Python 3.x 将HSV遮罩拆分为多个矩形
我从图像中创建了一个HSV遮罩。结果如下: 我的目标是绘制适合遮罩高度或宽度的多重矩形,如下所示: 我遇到了2道题Python 3.x 将HSV遮罩拆分为多个矩形,python-3.x,opencv,image-processing,Python 3.x,Opencv,Image Processing,我从图像中创建了一个HSV遮罩。结果如下: 我的目标是绘制适合遮罩高度或宽度的多重矩形,如下所示: 我遇到了2道题 我不知道如何在创建矩形的掩码中定位起点和终点。若我使用for循环逐行扫描遮罩,它可能会将遮罩分成两部分。 有时,在一张图像中还包含两个不同的遮罩。如何绘制矩形? 任何人都可以给我一些建议吗?您可以使用cv2.findContour()搜索您的最大轮廓(十字形状)。它返回轮廓的坐标数组。然后,您可以搜索具有最高X坐标(即最右点)、最低X坐标(最左点)、最高Y坐标(即最底点)和
任何人都可以给我一些建议吗?您可以使用
cv2.findContour()
搜索您的最大轮廓(十字形状)。它返回轮廓的坐标数组。然后,您可以搜索具有最高X坐标(即最右点)、最低X坐标(最左点)、最高Y坐标(即最底点)和最低Y坐标(即最高点)的轮廓点。在获得所有4个值后,可以再次搜索轮廓中具有该值的所有点,并将它们附加到4个不同的列表中,稍后对它们进行排序,以便将这些点作为淹没在底图上:
从这一点开始,我将列表转换为numpy数组,因为这对我来说更容易。你可以用任何其他方法
然后,问题就在于你想要多少个矩形,以及你想如何显示它们。在我的示例代码中,您必须输入需要多少相同大小的矩形,最后一个是剩余矩形的大小。我首先在Y坐标(绿色)上显示矩形,然后在X坐标上显示矩形,X坐标分为两段(左侧和右侧),因为它们的距离略有不同,我不想在Y坐标矩形上绘制,因为它们没有在示例图像上绘制。您可以根据需要更改写入矩形的逻辑。希望它能对你有所帮助,或者给你一个如何处理的想法。干杯
示例代码:
import cv2
import numpy as np
# Read image and search for contours.
img = cv2.imread('cross.png')
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
_, threshold = cv2.threshold(gray, 100, 255, cv2.THRESH_BINARY)
_, contours, hierarchy = cv2.findContours(threshold,cv2.RETR_TREE,cv2.CHAIN_APPROX_NONE)
# Select the biggest contour (if you wish to segmentize only the cross-like contour).
cnt = max(contours, key=cv2.contourArea)
# Create empty lists for appending key points.
top_vertical = []
bottom_vertical = []
left_horizontal = []
right_horizontal = []
# Setting the starter values for N, S, E, W.
top = 10000
bottom = 0
left = 10000
right = 0
# Loop to get highest key values of N, S, E, W.
for i in cnt:
y = int(i[:,1])
x = int(i[:, 0])
if x < left:
left = int(x)
if x > right:
right = int(x)
if y < top:
top = int(y)
if y > bottom:
bottom = int(y)
# Loop for appending all points containing key values of N, S, E, W.
for i in cnt:
if int(i[:,1]) == top:
up = (int(i[:,0]), int(i[:,1]))
top_vertical.append(up)
if int(i[:,1]) == bottom:
down = (int(i[:,0]), int(i[:,1]))
bottom_vertical.append(down)
if int(i[:,0]) == left:
l = (int(i[:,0]), int(i[:,1]))
left_horizontal.append(l)
if int(i[:,0]) == right:
r = (int(i[:,0]), int(i[:,1]))
right_horizontal.append(r)
# Sorting the lists.
top_vertical.sort(key=lambda tup: tup[0])
bottom_vertical.sort(key=lambda tup: tup[0])
left_horizontal.sort(key=lambda tup: tup[1])
right_horizontal.sort(key=lambda tup: tup[1])
