刚刚从GeeksforGeeks复制了这个。所以不要指望我能够回答你的问题。我希望这有帮助。
我就是这么做的:
loc = np.where( res >= threshhold)
print(len(loc[0])) #match occurences count
loc 是二维数组。
导入时间 导入CV2 将 numpy 导入为 np 从 PIL 导入 ImageGrab
而真实:
count = 0
stop = 0
img = ImageGrab.grab()
img_np = np.array(img)
gray = cv2.cvtColor(img_np, cv2.COLOR_BGR2GRAY)
frame = cv2.cvtColor(img_np, cv2.COLOR_BGR2RGB)
Template = cv2.imread('image.png' ,0)
w, h = Template.shape[::-1]
res = cv2.matchTemplate(gray, Template, cv2.TM_CCOEFF_NORMED)
threshold = 0.90
loc = np.where(res >= threshold)
font = cv2.FONT_HERSHEY_SIMPLEX
for pt in zip(*loc[::-1]):
cv2.rectangle(frame, pt, (pt[0] + w, pt[1] + h), (0,0,255) ,2)
count = count + 1
print(count)
stop = 1
cv2.imshow('frame',frame)
if (stop == 1):
break