我有多个png文件,我正在尝试获取多边形轮廓坐标。 那是简化的坐标,只有每个外角(不是凸壳多边形)。
目前要执行此操作的程序是python和opencv。 但是另一个程序是可以的,我确实尝试过使用npm软件包,imagemagick,potrace和Lua修复此问题。 在“根据图像构建多边形”过程中,它将用作shell命令。
这是python下的最后一次测试。
现在的问题是,在下面的示例中某些边缘“不正确”。
我执行了以下步骤
ret, mask = cv2.threshold(img[:, :, 3], 0, 255, cv2.THRESH_BINARY)
contours, hierarchy = cv2.findContours(mask, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
问题是两个孔,缺少1个像素和1个像素。
# https://opensource.com/article/19/5/python-3-default-mac#what-to-do
# https://solarianprogrammer.com/2019/10/21/install-opencv-python-macos/
# https://docs.opencv.org/master/d4/d73/tutorial_py_contours_begin.html
# https://stackoverflow.com/questions/25733694/process-image-to-find-external-contour
# https://docs.opencv.org/3.4/dd/d49/tutorial_py_contour_features.html
# https://stackoverflow.com/questions/39823221/imagemagick-find-coordinates-of-outline-of-transparent-png-not-border
import numpy as np
import cv2
img = cv2.imread('../temp/bord.png', cv2.IMREAD_UNCHANGED)
# make black and white
ret, mask = cv2.threshold(img[:, :, 3], 0, 255, cv2.THRESH_BINARY)
# find the external contour
contours, hierarchy = cv2.findContours(mask, cv2.RETR_EXTERNAL, cv2.CHAIN_APPROX_SIMPLE)
# at this point I want to have the correct contours to process them inside a other program
# print(contours)
# start debugging
#save image
cv2.imwrite('../temp/bord_converted.png',mask)
#create an empty image for contours
img_contours = np.zeros(img.shape)
# draw the contours on the empty image
cv2.drawContours(img_contours, contours, -1, (0,255,0), 1)
cv2.imwrite('../temp/bord_contour.jpg',img_contours)
编辑
我尝试过的其他事情:
行进方格程序
Golang
https://github.com/zx9597446/marchingsquare/issues/1 那给了我另一个问题,但是正确的密码
Npm https://github.com/scottglz/image-outline 那给了我几乎与上面相同的问题
imagemagick
正在尝试一些将png转换为黑白的操作,并返回轮廓。
convert "$IMAGE" -matte -bordercolor none -border 1 -alpha extract -edge 1 -threshold 50% -depth 8 txt: | awk -F: '/white/{print $1}'
种族
但是所有输出都有东西,所以我不能使用它。
potrace --progress -b svg --blacklevel 0 --turdsize 0 --longcurve --opttolerance 0 --unit 1 --turnpolicy white --alphamax 0 --scale 1 --group --flat ../temp/bordout.bmp -o ../temp/bordout.svg
potrace --progress -b eps -c --blacklevel 0 --turdsize 0 --longcurve --opttolerance 0 --unit 1 --turnpolicy white --alphamax 0 --scale 1 --flat ../temp/bordout.bmp -o ../temp/bordout.eps
potrace --progress -b pdf -c --blacklevel 0 --turdsize 0 --longcurve --opttolerance 0 --unit 1 --turnpolicy white --alphamax 0 --scale 1 --flat ../temp/bordout.bmp -o ../temp/bordout.pdf
potrace --progress -b pdfpage --blacklevel 0 --turdsize 0 --longcurve --opttolerance 0 --unit 1 --turnpolicy white --alphamax 0 --scale 1 --flat ../temp/bordout.bmp -o ../temp/bordout.pdfpage
potrace --progress -b ps -c --blacklevel 0 --turdsize 0 --longcurve --opttolerance 0 --unit 1 --turnpolicy white --alphamax 0 --scale 1 --flat ../temp/bordout.bmp -o ../temp/bordout.ps
