Python 如何确定一个点位于线段上的两个其他点之间？
假设有一个二维平面，平面上有两个点（称为a和b），每个点由一个x整数和一个y整数表示 如何确定另一个点c是否位于由a和b定义的线段上Python 如何确定一个点位于线段上的两个其他点之间？,python,math,geometry,Python,Math,Geometry,假设有一个二维平面，平面上有两个点（称为a和b），每个点由一个x整数和一个y整数表示 如何确定另一个点c是否位于由a和b定义的线段上 我大部分使用python，但任何语言中的示例都会有帮助。检查b-a和c-a的叉积是否为0：这意味着所有点都是共线的。如果是，检查c的坐标是否在a和b之间。使用x或y坐标，只要a和b在该轴上是分开的（或者两者都相同） def开启（a、b、c）： “返回真iff点c与从a到b的线段相交。” #（或所有3个点重合的退化情况） 返回（共线（a、b、c） 和（在（a.x，c
我大部分使用python，但任何语言中的示例都会有帮助。检查
b-ac-a0cababdef开启（a、b、c）：
“返回真iff点c与从a到b的线段相交。”
#（或所有3个点重合的退化情况）
返回（共线（a、b、c）
和（在（a.x，c.x，b.x）内，如果a.x！=b.x else
在（a.y，c.y，b.y））范围内
def共线（a、b、c）：
如果a、b和c都位于同一条线上，则返回true
返回（b.x-a.x）*（c.y-a.y）=（c.x-a.x）*（b.y-a.y）
（p、q、r）内的def：
“如果q在p和r之间（包括p和r），则返回真值。”
返回p检查（b-a）和（c-a）的叉积是否为0，正如Darius Bacon告诉您的那样，点a、b和c是否对齐
但是，当您想知道c是否在a和b之间时，您还必须检查（b-a）和（c-a）的点积是否为正，并且小于a和b之间距离的平方
在非优化伪代码中：
def isBetween(a, b, c):
crossproduct = (c.y - a.y) * (b.x - a.x) - (c.x - a.x) * (b.y - a.y)
# compare versus epsilon for floating point values, or != 0 if using integers
if abs(crossproduct) > epsilon:
return False
dotproduct = (c.x - a.x) * (b.x - a.x) + (c.y - a.y)*(b.y - a.y)
if dotproduct < 0:
return False
squaredlengthba = (b.x - a.x)*(b.x - a.x) + (b.y - a.y)*(b.y - a.y)
if dotproduct > squaredlengthba:
return False
return True
# epsilon = small constant
def isBetween(a, b, c):
lengthca2 = (c.x - a.x)*(c.x - a.x) + (c.y - a.y)*(c.y - a.y)
lengthba2 = (b.x - a.x)*(b.x - a.x) + (b.y - a.y)*(b.y - a.y)
if lengthca2 > lengthba2: return False
dotproduct = (c.x - a.x)*(b.x - a.x) + (c.y - a.y)*(b.y - a.y)
if dotproduct < 0.0: return False
if abs(dotproduct*dotproduct - lengthca2*lengthba2) > epsilon: return False
return True
def介于（a、b、c）之间：
叉积=（c.y-a.y）*（b.x-a.x）-（c.x-a.x）*（b.y-a.y）
#比较浮点值与ε，或！=如果使用整数，则为0
如果abs（叉积）>ε：
返回错误
点积=（c.x-a.x）*（b.x-a.x）+（c.y-a.y）*（b.y-a.y）
如果dotproduct<0：
返回错误
平方长度ba=（b.x-a.x）*（b.x-a.x）+（b.y-a.y）*（b.y-a.y）
如果dotproduct>squaredlengthba：
返回错误
返回真值
在（c-a）和（b-a）之间的标量积必须等于它们长度的乘积（这意味着向量（c-a）和（b-a）对齐且方向相同）。此外，（c-a）的长度必须小于或等于（b-a）。伪代码：
def isBetween(a, b, c):
crossproduct = (c.y - a.y) * (b.x - a.x) - (c.x - a.x) * (b.y - a.y)
# compare versus epsilon for floating point values, or != 0 if using integers
if abs(crossproduct) > epsilon:
return False
dotproduct = (c.x - a.x) * (b.x - a.x) + (c.y - a.y)*(b.y - a.y)
if dotproduct < 0:
return False
squaredlengthba = (b.x - a.x)*(b.x - a.x) + (b.y - a.y)*(b.y - a.y)
if dotproduct > squaredlengthba:
return False
return True
