Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
2.7k views
in Technique[技术] by (71.8m points)

python - opencv rectangle with dotted or dashed lines

I have a line of code here that uses the python binding for opencv:

cv2.rectangle(img, (box[1], box[0]), (box[3], box[2]), (255,0,0), 4)

This draws a red rectangle on image img of thickness 4.

But is there a way the lines of the rectangles can be stylized? Not too much. Just dotted, or dashed, that's it really.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)
import cv2
import numpy as np
def drawline(img,pt1,pt2,color,thickness=1,style='dotted',gap=20):
    dist =((pt1[0]-pt2[0])**2+(pt1[1]-pt2[1])**2)**.5
    pts= []
    for i in  np.arange(0,dist,gap):
        r=i/dist
        x=int((pt1[0]*(1-r)+pt2[0]*r)+.5)
        y=int((pt1[1]*(1-r)+pt2[1]*r)+.5)
        p = (x,y)
        pts.append(p)

    if style=='dotted':
        for p in pts:
            cv2.circle(img,p,thickness,color,-1)
    else:
        s=pts[0]
        e=pts[0]
        i=0
        for p in pts:
            s=e
            e=p
            if i%2==1:
                cv2.line(img,s,e,color,thickness)
            i+=1

def drawpoly(img,pts,color,thickness=1,style='dotted',):
    s=pts[0]
    e=pts[0]
    pts.append(pts.pop(0))
    for p in pts:
        s=e
        e=p
        drawline(img,s,e,color,thickness,style)

def drawrect(img,pt1,pt2,color,thickness=1,style='dotted'):
    pts = [pt1,(pt2[0],pt1[1]),pt2,(pt1[0],pt2[1])] 
    drawpoly(img,pts,color,thickness,style)

im = np.zeros((800,800,3),dtype='uint8')
s=(234,222)
e=(500,700)
drawrect(im,s,e,(0,255,255),1,'dotted')

cv2.imshow('im',im)
cv2.waitKey()      

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...