potrace --progress -b pgm --blacklevel 0 --turdsize 0 --longcurve --opttolerance 0 --unit 1 --turnpolicy white --alphamax 0 --scale 1 --flat ../temp/bordout.bmp -o ../temp/bordout.pgm
potrace --progress -b dxf --blacklevel 0 --turdsize 0 --longcurve --opttolerance 0 --unit 1 --turnpolicy white --alphamax 0 --scale 1 --flat ../temp/bordout.bmp -o ../temp/bordout.dxf
potrace --progress -b geojson --blacklevel 0 --turdsize 0 --longcurve --opttolerance 0 --unit 1 --turnpolicy white --alphamax 0 --scale 1 --flat ../temp/bordout.bmp -o ../temp/bordout.geojson
potrace --progress -b gimppath --blacklevel 0 --turdsize 0 --longcurve --opttolerance 0 --unit 1 --turnpolicy white --alphamax 0 --scale 1 --flat ../temp/bordout.bmp -o ../temp/bordout.gimppath
potrace --progress -b xfig --blacklevel 0 --turdsize 0 --longcurve --opttolerance 0 --unit 1 --turnpolicy white --alphamax 0 --scale 1 --flat ../temp/bordout.bmp -o ../temp/bordout.xfig
例如,SVG输出作为图片看起来是正确的,但是我无法将其转换为x,y点数组多边形。
<path d="M121 132 l-121 0 0 -66 0 -66 121 0 121 0 0 66 0 66 -121 0z m0 -1
l120 0 0 -22 0 -23 -10 0 -11 0 0 -42 0 -43 -5 0 -5 0 0 43 0 42 -89 0 -89 0
0 -42 0 -43 -5 0 -5 0 0 43 0 42 -10 0 -11 0 0 23 0 22 120 0z M121 130 l-119
0 0 -21 0 -22 11 0 10 0 0 -42 0 -43 4 0 4 0 0 43 0 42 90 0 90 0 0 -42 0 -43
4 0 4 0 0 43 0 42 11 0 10 0 0 22 0 21 -119 0z"/>
</g>
例如,使用https://github.com/Phrogz/svg-path-to-polygons会给我
[
[
[ 121, 132 ], [ 0, 132 ],
[ 0, 66 ], [ 0, 0 ],
[ 121, 0 ], [ 242, 0 ],
[ 242, 66 ], [ 242, 132 ],
[ 121, 132 ], [ 121, 132 ],
closed: true
],
[
[ 121, 131 ], [ 241, 131 ], [ 241, 109 ],
[ 241, 86 ], [ 231, 86 ], [ 220, 86 ],
[ 220, 44 ], [ 220, 1 ], [ 215, 1 ],
[ 210, 1 ], [ 210, 44 ], [ 210, 86 ],
[ 121, 86 ], [ 32, 86 ], [ 32, 44 ],
[ 32, 1 ], [ 27, 1 ], [ 22, 1 ],
[ 22, 44 ], [ 22, 86 ], [ 12, 86 ],
[ 1, 86 ], [ 1, 109 ], [ 1, 131 ],
[ 121, 131 ], [ 121, 131 ], closed: true
],
[
[ 121, 130 ], [ 2, 130 ], [ 2, 109 ],
[ 2, 87 ], [ 13, 87 ], [ 23, 87 ],
[ 23, 45 ], [ 23, 2 ], [ 27, 2 ],
[ 31, 2 ], [ 31, 45 ], [ 31, 87 ],
[ 121, 87 ], [ 211, 87 ], [ 211, 45 ],
[ 211, 2 ], [ 215, 2 ], [ 219, 2 ],
[ 219, 45 ], [ 219, 87 ], [ 230, 87 ],
[ 240, 87 ], [ 240, 109 ], [ 240, 130 ],
[ 121, 130 ], [ 121, 130 ], closed: true
]
]
编辑2
当我使用SVG解决方案时,输出现在会给我一个可读的点列表
<polygon fill="none" points="0,0 0,44 20,44 21,45 21,129 30,129 30,45 31,44 208,44 209,45 209,129 218,129 218,45 219,44 239,44 239,0" stroke="black" stroke-linecap="round" stroke-linejoin="miter" />
但是当我使用该坐标列表时,它不是100%正确的。 角落仍然不正确。
firefox的输出(放大)将显示以下内容:
还有我将使用坐标列表的程序(不是SVG)
love.graphics.polygon("line",0,0,0,44,20,44,21,45,21,129,30,129,30,45,31,44,208,44,209,45,209,129,218,129,218,45,219,44,239,44,239,0)
将输出:
编辑3
使用最后一个python脚本不要制作多边形。
答案 0 :(得分:0)