# epsilon = small constant
def isBetween(a, b, c):
lengthca2 = (c.x - a.x)*(c.x - a.x) + (c.y - a.y)*(c.y - a.y)
lengthba2 = (b.x - a.x)*(b.x - a.x) + (b.y - a.y)*(b.y - a.y)
if lengthca2 > lengthba2: return False
dotproduct = (c.x - a.x)*(b.x - a.x) + (c.y - a.y)*(b.y - a.y)
if dotproduct < 0.0: return False
if abs(dotproduct*dotproduct - lengthca2*lengthba2) > epsilon: return False
return True
#ε=小常数
def介于（a、b、c）之间：
长度ca2=（c.x-a.x）*（c.x-a.x）+（c.y-a.y）*（c.y-a.y）
长度ba2=（b.x-a.x）*（b.x-a.x）+（b.y-a.y）*（b.y-a.y）
如果lengthca2>lengthba2：返回False
点积=（c.x-a.x）*（b.x-a.x）+（c.y-a.y）*（b.y-a.y）
如果dotproduct<0.0：返回False
如果abs（dotproduct*dotproduct-lengthca2*lengthba2）>ε：返回False
返回真值
还有另一种方法：
- 假设两点是A（x1，y1）和B（x2，y2）
- 通过这些点的直线方程为（x-x1）/（y-y1）=（x2-x1）/（y2-y1）。。（只是使坡度相等）
点C（x3，y3）将位于A和B之间，如果：
- x3，y3满足上述方程
- x3位于x1和x2之间，y3位于y1和y2之间（普通检查）
使用更几何的方法，计算以下距离：
ab = sqrt((a.x-b.x)**2 + (a.y-b.y)**2)
ac = sqrt((a.x-c.x)**2 + (a.y-c.y)**2)
bc = sqrt((b.x-c.x)**2 + (b.y-c.y)**2)
并测试ac+bc是否等于ab：
is_on_segment = abs(ac + bc - ab) < EPSILON
c = ma + (1 - m)b, where 0 <= m <= 1
是否在\u段=abs（ac+bc-ab）
这是因为有三种可能性：
- 这三个点形成一个三角形=>ac+bc>ab
- 它们是共线的，c在ab段=>ac+bc>ab
- 它们是共线的，c位于ab段=>ac+bc=ab
以下是我的做法：
def distance(a,b):
return sqrt((a.x - b.x)**2 + (a.y - b.y)**2)
def is_between(a,c,b):
return distance(a,c) + distance(c,b) == distance(a,b)
这是我在学校的表现。我忘了为什么这不是个好主意
#!/usr/bin/env python
from __future__ import division
epsilon = 1e-6
class Point:
def __init__(self, x, y):
self.x, self.y = x, y
class LineSegment:
"""
>>> ls = LineSegment(Point(0,0), Point(2,4))
>>> Point(1, 2) in ls
True
>>> Point(.5, 1) in ls
True
>>> Point(.5, 1.1) in ls
False
>>> Point(-1, -2) in ls
False
>>> Point(.1, 0.20000001) in ls
True
>>> Point(.1, 0.2001) in ls
False
>>> ls = LineSegment(Point(1, 1), Point(3, 5))
>>> Point(2, 3) in ls
True
>>> Point(1.5, 2) in ls
True
>>> Point(0, -1) in ls
False
>>> ls = LineSegment(Point(1, 2), Point(1, 10))
>>> Point(1, 6) in ls
True
>>> Point(1, 1) in ls
False
>>> Point(2, 6) in ls
False
>>> ls = LineSegment(Point(-1, 10), Point(5, 10))
>>> Point(3, 10) in ls
True
>>> Point(6, 10) in ls
False
>>> Point(5, 10) in ls
True
>>> Point(3, 11) in ls
False
"""
def __init__(self, a, b):
if a.x > b.x:
a, b = b, a
(self.x0, self.y0, self.x1, self.y1) = (a.x, a.y, b.x, b.y)
self.slope = (self.y1 - self.y0) / (self.x1 - self.x0) if self.x1 != self.x0 else None
def __contains__(self, c):
return (self.x0 <= c.x <= self.x1 and
