import numpy as np
def closestDistanceBetweenLines(a0,a1,b0,b1,clampAll=False,clampA0=False,clampA1=False,clampB0=False,clampB1=False):
''' Given two lines defined by numpy.array pairs (a0,a1,b0,b1)
Return the closest points on each segment and their distance
'''
# If clampAll=True, set all clamps to True
if clampAll:
clampA0=True
clampA1=True
clampB0=True
clampB1=True
# Calculate denomitator
A = a1 - a0
B = b1 - b0
magA = np.linalg.norm(A)
magB = np.linalg.norm(B)
_A = A / magA
_B = B / magB
cross = np.cross(_A, _B);
denom = np.linalg.norm(cross)**2
# If lines are parallel (denom=0) test if lines overlap.
# If they don't overlap then there is a closest point solution.
# If they do overlap, there are infinite closest positions, but there is a closest distance
if not denom:
d0 = np.dot(_A,(b0-a0))
# Overlap only possible with clamping
if clampA0 or clampA1 or clampB0 or clampB1:
d1 = np.dot(_A,(b1-a0))
# Is segment B before A?
if d0 <= 0 >= d1:
if clampA0 and clampB1:
if np.absolute(d0) < np.absolute(d1):
return a0,b0,np.linalg.norm(a0-b0)
return a0,b1,np.linalg.norm(a0-b1)
# Is segment B after A?
elif d0 >= magA <= d1:
if clampA1 and clampB0:
if np.absolute(d0) < np.absolute(d1):
return a1,b0,np.linalg.norm(a1-b0)
return a1,b1,np.linalg.norm(a1-b1)
# Segments overlap, return distance between parallel segments
return None,None,np.linalg.norm(((d0*_A)+a0)-b0)
# Lines criss-cross: Calculate the projected closest points
t = (b0 - a0);
detA = np.linalg.det([t, _B, cross])
detB = np.linalg.det([t, _A, cross])
t0 = detA/denom;
t1 = detB/denom;
pA = a0 + (_A * t0) # Projected closest point on segment A
pB = b0 + (_B * t1) # Projected closest point on segment B
# Clamp projections
if clampA0 or clampA1 or clampB0 or clampB1:
if clampA0 and t0 < 0:
pA = a0
elif clampA1 and t0 > magA:
pA = a1
if clampB0 and t1 < 0:
pB = b0
elif clampB1 and t1 > magB:
pB = b1
# Clamp projection A
if (clampA0 and t0 < 0) or (clampA1 and t0 > magA):
dot = np.dot(_B,(pA-b0))
if clampB0 and dot < 0:
dot = 0
elif clampB1 and dot > magB:
dot = magB
pB = b0 + (_B * dot)
# Clamp projection B
if (clampB0 and t1 < 0) or (clampB1 and t1 > magB):
dot = np.dot(_A,(pB-a0))
if clampA0 and dot < 0:
dot = 0
elif clampA1 and dot > magA:
dot = magA
pA = a0 + (_A * dot)
return pA,pB,np.linalg.norm(pA-pB)
#!/usr/bin/env python
"""Calculate the distance between line segments."""
import math
class Point(object):
"""A two dimensional point."""
def __init__(self, x, y):
self.x = float(x)
self.y = float(y)
class LineSegment(object):
"""A line segment in a two dimensional space."""
def __init__(self, p1, p2):
assert isinstance(p1, Point), \
"p1 is not of type Point, but of %r" % type(p1)
assert isinstance(p2, Point), \
"p2 is not of type Point, but of %r" % type(p2)
self.p1 = p1
self.p2 = p2
def segments_distance(segment1, segment2):
"""Calculate the distance between two line segments in the plane.
>>> a = LineSegment(Point(1,0), Point(2,0))
>>> b = LineSegment(Point(0,1), Point(0,2))
>>> "%0.2f" % segments_distance(a, b)
'1.41'
>>> c = LineSegment(Point(0,0), Point(5,5))
>>> d = LineSegment(Point(2,2), Point(4,4))
>>> e = LineSegment(Point(2,2), Point(7,7))
>>> "%0.2f" % segments_distance(c, d)
'0.00'
>>> "%0.2f" % segments_distance(c, e)
'0.00'
"""
if segments_intersect(segment1, segment2):
return 0
# try each of the 4 vertices w/the other segment
distances = []
distances.append(point_segment_distance(segment1.p1, segment2))
distances.append(point_segment_distance(segment1.p2, segment2))
distances.append(point_segment_distance(segment2.p1, segment1))
distances.append(point_segment_distance(segment2.p2, segment1))
return min(distances)
def segments_intersect(segment1, segment2):
"""Check if two line segments in the plane intersect.
