I have a code that allows me to open an image and click to add points on the picture and it shows me their coordinates as shown below:
The coordinates are already being displayed. First x-y coordinates: (131,133) Second : (28,242) Third: (99,328) Fourth: (111,321) ...
I need to find the linear distance between 2 successive points. That is:
- Distance between Second and First coordinates,
- Distance between Third and Second coordinates,
- Distance between Fourth and Third coordinates, ...
Example: (131,133) & (28,242)
Distance using √[(x₂ - x₁)² (y₂ - y₁)²].
Can someone help, please? Thanks!
Code:
import cv2
[print(i) for i in dir(cv2) if 'EVENT' in i]
# importing the module
import cv2
# function to display the coordinates of
# of the points clicked on the image
def click_event(event, x, y, flags, params):
# checking for left mouse clicks
if event == cv2.EVENT_LBUTTONDOWN:
# displaying the coordinates
# on the Shell
print(x, ' ', y)
# displaying the coordinates
# on the image window
font = cv2.FONT_HERSHEY_SIMPLEX
cv2.putText(img, str(x) ','
str(y), (x, y), font,
1, (255, 0, 0), 2)
cv2.imshow('image', img)
# checking for right mouse clicks
if event == cv2.EVENT_RBUTTONDOWN:
# displaying the coordinates
# on the Shell
print(x, ' ', y)
# displaying the coordinates
# on the image window
font = cv2.FONT_HERSHEY_SIMPLEX
b = img[y, x, 0]
g = img[y, x, 1]
r = img[y, x, 2]
cv2.putText(img, str(b) ','
str(g) ',' str(r),
(x, y), font, 1,
(255, 255, 0), 2)
cv2.imshow('image', img)
# driver function
if __name__ == "__main__":
# reading the image
img = cv2.imread('shirt.jpg', 1)
img = cv2.resize(img, (0, 0), None, 0.2, 0.2)
# displaying the image
cv2.imshow('image', img)
# setting mouse hadler for the image
# and calling the click_event() function
cv2.setMouseCallback('image', click_event)
# wait for a key to be pressed to exit
cv2.waitKey(0)
# close the window
cv2.destroyAllWindows()
CodePudding user response:
You can create a function that accepts 2 points as parameters and returns the distance between them:
def distanceCalculate(p1,p2): # p1 and p2 in format (x1,y1) and (x2,y2) tuples
dis=((p2[0]-p1[0])**2 (p2[1]-p1[1])**2)**0.5
dis=abs(dis) #removing negative sign.
return dis
EDIT: 3rd line 'dis=abs(dis) ' dosen't change anything. Can be ignored