min(self.y0, self.y1) <= c.y <= max(self.y0, self.y1) and
(not self.slope or -epsilon < (c.y - self.y(c.x)) < epsilon))
def y(self, x):
return self.slope * (x - self.x0) + self.y0
if __name__ == '__main__':
import doctest
doctest.testmod()
编辑：
@Darius Bacon：这包含了为什么下面的代码不是一个好主意的解释
#!/usr/bin/env python
from __future__ import division
epsilon = 1e-6
class Point:
def __init__(self, x, y):
self.x, self.y = x, y
class LineSegment:
"""
>>> ls = LineSegment(Point(0,0), Point(2,4))
>>> Point(1, 2) in ls
True
>>> Point(.5, 1) in ls
True
>>> Point(.5, 1.1) in ls
False
>>> Point(-1, -2) in ls
False
>>> Point(.1, 0.20000001) in ls
True
>>> Point(.1, 0.2001) in ls
False
>>> ls = LineSegment(Point(1, 1), Point(3, 5))
>>> Point(2, 3) in ls
True
>>> Point(1.5, 2) in ls
True
>>> Point(0, -1) in ls
False
>>> ls = LineSegment(Point(1, 2), Point(1, 10))
>>> Point(1, 6) in ls
True
>>> Point(1, 1) in ls
False
>>> Point(2, 6) in ls
False
>>> ls = LineSegment(Point(-1, 10), Point(5, 10))
>>> Point(3, 10) in ls
True
>>> Point(6, 10) in ls
False
>>> Point(5, 10) in ls
True
>>> Point(3, 11) in ls
False
"""
def __init__(self, a, b):
if a.x > b.x:
a, b = b, a
(self.x0, self.y0, self.x1, self.y1) = (a.x, a.y, b.x, b.y)
self.slope = (self.y1 - self.y0) / (self.x1 - self.x0) if self.x1 != self.x0 else None
def __contains__(self, c):
return (self.x0 <= c.x <= self.x1 and
min(self.y0, self.y1) <= c.y <= max(self.y0, self.y1) and
(not self.slope or -epsilon < (c.y - self.y(c.x)) < epsilon))
def y(self, x):
return self.slope * (x - self.x0) + self.y0
if __name__ == '__main__':
import doctest
doctest.testmod()
#/usr/bin/env python
来自未来进口部
ε=1e-6
课程点：
定义初始化（self，x，y）：
self.x，self.y=x，y
类别线段：
"""
>>>ls=线段（点（0,0），点（2,4））
>>>ls中的点（1,2）
真的
>>>ls中的点（.5，1）
真的
>>>ls中的点（.5，1.1）
假的
>>>ls中的点（-1，-2）
假的
>>>ls中的点（.1，0.2000001）
真的
>>>ls中的点（.1，0.2001）
假的
>>>ls=线段（点（1，1），点（3，5））
>>>ls中的点（2,3）
真的
>>>ls中的点（1.5,2）
真的
>>>ls中的点（0，-1）
假的
>>>ls=线段（点（1，2），点（1，10））
>>>ls中的点（1,6）
真的
>>>ls中的点（1，1）
假的
>>>ls中的点（2,6）
假的
>>>ls=线段（点（-1,10），点（5,10））
>>>ls中的点（3,10）
真的
>>>ls中的点（6,10）
假的
>>>ls中的点（5,10）
真的
>>>ls中的点（3,11）
假的
"""
定义初始化（self，a，b）：
如果a.x>b.x：
a、 b=b，a
（self.x0，self.y0，self.x1，self.y1）=（a.x，a.y，b.x，b.y）
self.slope=（self.y1-self.y0）/（self.x1-self.x0）如果self.x1！=self.x0其他无
def___;包含____;（self，c）：
返回（self.x0段的长度不重要，因此不需要使用平方根，应避免使用平方根，因为我们可能会失去一些精度
class Point:
def __init__(self, x, y):
self.x = x
self.y = y
class Segment:
def __init__(self, a, b):
self.a = a
self.b = b
def is_between(self, c):
