使用另一个线集合裁剪线集合
Crop a collection of lines using another collection of lines
首先，这是我打算做什么的快速图形描述。我将使用 Python，但您可以在您的答案中随意使用伪代码。
我有 2 个 2D 段集合，存储如下：[ [start_point, end_point], [...] ].
第一步，我必须检测 blue 集合中与 black 集合冲突的每个片段。为此，我使用 LeMothe 的 line/line 交集算法。
然后是我的第一个问题：有一段 AB 与我在 C 中的线相交，我不知道如何确定使用代码是否必须 trim 我的段作为 AC 或 CB ，即：我不知道哪一部分我需要保留我的片段，我需要删除哪个片段。
那么对于第二步，我真的不知道如何实现。
如有任何帮助，我们将不胜感激，在此先致谢！
第二步很简单，一旦您弄清楚要保留什么，不保留什么，您只需要跟踪您剪辑的片段并查看它们最初连接的位置（例如，假设这些片段是有序的并形成一个连接线）。
另一方面，考虑到你的黑线实际上是一条线而不是多边形，在你的第一步中，选择 "outside" 和 "inside" 似乎完全是任意的；是否可以将其关闭为多边形？否则，您可能需要人为地创建两个多边形（线的每一侧各一个），然后在这些多边形内部进行裁剪。您可以使用类似 Cyrus 和 Beck 线裁剪算法的算法（请参阅本教程以获取概述：https://www.tutorialspoint.com/computer_graphics/viewing_and_clipping.htm）
随意使用下面的任何代码作为起点（您有一个相交函数和一些有用的 类）。实现 Sutherland 和 Hodgman。
class Point2(object):
"""Structure for a 2D point"""
def __init__(self, x=0, y=0):
self.x = x
self.y = y
def __copy__(self):
return self.__class__(self.x, self.y)
copy = __copy__
def __repr__(self):
return 'Point2(%d, %d)' % (self.x, self.y)
def __getitem__(self, key):
return (self.x, self.y)[key]
def __setitem__(self, key, value):
l = [self.x, self.y]
l[key] = value
self.x, self.y = l
def __eq__(self, other):
if isinstance(other, Point2):
return self.x == other.x and \
self.y == other.y
else:
assert hasattr(other, '__len__') and len(other) == 2
return self.x == other[0] and \
self.y == other[1]
def __ne__(self, other):
return not self.__eq__(other)
def __nonzero__(self):
return self.x != 0 or self.y != 0
def __len__(self):
return 2
class Line2(object):
"""Structure for a 2D line"""
def __init__(self,pt1,pt2):
self.pt1,self.pt2=pt1,pt2
def __repr__(self):
return 'Line2(%s, %s)' % (self.pt1, self.pt2)
class Polygon2(object):
def __init__(self,points):
self.points = points
def __repr__(self):
return '[\n %s\n]' % '\n '.join([str(i) for i in self.points])
def lines(self):
lines = []
e = self.points[-1].copy()
for p in self.points:
lines.append(Line2(e,p))
e = p.copy()
return lines
#return [Line2(a,b) for a,b in zip(self.points,self.points[1:]+[self.points[0]])]
def __copy__(self):
return self.__class__(list(self.points))
copy = __copy__
class Renderer(object):
"""Rendering algorithm implementations"""
def __init__(self,world,img,color=1):
self.world,self.img,self.color=world,img,color
def transform(self,s,r,m,n):
"""Homogeneous transformation operations"""
for i in self.world.points():
j = Matrix3.new_translate(m, n)*Matrix3.new_rotate(r)*Matrix3.new_scale(s)*i
i.x,i.y = j.x,j.y
def clip(self,a,b,c,d):
"""Clipping for the world window defined by a,b,c,d"""
self.clip_lines(a, b, c, d)
self.clip_polygons(a, b, c, d)
def shift(self,a,b,c,d):
"""Shift the world window"""
for i in self.world.points():
i.x -= a
i.y -= b
def clip_lines(self,a,b,c,d):
"""Clipping for lines (i.e. open polygons)"""
clipped = []
for i in self.world.lines:
clipped += [self.clip_lines_cohen_sutherland(i.pt1, i.pt2, a, b, c, d)]
self.world.lines = [i for i in clipped if i]
def clip_polygons(self,a,b,c,d):
"""Clipping for polygons"""
polygons = []
for polygon in self.world.polygons:
new_polygon = self.clip_polygon_sutherland_hodgman(polygon, a, b, c, d)
polygons.append(new_polygon)
self.world.polygons = polygons