# Optional drawing of key points.
'''cv2.circle(img,(top_vertical[0]), 4, (0,0,255), -1)
cv2.circle(img,(top_vertical[-1]), 4, (0,0,255), -1)
cv2.circle(img,(bottom_vertical[0]), 4, (0,0,255), -1)
cv2.circle(img,(bottom_vertical[-1]), 4, (0,0,255), -1)
cv2.circle(img,(left_horizontal[0]), 4, (0,0,255), -1)
cv2.circle(img,(left_horizontal[-1]), 4, (0,0,255), -1)
cv2.circle(img,(right_horizontal[0]), 4, (0,0,255), -1)
cv2.circle(img,(right_horizontal[-1]), 4, (0,0,255), -1)'''
# Transforming lists to arrays.
top_vertical = np.array(top_vertical)
bottom_vertical = np.array(bottom_vertical)
left_horizontal = np.array(left_horizontal)
right_horizontal = np.array(right_horizontal)
# Calculating height and weight of the contour.
distance_y = bottom - top
distance_x = right - left
# Inputs for the number of same size segments.
a = input('Input the number of same size segments in Y coordinate: ')
b = input('Input the number of same size segments in left X coordinate: ')
c = input('Input the number of same size segments in right X coordinate: ')
# Calculation of area per segment and limit for the lenght of combined segments (height and weight) .
segment_y = distance_y/int(a)
segment_x_reference = int(top_vertical[0,0]) - int(left_horizontal[0,0])
segment_x = segment_x_reference/int(b)
segment_x_right_reference = int(right_horizontal[0,0]) - int(top_vertical[-1,0])
segment_x_right = segment_x_right_reference/int(c)
# Drawing rectangles on the Y axis.
for i in range(1,20):
sq = int(segment_y)*i
if sq < distance_y:
cv2.rectangle(img,(top_vertical[0,0], top_vertical[0,1]),((top_vertical[-1,0]),top_vertical[0,1] + sq),(0,255,0),1)
else:
sq = distance_y
cv2.rectangle(img,(top_vertical[0,0], top_vertical[0,1]),((top_vertical[-1,0]),sq),(0,255,0),1)
break
# Drawing rectangles on the left side of X axis.
for i in range(1,20):
sq = int(segment_x)*i
if sq < segment_x_reference:
cv2.rectangle(img,(left_horizontal[0,0], left_horizontal[0,1]),((left_horizontal[0,0])+sq, left_horizontal[-1,1]),(255,0,0),1)
else:
sq = segment_x_reference
cv2.rectangle(img,(left_horizontal[0,0], left_horizontal[0,1]),((left_horizontal[0,0])+sq, left_horizontal[-1,1]),(255,0,0),1)
break
# Drawing rectangles on the right side of X axis.
for i in range(1,20):
sq = int(segment_x_right)*i
if sq < segment_x_right_reference:
cv2.rectangle(img,(right_horizontal[0,0], right_horizontal[0,1]),((right_horizontal[0,0])-sq, right_horizontal[-1,1]),(255,0,0),1)
else:
sq = segment_x_right_reference
cv2.rectangle(img,(right_horizontal[0,0], right_horizontal[0,1]),((right_horizontal[0,0])-sq, right_horizontal[-1,1]),(255,0,0),1)
break
# Displaying result.
cv2.imshow('img', img)
您的矩形是否有特定的尺寸需要考虑,或者每个遮罩中是否有最佳数量的矩形需要绘制?激发灵感的帖子+1。这种方法也适用于不规则形状的轮廓吗?谢谢!我认为这是行不通的,因为它假设外墙是直线,并且它有一个十字架形状。如果你改变了极值点的阈值,如果它有多个交叉点(如t形)或没有交叉点(如普通正方形)等,它可能适用于不规则形状。但这可能需要大量工作:)很好的解决方案。但是,如果交叉形状围绕图像中的某个噪声点,则可能会影响4个值(上、右、左、下)的提取。你有什么办法解决这个问题吗?嗯……如果只有一两个像素,你可以将“if int(i[:,1])==top”改为“if int(i[:,1])>=top-2”等等,或者你可以制作附加列表来附加top、left、right、bottom的值,并选择具有最高值且在列表中出现一次或两次以上的值。
import cv2
import numpy as np
# Read image and search for contours.
img = cv2.imread('cross.png')
gray = cv2.cvtColor(img, cv2.COLOR_BGR2GRAY)
_, threshold = cv2.threshold(gray, 100, 255, cv2.THRESH_BINARY)
_, contours, hierarchy = cv2.findContours(threshold,cv2.RETR_TREE,cv2.CHAIN_APPROX_NONE)
# Select the biggest contour (if you wish to segmentize only the cross-like contour).
cnt = max(contours, key=cv2.contourArea)
# Create empty lists for appending key points.
top_vertical = []
bottom_vertical = []
left_horizontal = []
right_horizontal = []