# Check if slope of a to c is the same as a to b ;
# that is, when moving from a.x to c.x, c.y must be proportionally
# increased than it takes to get from a.x to b.x .
# Then, c.x must be between a.x and b.x, and c.y must be between a.y and b.y.
# => c is after a and before b, or the opposite
# that is, the absolute value of cmp(a, b) + cmp(b, c) is either 0 ( 1 + -1 )
# or 1 ( c == a or c == b)
a, b = self.a, self.b
return ((b.x - a.x) * (c.y - a.y) == (c.x - a.x) * (b.y - a.y) and
abs(cmp(a.x, c.x) + cmp(b.x, c.x)) <= 1 and
abs(cmp(a.y, c.y) + cmp(b.y, c.y)) <= 1)
好的，很多人提到线性代数（向量的叉积），这在实（即连续或浮点）空间中有效，但问题明确指出两点表示为整数，因此叉积不是正确的解，尽管它可以给出近似解
正确的解决方案是在两点之间使用，并查看第三点是否为其中一点
def is_on(a, b, c):
"""Is c on the line segment ab?"""
def _is_zero( val ):
return -epsilon < val < epsilon
x1 = a.x - b.x
x2 = c.x - b.x
y1 = a.y - b.y
y2 = c.y - b.y
if _is_zero(x1) and _is_zero(y1):
# a and b are the same point:
# so check that c is the same as a and b
return _is_zero(x2) and _is_zero(y2)
if _is_zero(x1):
# a and b are on same vertical line
m2 = y2 * 1.0 / y1
return _is_zero(x2) and 0 <= m2 <= 1
elif _is_zero(y1):
# a and b are on same horizontal line
m1 = x2 * 1.0 / x1
return _is_zero(y2) and 0 <= m1 <= 1
else:
m1 = x2 * 1.0 / x1
if m1 < 0 or m1 > 1:
return False
m2 = y2 * 1.0 / y1
return _is_zero(m2 - m1)
Boolean Contains(PointF from, PointF to, PointF pt, double epsilon)
{
double segmentLengthSqr = (to.X - from.X) * (to.X - from.X) + (to.Y - from.Y) * (to.Y - from.Y);
double r = ((pt.X - from.X) * (to.X - from.X) + (pt.Y - from.Y) * (to.Y - from.Y)) / segmentLengthSqr;
if(r<0 || r>1) return false;
double sl = ((from.Y - pt.Y) * (to.X - from.X) - (from.X - pt.X) * (to.Y - from.Y)) / System.Math.Sqrt(segmentLengthSqr);
return -epsilon <= sl && sl <= epsilon;
}
is_on = (a,b,c) ->
# "Return true if point c intersects the line segment from a to b."
# (or the degenerate case that all 3 points are coincident)
return (collinear(a,b,c) and withincheck(a,b,c))
withincheck = (a,b,c) ->
if a[0] != b[0]
within(a[0],c[0],b[0])
else
within(a[1],c[1],b[1])
collinear = (a,b,c) ->
# "Return true if a, b, and c all lie on the same line."
((b[0]-a[0])*(c[1]-a[1]) < (c[0]-a[0])*(b[1]-a[1]) + 1000) and ((b[0]-a[0])*(c[1]-a[1]) > (c[0]-a[0])*(b[1]-a[1]) - 1000)
within = (p,q,r) ->
# "Return true if q is between p and r (inclusive)."
p <= q <= r or r <= q <= p
l1 + A(l2 - l1)
x = l1.x + A(l2.x - l1.x)
y = l1.y + A(l2.y - l1.y)
// Vec2 is a simple x/y struct - it could very well be named Point for this use
bool isBetween(double a, double b, double c) {
// return if c is between a and b
double larger = (a >= b) ? a : b;
double smaller = (a != larger) ? a : b;
return c <= larger && c >= smaller;
}
bool pointOnLine(Vec2<double> p, Vec2<double> l1, Vec2<double> l2) {
if(l2.x - l1.x == 0) return isBetween(l1.y, l2.y, p.y); // vertical line
if(l2.y - l1.y == 0) return isBetween(l1.x, l2.x, p.x); // horizontal line
double Ax = (p.x - l1.x) / (l2.x - l1.x);
double Ay = (p.y - l1.y) / (l2.y - l1.y);
// We want Ax == Ay, so check if the difference is very small (floating
// point comparison is fun!)