def clip_polygon_sutherland_hodgman(self,polygon,xmin,ymin,xmax,ymax):
edges = [Line2(Point2(xmax,ymax),Point2(xmin,ymax)), #top
Line2(Point2(xmin,ymax),Point2(xmin,ymin)), #left
Line2(Point2(xmin,ymin),Point2(xmax,ymin)), #bottom
Line2(Point2(xmax,ymin),Point2(xmax,ymax)), #right
]
def is_inside(pt,line):
# uses the determinant of the vectors (AB,AQ), Q(X,Y) is the query
# left is inside
det = (line.pt2.x-line.pt1.x)*(pt.y-line.pt1.y) - (line.pt2.y-line.pt1.y)*(pt.x-line.pt1.x)
return det>=0
def intersect(pt0,pt1,line):
x1,x2,x3,x4 = pt0.x,pt1.x,line.pt1.x,line.pt2.x
y1,y2,y3,y4 = pt0.y,pt1.y,line.pt1.y,line.pt2.y
x = ((x1*y2-y1*x2)*(x3-x4)-(x1-x2)*(x3*y4-y3*x4)) / ((x1-x2)*(y3-y4)-(y1-y2)*(x3-x4))
y = ((x1*y2-y1*x2)*(y3-y4)-(y1-y2)*(x3*y4-y3*x4)) / ((x1-x2)*(y3-y4)-(y1-y2)*(x3-x4))
return Point2(int(x),int(y))
polygon_new = polygon.copy()
for edge in edges:
polygon_copy = polygon_new.copy()
polygon_new = Polygon2([])
s = polygon_copy.points[-1]
for p in polygon_copy.points:
if is_inside(s,edge) and is_inside(p,edge):
polygon_new.points.append(p)
elif is_inside(s,edge) and not is_inside(p,edge):
polygon_new.points.append(intersect(s,p,edge))
elif not is_inside(s,edge) and not is_inside(p,edge):
pass
else:
polygon_new.points.append(intersect(s,p,edge))
polygon_new.points.append(p)
s = p
return polygon_new
def clip_lines_cohen_sutherland(self,pt0,pt1,xmin,ymin,xmax,ymax):
"""Cohen-Sutherland clipping algorithm for line pt0 to pt1 and clip rectangle with diagonal from (xmin,ymin) to (xmax,ymax)."""
TOP = 1
BOTTOM = 2
RIGHT = 4
LEFT = 8
def ComputeOutCode(pt):
code = 0
if pt.y > ymax: code += TOP
elif pt.y < ymin: code += BOTTOM
if pt.x > xmax: code += RIGHT
elif pt.x < xmin: code += LEFT
return code
accept = False
outcode0, outcode1 = ComputeOutCode(pt0), ComputeOutCode(pt1)
while True:
if outcode0==outcode1==0:
accept=True
break
elif outcode0&outcode1:
accept=False
break
else:
#Failed both tests, so calculate the line segment to clip from an outside point to an intersection with clip edge.
outcodeOut = outcode0 if not outcode0 == 0 else outcode1
if TOP & outcodeOut:
x = pt0.x + (pt1.x - pt0.x) * (ymax - pt0.y) / (pt1.y - pt0.y)
y = ymax
elif BOTTOM & outcodeOut:
x = pt0.x + (pt1.x - pt0.x) * (ymin - pt0.y) / (pt1.y - pt0.y)
y = ymin
elif RIGHT & outcodeOut:
y = pt0.y + (pt1.y - pt0.y) * (xmax - pt0.x) / (pt1.x - pt0.x);
x = xmax;
elif LEFT & outcodeOut:
y = pt0.y + (pt1.y - pt0.y) * (xmin - pt0.x) / (pt1.x - pt0.x);
x = xmin;
if outcodeOut == outcode0:
pt0 = Point2(x,y)
outcode0 = ComputeOutCode(pt0)
else:
pt1 = Point2(x,y)
outcode1 = ComputeOutCode(pt1);
if accept:
return Line2(pt0,pt1)
else:
return False
我认为您需要做的是找到一条从蓝色中心 object 到相关线段的直线。如果从中心到线段 AB 或 BC 的新线在通往蓝线段的途中碰到黑线，则该线段位于外部并被修剪。您可能希望在 A 和 B 之间或 B 和 C 之间的某个点进行检查，以免碰到交点。
至于 python 方面，我建议定义一条带有一些中点属性的线 object class 和一个由具有center 属性，（实际上想到了，那么一条线将算作一个形状，所以你可以将线设为 child class 形状 class 并保留代码），这样您可以创建比较两行的方法，作为每行 object 的一部分。
line_a = Line((4,2),(6,9))
line_b = Line((8,1),(2,10))
line_a.intersects(line.b) #Could return Boolean, or the point of intersection
在我看来，这感觉是解决这个问题的一种非常舒适的方式，因为它可以让您跟踪一切都在做什么。
首先，这是我打算做什么的快速图形描述。我将使用 Python，但您可以在您的答案中随意使用伪代码。
我有 2 个 2D 段集合，存储如下：[ [start_point, end_point], [...] ].