# Setting the starter values for N, S, E, W.
top = 10000
bottom = 0
left = 10000
right = 0
# Loop to get highest key values of N, S, E, W.
for i in cnt:
y = int(i[:,1])
x = int(i[:, 0])
if x < left:
left = int(x)
if x > right:
right = int(x)
if y < top:
top = int(y)
if y > bottom:
bottom = int(y)
# Loop for appending all points containing key values of N, S, E, W.
for i in cnt:
if int(i[:,1]) == top:
up = (int(i[:,0]), int(i[:,1]))
top_vertical.append(up)
if int(i[:,1]) == bottom:
down = (int(i[:,0]), int(i[:,1]))
bottom_vertical.append(down)
if int(i[:,0]) == left:
l = (int(i[:,0]), int(i[:,1]))
left_horizontal.append(l)
if int(i[:,0]) == right:
r = (int(i[:,0]), int(i[:,1]))
right_horizontal.append(r)
# Sorting the lists.
top_vertical.sort(key=lambda tup: tup[0])
bottom_vertical.sort(key=lambda tup: tup[0])
left_horizontal.sort(key=lambda tup: tup[1])
right_horizontal.sort(key=lambda tup: tup[1])
# Optional drawing of key points.
'''cv2.circle(img,(top_vertical[0]), 4, (0,0,255), -1)
cv2.circle(img,(top_vertical[-1]), 4, (0,0,255), -1)
cv2.circle(img,(bottom_vertical[0]), 4, (0,0,255), -1)
cv2.circle(img,(bottom_vertical[-1]), 4, (0,0,255), -1)
cv2.circle(img,(left_horizontal[0]), 4, (0,0,255), -1)
cv2.circle(img,(left_horizontal[-1]), 4, (0,0,255), -1)
cv2.circle(img,(right_horizontal[0]), 4, (0,0,255), -1)
cv2.circle(img,(right_horizontal[-1]), 4, (0,0,255), -1)'''
# Transforming lists to arrays.
top_vertical = np.array(top_vertical)
bottom_vertical = np.array(bottom_vertical)
left_horizontal = np.array(left_horizontal)
right_horizontal = np.array(right_horizontal)
# Calculating height and weight of the contour.
distance_y = bottom - top
distance_x = right - left
# Inputs for the number of same size segments.
a = input('Input the number of same size segments in Y coordinate: ')
b = input('Input the number of same size segments in left X coordinate: ')
c = input('Input the number of same size segments in right X coordinate: ')
# Calculation of area per segment and limit for the lenght of combined segments (height and weight) .
segment_y = distance_y/int(a)
segment_x_reference = int(top_vertical[0,0]) - int(left_horizontal[0,0])
segment_x = segment_x_reference/int(b)
segment_x_right_reference = int(right_horizontal[0,0]) - int(top_vertical[-1,0])
segment_x_right = segment_x_right_reference/int(c)
# Drawing rectangles on the Y axis.
for i in range(1,20):
sq = int(segment_y)*i
if sq < distance_y:
cv2.rectangle(img,(top_vertical[0,0], top_vertical[0,1]),((top_vertical[-1,0]),top_vertical[0,1] + sq),(0,255,0),1)
else:
sq = distance_y
cv2.rectangle(img,(top_vertical[0,0], top_vertical[0,1]),((top_vertical[-1,0]),sq),(0,255,0),1)
break
# Drawing rectangles on the left side of X axis.
for i in range(1,20):
sq = int(segment_x)*i
if sq < segment_x_reference:
cv2.rectangle(img,(left_horizontal[0,0], left_horizontal[0,1]),((left_horizontal[0,0])+sq, left_horizontal[-1,1]),(255,0,0),1)
else:
sq = segment_x_reference
cv2.rectangle(img,(left_horizontal[0,0], left_horizontal[0,1]),((left_horizontal[0,0])+sq, left_horizontal[-1,1]),(255,0,0),1)
break
# Drawing rectangles on the right side of X axis.
for i in range(1,20):
sq = int(segment_x_right)*i
if sq < segment_x_right_reference:
cv2.rectangle(img,(right_horizontal[0,0], right_horizontal[0,1]),((right_horizontal[0,0])-sq, right_horizontal[-1,1]),(255,0,0),1)
else:
sq = segment_x_right_reference
cv2.rectangle(img,(right_horizontal[0,0], right_horizontal[0,1]),((right_horizontal[0,0])-sq, right_horizontal[-1,1]),(255,0,0),1)
break
# Displaying result.
cv2.imshow('img', img)
Input the number of same size segments in Y coordinate: 5
Input the number of same size segments in left X coordinate: 2
Input the number of same size segments in right X coordinate: 2