>>> segments_intersect(LineSegment(Point(0,0), Point(1,0)), \
LineSegment(Point(0,0), Point(1,0)))
True
"""
dx1 = segment1.p2.x - segment1.p1.x
dy1 = segment1.p2.y - segment1.p2.y
dx2 = segment2.p2.x - segment2.p1.x
dy2 = segment2.p2.y - segment2.p1.y
delta = dx2 * dy1 - dy2 * dx1
if delta == 0: # parallel segments
# TODO: Could be (partially) identical!
return False
s = (dx1 * (segment2.p1.y - segment1.p1.y) +
dy1 * (segment1.p1.x - segment2.p1.x)) / delta
t = (dx2 * (segment1.p1.y - segment2.p1.y) +
dy2 * (segment2.p1.x - segment1.p1.x)) / (-delta)
return (0 <= s <= 1) and (0 <= t <= 1)
def point_segment_distance(point, segment):
"""
>>> a = LineSegment(Point(1,0), Point(2,0))
>>> b = LineSegment(Point(2,0), Point(0,2))
>>> point_segment_distance(Point(0,0), a)
1.0
>>> "%0.2f" % point_segment_distance(Point(0,0), b)
'1.41'
"""
assert isinstance(point, Point), \
"point is not of type Point, but of %r" % type(point)
dx = segment.p2.x - segment.p1.x
dy = segment.p2.y - segment.p1.y
if dx == dy == 0: # the segment's just a point
return math.hypot(point.x - segment.p1.x, point.y - segment.p1.y)
if dx == 0:
if (point.y <= segment.p1.y or point.y <= segment.p2.y) and \
(point.y >= segment.p2.y or point.y >= segment.p2.y):
return abs(point.x - segment.p1.x)
if dy == 0:
if (point.x <= segment.p1.x or point.x <= segment.p2.x) and \
(point.x >= segment.p2.x or point.x >= segment.p2.x):
return abs(point.y - segment.p1.y)
# Calculate the t that minimizes the distance.
t = ((point.x - segment.p1.x) * dx + (point.y - segment.p1.y) * dy) / \
(dx * dx + dy * dy)
# See if this represents one of the segment's
# end points or a point in the middle.
if t < 0:
dx = point.x - segment.p1.x
dy = point.y - segment.p1.y
elif t > 1:
dx = point.x - segment.p2.x
dy = point.y - segment.p2.y
else:
near_x = segment.p1.x + t * dx
near_y = segment.p1.y + t * dy
dx = point.x - near_x
dy = point.y - near_y
return math.hypot(dx, dy)
if __name__ == '__main__':
import doctest
doctest.testmod()
9 回答
这是2维吗？如果是这样，答案就是A点和线段CD，B和CD，C和AB或D和AB之间的距离最短 . 所以这是一个相当简单的“点和线之间的距离”计算（如果距离都相同，那么线是平行的） .
This site explains the algorithm for distance between a point and a line pretty well.
它在3个维度上稍微有点棘手，因为线条不一定在同一个平面上，但这似乎不是这里的情况？
这是我在python中的解决方案 . 使用3d点，你可以简化2d .
[编辑1]如果要将结果限制为线段，我添加了一个钳位选项
[编辑2]作为D.A.指出，因为两条线是平行的并不意味着它们之间不能有距离 . 所以我编辑了代码来处理这种情况 . 我还使钳位条件更加通用，因此每个段都可以夹在两侧 .
[编辑3]解决了一个错误jhutar指出，当两条线都有条件并且投影结果超出线段时可能会发生 .
测试示例与图片，以帮助可视化:)
取自this example，其中还附带了一个简单的解释，为什么它的工作原理和VB代码一样（比你需要的更多，所以我简化为翻译成Python时 - 注意：我已翻译，但没有经过测试，所以一个错字可能已经滑落......）：
Distance between Lines and Segments with their Closest Point of Approach
为了计算2个2D线段之间的最小距离，您必须使用4个 endpoints 中的每一个连续执行从 endpoints 到其他线路检查的4个垂直距离 . 但是，如果您发现在4种情况中的任何一种情况下绘制的垂直线与线段不相交，则必须执行4个额外的 endpoints 到 endpoints 距离检查以找到最短距离 .
是否有一个更加优雅的解决方案，我不知道 .
我的解决方案是Fnord解决方案的翻译 . 我在javascript和C.
在Javascript中 . 您需要包含mathjs .
在纯C中
请注意，上述解决方案是正确的，假设线段不相交！如果线段相交，则很明显它们的距离应为0.因此，有必要进行一次最终检查：假设A点和CD点之间的距离d（A，CD）是上述4个检查中的最小值 . 由迪恩 . 然后从A点沿AB段走一小步 . 表示这一点E.如果d（E，CD）<d（A，CD），则段必须相交！请注意，这绝不是斯蒂芬解决的问题 .
这个解决方案实质上是来自Alex Martelli的解决方案，但我添加了一个Point和一个LineSegment类，使阅读更容易 . 我还调整了格式并添加了一些测试 .
线段交叉是错误的，但对于线段距离的计算似乎无关紧要 . 如果您对正确的线段交叉点感兴趣，请查看此处：How do you detect whether or not two line segments intersect?
除非它们是平行的，否则所有2D线最终都会满足 . 了解所教的内容，以便您理解它，而不是试图欺骗这个特定的q .