第一步，我必须检测 blue 集合中与 black 集合冲突的每个片段。为此，我使用 LeMothe 的 line/line 交集算法。
然后是我的第一个问题：有一段 AB 与我在 C 中的线相交，我不知道如何确定使用代码是否必须 trim 我的段作为 AC 或 CB ，即：我不知道哪一部分我需要保留我的片段，我需要删除哪个片段。
那么对于第二步，我真的不知道如何实现。
如有任何帮助，我们将不胜感激，在此先致谢！
第二步很简单，一旦您弄清楚要保留什么，不保留什么，您只需要跟踪您剪辑的片段并查看它们最初连接的位置（例如，假设这些片段是有序的并形成一个连接线）。 另一方面，考虑到你的黑线实际上是一条线而不是多边形，在你的第一步中，选择 "outside" 和 "inside" 似乎完全是任意的；是否可以将其关闭为多边形？否则，您可能需要人为地创建两个多边形（线的每一侧各一个），然后在这些多边形内部进行裁剪。您可以使用类似 Cyrus 和 Beck 线裁剪算法的算法（请参阅本教程以获取概述：https://www.tutorialspoint.com/computer_graphics/viewing_and_clipping.htm）
随意使用下面的任何代码作为起点（您有一个相交函数和一些有用的 类）。实现 Sutherland 和 Hodgman。
class Point2(object):
"""Structure for a 2D point"""
def __init__(self, x=0, y=0):
self.x = x
self.y = y
def __copy__(self):
return self.__class__(self.x, self.y)
copy = __copy__
def __repr__(self):
return 'Point2(%d, %d)' % (self.x, self.y)
def __getitem__(self, key):
return (self.x, self.y)[key]
def __setitem__(self, key, value):
l = [self.x, self.y]
l[key] = value
self.x, self.y = l
def __eq__(self, other):
if isinstance(other, Point2):
return self.x == other.x and \
self.y == other.y
else:
assert hasattr(other, '__len__') and len(other) == 2
return self.x == other[0] and \
self.y == other[1]
def __ne__(self, other):
return not self.__eq__(other)
def __nonzero__(self):
return self.x != 0 or self.y != 0
def __len__(self):
return 2
class Line2(object):
"""Structure for a 2D line"""
def __init__(self,pt1,pt2):
self.pt1,self.pt2=pt1,pt2
def __repr__(self):
return 'Line2(%s, %s)' % (self.pt1, self.pt2)
class Polygon2(object):
def __init__(self,points):
self.points = points
def __repr__(self):
return '[\n %s\n]' % '\n '.join([str(i) for i in self.points])
def lines(self):
lines = []
e = self.points[-1].copy()
for p in self.points:
lines.append(Line2(e,p))
e = p.copy()
return lines
#return [Line2(a,b) for a,b in zip(self.points,self.points[1:]+[self.points[0]])]
def __copy__(self):
return self.__class__(list(self.points))
copy = __copy__
class Renderer(object):
"""Rendering algorithm implementations"""
def __init__(self,world,img,color=1):
self.world,self.img,self.color=world,img,color
def transform(self,s,r,m,n):
"""Homogeneous transformation operations"""
for i in self.world.points():
j = Matrix3.new_translate(m, n)*Matrix3.new_rotate(r)*Matrix3.new_scale(s)*i
i.x,i.y = j.x,j.y
def clip(self,a,b,c,d):
"""Clipping for the world window defined by a,b,c,d"""
self.clip_lines(a, b, c, d)
self.clip_polygons(a, b, c, d)
def shift(self,a,b,c,d):
"""Shift the world window"""
for i in self.world.points():
i.x -= a
i.y -= b
def clip_lines(self,a,b,c,d):
"""Clipping for lines (i.e. open polygons)"""
clipped = []
for i in self.world.lines:
clipped += [self.clip_lines_cohen_sutherland(i.pt1, i.pt2, a, b, c, d)]
self.world.lines = [i for i in clipped if i]
def clip_polygons(self,a,b,c,d):
"""Clipping for polygons"""
polygons = []
for polygon in self.world.polygons:
new_polygon = self.clip_polygon_sutherland_hodgman(polygon, a, b, c, d)
polygons.append(new_polygon)
self.world.polygons = polygons
def clip_polygon_sutherland_hodgman(self,polygon,xmin,ymin,xmax,ymax):
