[英]How to draw a line on an image in OpenCV?
Just calculate for 2 points outside.只需计算2分外。 opencv's Line is fine with eg (-10,-10) for a point.
opencv 的 Line 很好,例如 (-10,-10) 一个点。
import cv2 # python-opencv
import numpy as np
width, height = 800, 600
x1, y1 = 0, 0
x2, y2 = 200, 400
image = np.ones((height, width)) * 255
line_thickness = 2
cv2.line(image, (x1, y1), (x2, y2), (0, 255, 0), thickness=line_thickness)
http://docs.opencv.org/2.4/modules/core/doc/drawing_functions.html#cv2.line http://docs.opencv.org/2.4/modules/core/doc/drawing_functions.html#cv2.line
Take a look to the following solution, I firstly convert a line in polar equations to cartesian and then I use numpy.vectorize()
to generate a vector that allows me to get represent the line in any point of the space.看看下面的解决方案,我首先将极坐标方程中的一条线转换为笛卡尔,然后我使用
numpy.vectorize()
生成一个向量,该向量允许我在空间的任何点上表示该线。
import cv2
import numpy as np
img_size = (200,200)
img = np.ones(img_size) * 255
# polar equation
theta = np.linspace(0, np.pi, 1000)
r = 1 / (np.sin(theta) - np.cos(theta))
# polar to cartesian
def polar2cart(r, theta):
x = r * np.cos(theta)
y = r * np.sin(theta)
return x, y
x,y = polar2cart(r, theta)
x1, x2, y1, y2 = x[0], x[1], y[0], y[1]
# line equation y = f(X)
def line_eq(X):
m = (y2 - y1) / (x2 - x1)
return m * (X - x1) + y1
line = np.vectorize(line_eq)
x = np.arange(0, img_size[0])
y = line(x).astype(np.uint)
cv2.line(img, (x[0], y[0]), (x[-1], y[-1]), (0,0,0))
cv2.imshow("foo",img)
cv2.waitKey()
Result:结果:
You can see how to do this in the Hough Line Transform tutorial .您可以在霍夫线变换教程中了解如何执行此操作。
import cv2
import numpy as np
img = cv2.imread('dave.jpg')
gray = cv2.cvtColor(img,cv2.COLOR_BGR2GRAY)
edges = cv2.Canny(gray,50,150,apertureSize = 3)
lines = cv2.HoughLines(edges,1,np.pi/180,200)
for rho,theta in lines[0]:
a = np.cos(theta)
b = np.sin(theta)
x0 = a*rho
y0 = b*rho
x1 = int(x0 + 1000*(-b))
y1 = int(y0 + 1000*(a))
x2 = int(x0 - 1000*(-b))
y2 = int(y0 - 1000*(a))
cv2.line(img,(x1,y1),(x2,y2),(0,0,255),2)
cv2.imwrite('houghlines3.jpg',img)
This is one way to solve the problem of drawing infinite line segment in OpenCV with two given points.这是解决在 OpenCV 中用两个给定点绘制无限线段问题的一种方法。
### function to find slope
def slope(p1,p2):
x1,y1=p1
x2,y2=p2
if x2!=x1:
return((y2-y1)/(x2-x1))
else:
return 'NA'
### main function to draw lines between two points
def drawLine(image,p1,p2):
x1,y1=p1
x2,y2=p2
### finding slope
m=slope(p1,p2)
### getting image shape
h,w=image.shape[:2]
if m!='NA':
### here we are essentially extending the line to x=0 and x=width
### and calculating the y associated with it
##starting point
px=0
py=-(x1-0)*m+y1
##ending point
qx=w
qy=-(x2-w)*m+y2
else:
### if slope is zero, draw a line with x=x1 and y=0 and y=height
px,py=x1,0
qx,qy=x1,h
cv2.line(image, (int(px), int(py)), (int(qx), int(qy)), (0, 255, 0), 2)
return image
You can use p1 and p2 according to your requirement and call the function drawLine
.您可以根据需要使用 p1 和 p2 并调用函数
drawLine
。
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