OpenCV python:如何使用渐变和第一个点画一条线?
OpenCV python: How do I draw a line using the gradient and the first point?
我正在尝试使用带有 opencv 的实时提要画一条线。我正在使用一帧并存储 x,y 坐标。我使用下一帧的点的 x、y 坐标来计算梯度 ((y2-y1)/(x2-x1))。我想从第一个坐标直接通过第二个坐标画一条直线,然后继续过去,这会画出一条轨迹。我现在可以使用 cv2.line()
在两点之间画一条直线。我的代码如下。任何建议都会很棒!谢谢
import numpy as np
import math
import matplotlib.pyplot as plt
lower_red = np.array([-10,160,160])
upper_red = np.array([10,255,255])
oX, oY = 0,0
cap = cv2.VideoCapture(0)
if not cap.isOpened():
print("Cannot open camera")
exit()
while(1):
ret, frame = cap.read()
if not ret:
print("Can't receive frame (stream end?). Exiting ...")
break
hsv = cv2.cvtColor(frame, cv2.COLOR_BGR2HSV)
mask = cv2.inRange(hsv, lower_red, upper_red)
#ret, thresh = cv2.threshold(mask, 80, 255, cv2.THRESH_BINARY)
contours, _ = cv2.findContours(mask, cv2.RETR_TREE, cv2.CHAIN_APPROX_NONE)
if len(contours) != 0:
c = max(contours, key = cv2.contourArea)
x1, y1, w, h = cv2.boundingRect(c)
x2, y2 = x1 + w, y1 + h
cv2.rectangle(frame, (x1, y1), (x2, y2), (0, 255, 0), 2)
x3, y3 = round((x1+x2)/2), round((y1+y2)/2)
cv2.circle(frame, (x3,y3), 4, (255,0,0), 2)
#print(x3, y3)
if oX and oY != 0:
try:
angle = (x3-oX)/(y3-oY)
cv2.line(frame,(oX,oY),(x3, y3),(0,255,255),2)
except ZeroDivisionError:
oX, oY = x3, y3
oX, oY = x3, y3
cv2.imshow('frame', frame)
cv2.imshow('mask', mask)
if cv2.waitKey(1) == ord('q'):
break
# When everything done, release the capture
cap.release()
cv2.destroyAllWindows()
从这个 answer 中找到了一个解决方案,它在 c 中,已转换为 python 并针对您的用例进行了修改。
给定两个点在图像中绘制无限线的解决方案。
def slope(x1,y1,x2,y2):
###finding slope
if x2!=x1:
return((y2-y1)/(x2-x1))
else:
return 'NA'
def drawLine(image,x1,y1,x2,y2):
m=slope(x1,y1,x2,y2)
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)
您可以根据您的用例将 (px,py)
替换为 (x1,y1)
或将 (qx,qy)
替换为 (x2,y2)
。
我正在尝试使用带有 opencv 的实时提要画一条线。我正在使用一帧并存储 x,y 坐标。我使用下一帧的点的 x、y 坐标来计算梯度 ((y2-y1)/(x2-x1))。我想从第一个坐标直接通过第二个坐标画一条直线,然后继续过去,这会画出一条轨迹。我现在可以使用 cv2.line()
在两点之间画一条直线。我的代码如下。任何建议都会很棒!谢谢
import numpy as np
import math
import matplotlib.pyplot as plt
lower_red = np.array([-10,160,160])
upper_red = np.array([10,255,255])
oX, oY = 0,0
cap = cv2.VideoCapture(0)
if not cap.isOpened():
print("Cannot open camera")
exit()
while(1):
ret, frame = cap.read()
if not ret:
print("Can't receive frame (stream end?). Exiting ...")
break
hsv = cv2.cvtColor(frame, cv2.COLOR_BGR2HSV)
mask = cv2.inRange(hsv, lower_red, upper_red)
#ret, thresh = cv2.threshold(mask, 80, 255, cv2.THRESH_BINARY)
contours, _ = cv2.findContours(mask, cv2.RETR_TREE, cv2.CHAIN_APPROX_NONE)
if len(contours) != 0:
c = max(contours, key = cv2.contourArea)
x1, y1, w, h = cv2.boundingRect(c)
x2, y2 = x1 + w, y1 + h
cv2.rectangle(frame, (x1, y1), (x2, y2), (0, 255, 0), 2)
x3, y3 = round((x1+x2)/2), round((y1+y2)/2)
cv2.circle(frame, (x3,y3), 4, (255,0,0), 2)
#print(x3, y3)
if oX and oY != 0:
try:
angle = (x3-oX)/(y3-oY)
cv2.line(frame,(oX,oY),(x3, y3),(0,255,255),2)
except ZeroDivisionError:
oX, oY = x3, y3
oX, oY = x3, y3
cv2.imshow('frame', frame)
cv2.imshow('mask', mask)
if cv2.waitKey(1) == ord('q'):
break
# When everything done, release the capture
cap.release()
cv2.destroyAllWindows()
从这个 answer 中找到了一个解决方案,它在 c 中,已转换为 python 并针对您的用例进行了修改。
给定两个点在图像中绘制无限线的解决方案。
def slope(x1,y1,x2,y2):
###finding slope
if x2!=x1:
return((y2-y1)/(x2-x1))
else:
return 'NA'
def drawLine(image,x1,y1,x2,y2):
m=slope(x1,y1,x2,y2)
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)
您可以根据您的用例将 (px,py)
替换为 (x1,y1)
或将 (qx,qy)
替换为 (x2,y2)
。