edges = [Line2(Point2(xmax,ymax),Point2(xmin,ymax)), #top
Line2(Point2(xmin,ymax),Point2(xmin,ymin)), #left
Line2(Point2(xmin,ymin),Point2(xmax,ymin)), #bottom
Line2(Point2(xmax,ymin),Point2(xmax,ymax)), #right
]
def is_inside(pt,line):
# uses the determinant of the vectors (AB,AQ), Q(X,Y) is the query
# left is inside
det = (line.pt2.x-line.pt1.x)*(pt.y-line.pt1.y) - (line.pt2.y-line.pt1.y)*(pt.x-line.pt1.x)
return det>=0
def intersect(pt0,pt1,line):
x1,x2,x3,x4 = pt0.x,pt1.x,line.pt1.x,line.pt2.x
y1,y2,y3,y4 = pt0.y,pt1.y,line.pt1.y,line.pt2.y
x = ((x1*y2-y1*x2)*(x3-x4)-(x1-x2)*(x3*y4-y3*x4)) / ((x1-x2)*(y3-y4)-(y1-y2)*(x3-x4))
y = ((x1*y2-y1*x2)*(y3-y4)-(y1-y2)*(x3*y4-y3*x4)) / ((x1-x2)*(y3-y4)-(y1-y2)*(x3-x4))
return Point2(int(x),int(y))
polygon_new = polygon.copy()
for edge in edges:
polygon_copy = polygon_new.copy()
polygon_new = Polygon2([])
s = polygon_copy.points[-1]
for p in polygon_copy.points:
if is_inside(s,edge) and is_inside(p,edge):
polygon_new.points.append(p)
elif is_inside(s,edge) and not is_inside(p,edge):
polygon_new.points.append(intersect(s,p,edge))
elif not is_inside(s,edge) and not is_inside(p,edge):
pass
else:
polygon_new.points.append(intersect(s,p,edge))
polygon_new.points.append(p)
s = p
return polygon_new
def clip_lines_cohen_sutherland(self,pt0,pt1,xmin,ymin,xmax,ymax):
"""Cohen-Sutherland clipping algorithm for line pt0 to pt1 and clip rectangle with diagonal from (xmin,ymin) to (xmax,ymax)."""
TOP = 1
BOTTOM = 2
RIGHT = 4
LEFT = 8
def ComputeOutCode(pt):
code = 0
if pt.y > ymax: code += TOP
elif pt.y < ymin: code += BOTTOM
if pt.x > xmax: code += RIGHT
elif pt.x < xmin: code += LEFT
return code
accept = False
outcode0, outcode1 = ComputeOutCode(pt0), ComputeOutCode(pt1)
while True:
if outcode0==outcode1==0:
accept=True
break
elif outcode0&outcode1:
accept=False
break
else:
#Failed both tests, so calculate the line segment to clip from an outside point to an intersection with clip edge.
outcodeOut = outcode0 if not outcode0 == 0 else outcode1
if TOP & outcodeOut:
x = pt0.x + (pt1.x - pt0.x) * (ymax - pt0.y) / (pt1.y - pt0.y)
y = ymax
elif BOTTOM & outcodeOut:
x = pt0.x + (pt1.x - pt0.x) * (ymin - pt0.y) / (pt1.y - pt0.y)
y = ymin
elif RIGHT & outcodeOut:
y = pt0.y + (pt1.y - pt0.y) * (xmax - pt0.x) / (pt1.x - pt0.x);
x = xmax;
elif LEFT & outcodeOut:
y = pt0.y + (pt1.y - pt0.y) * (xmin - pt0.x) / (pt1.x - pt0.x);
x = xmin;
if outcodeOut == outcode0:
pt0 = Point2(x,y)
outcode0 = ComputeOutCode(pt0)
else:
pt1 = Point2(x,y)
outcode1 = ComputeOutCode(pt1);
if accept:
return Line2(pt0,pt1)
else:
return False
我认为您需要做的是找到一条从蓝色中心 object 到相关线段的直线。如果从中心到线段 AB 或 BC 的新线在通往蓝线段的途中碰到黑线，则该线段位于外部并被修剪。您可能希望在 A 和 B 之间或 B 和 C 之间的某个点进行检查，以免碰到交点。
至于 python 方面，我建议定义一条带有一些中点属性的线 object class 和一个由具有center 属性，（实际上想到了，那么一条线将算作一个形状，所以你可以将线设为 child class 形状 class 并保留代码），这样您可以创建比较两行的方法，作为每行 object 的一部分。
line_a = Line((4,2),(6,9))
line_b = Line((8,1),(2,10))
line_a.intersects(line.b) #Could return Boolean, or the point of intersection
在我看来，这感觉是解决这个问题的一种非常舒适的方式，因为它可以让您跟踪一切都在做什么。