矢量图中的点,线和曲线可以按比例放大或缩小到任何分辨率,而不会出现锯齿。这样,您将看不到破损的角落。假设输出是SVG格式的矢量图。通过将每个轮廓转换为SVG多边形,可以很好地显示拐角。您可以参考here获得三种渲染角的选择。我还添加了一个函数add_pixel_fillers
来调整足够近的点。
import cv2
import svgwrite
img = cv2.imread("WFVso.png", cv2.IMREAD_UNCHANGED)
ret, mask = cv2.threshold(img[:, :, 3], 0, 255, cv2.THRESH_BINARY)
def add_pixel_fillers(img, cnt):
n_points = len(cnt)
for idx in range(n_points):
prev_pt = cnt[(idx+n_points+1) % n_points]
next_pt = cnt[(idx+1) % n_points]
if abs(cnt[idx][0]-next_pt[0])==1 and abs(cnt[idx][1]-next_pt[1])==1:
temp_x, temp_y = max(cnt[idx][0], next_pt[0]), min(cnt[idx][1], next_pt[1])
if img[temp_y, temp_x] == 255:
cnt[idx][0] = temp_x
cnt[idx][1] = temp_y
else:
temp_x, temp_y = min(cnt[idx][0], next_pt[0]), max(cnt[idx][1], next_pt[1])
if img[temp_y, temp_x] == 255:
cnt[idx][0] = temp_x
cnt[idx][1] = temp_y
return cnt
contours, hierarchy = cv2.findContours(mask, cv2.RETR_LIST, cv2.CHAIN_APPROX_SIMPLE)
h, w = width=img.shape[0], img.shape[1]
dwg = svgwrite.Drawing('test.svg', height=h, width=w, viewBox=(f'-10 -10 {h} {w}'))
for cnt in contours:
cnt = add_pixel_fillers(mask, cnt.squeeze().tolist())
dwg.add(dwg.polygon(
points=cnt,
stroke_linecap='round',
stroke='black',
fill='none',
stroke_linejoin='miter'
))
dwg.save()
示例输入的SVG输出为
<?xml version="1.0" encoding="utf-8" ?>
<svg baseProfile="full" height="100%" version="1.1" viewBox="-10 -10 130 240" width="100%" xmlns="http://www.w3.org/2000/svg" xmlns:ev="http://www.w3.org/2001/xml-events" xmlns:xlink="http://www.w3.org/1999/xlink">
<defs />
<polygon fill="none" points="0,0 0,44 21,44 21,45 21,129 30,129 31,44 31,44 209,44 209,45 209,129 218,129 219,44 219,44 239,44 239,0" stroke="black" stroke-linecap="round" stroke-linejoin="miter" />
</svg>
如何填充角落中缺少的像素?假设在给定的情况下,您只需要处理90度角,则只有有限数量的像素图案需要填充丢失的像素。有了这些样式,您就可以在scipy包中使用ndimage.correlate
函数来找出要填充像素的位置。
示例代码
import numpy as np
from scipy import ndimage
# Assume this is the contour you obtained in step 3
img = np.array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[1, 1, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0, 0, 1, 1, 1],
[0, 0, 1, 0, 0, 0, 1, 0, 0, 0],
[0, 0, 1, 0, 0, 0, 1, 0, 0, 0]])
# You can add the patterns here
patterns = [
np.array([[0,0,0,0,0],
[0,0,0,0,0],
[1,1,0,0,0],
[0,0,1,0,0],
[0,0,1,0,0]]),
np.array([[0,0,0,0,0],
[0,0,0,0,0],
[0,0,0,1,1],
[0,0,1,0,0],
[0,0,1,0,0]])]
missing_corners = np.zeros_like(img) #Result will be stored here
for patt in patterns:
result = ndimage.correlate(img, patt, mode="constant")
corners = np.floor(result/np.sum(patt)).astype(int)
missing_corners = (missing_corners + corners) % 2 #Can use binary OR
查看结果
print(missing_corners)
将显示缺失角的位置:
array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 1, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 1, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]])
对于更复杂的形状,我建议您在导出轮廓时使用drawSvg
之类的包来生成矢量绘图(例如SVG)。