我这里有一行代码,它使用了opencv的python绑定:
cv2.rectangle(img, (box[1], box[0]), (box[3], box[2]), (255,0,0), 4)
这会在厚度为img
的图像4
上绘制一个红色矩形。
但有没有办法可以将矩形线条风格化?不是太多。只是点缀,或虚线,这是真的。
答案 0 :(得分:10)
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()
答案 1 :(得分:3)
OpenCV(目前)不支持超出厚度和抗锯齿的线属性。
答案 2 :(得分:3)
答案 3 :(得分:0)
尝试这样的事情:
cv::Point P1, P2;
P1.y = 50;
P2.y = 50;
int dot_gap = 50;
int dot_width = 50;
for( int i=0 ; i < in_img.cols; (i = i+d_width) ) {
P1.x = i;
P2.x = i + dot_width;
cv::line(in_img, P1, P2, cv::Scalar(0, 255, 255), 2, cv::LINE_8);
}