return fabs(Ax - Ay) < 0.000001 && Ax >= 0.0 && Ax <= 1.0;
}
boolean liesOnSegment(Coordinate a, Coordinate b, Coordinate c) {
double dotProduct = (c.x - a.x) * (c.x - b.x) + (c.y - a.y) * (c.y - b.y);
return (dotProduct < 0);
}
public static bool IsOnSegment(this Segment2D @this, Point2D c, double tolerance)
{
var distanceSquared = tolerance*tolerance;
// Start of segment to test point vector
var v = new Vector2D( @this.P0, c ).To3D();
// Segment vector
var s = new Vector2D( @this.P0, @this.P1 ).To3D();
// Dot product of s
var ss = s*s;
// k is the scalar we multiply s by to get the projection of c onto s
// where we assume s is an infinte line
var k = v*s/ss;
// Convert our tolerance to the units of the scalar quanity k
var kd = tolerance / Math.Sqrt( ss );
// Check that the projection is within the bounds
if (k <= -kd || k >= (1+kd))
{
return false;
}
// Find the projection point
var p = k*s;
// Find the vector between test point and it's projection
var vp = (v - p);
// Check the distance is within tolerance.
return vp * vp < distanceSquared;
}
s * s
def dot(v,w): return v.x*w.x + v.y*w.y
def wedge(v,w): return v.x*w.y - v.y*w.x
def is_between(a,b,c):
v = a - b
w = b - c
return wedge(v,w) == 0 and dot(v,w) > 0
private bool _isPointOnLine( Vector2 ptLineStart, Vector2 ptLineEnd, Vector2 ptPoint )
{
bool bRes = false;
if((Mathf.Approximately(ptPoint.x, ptLineStart.x) || Mathf.Approximately(ptPoint.x, ptLineEnd.x)))
{
if(ptPoint.y > ptLineStart.y && ptPoint.y < ptLineEnd.y)
{
bRes = true;
}
}
else if((Mathf.Approximately(ptPoint.y, ptLineStart.y) || Mathf.Approximately(ptPoint.y, ptLineEnd.y)))
{
if(ptPoint.x > ptLineStart.x && ptPoint.x < ptLineEnd.x)
{
bRes = true;
}
}
return bRes;
}
public static double CalcDistanceBetween2Points(double x1, double y1, double x2, double y2)
{
return Math.Sqrt(Math.Pow (x1-x2, 2) + Math.Pow (y1-y2, 2));
}
public static bool PointLinesOnLine (double x, double y, double x1, double y1, double x2, double y2, double allowedDistanceDifference)
{
double dist1 = CalcDistanceBetween2Points(x, y, x1, y1);
double dist2 = CalcDistanceBetween2Points(x, y, x2, y2);
double dist3 = CalcDistanceBetween2Points(x1, y1, x2, y2);
return Math.Abs(dist3 - (dist1 + dist2)) <= allowedDistanceDifference;
}
function getLineDefinition($p1=array(0,0), $p2=array(0,0)){
$k = ($p1[1]-$p2[1])/($p1[0]-$p2[0]);
$q = $p1[1]-$k*$p1[0];
return array($k, $q);
}
function isPointOnLineSegment($line=array(array(0,0),array(0,0)), $pt=array(0,0)){
// GET THE LINE DEFINITION y = k.x + q AS array(k, q)
$def = getLineDefinition($line[0], $line[1]);
// use the line definition to find y for the x of your point
$y = $def[0]*$pt[0]+$def[1];
$yMin = min($line[0][1], $line[1][1]);
$yMax = max($line[0][1], $line[1][1]);
// exclude y values that are outside this segments bounds
if($y>$yMax || $y<$yMin) return false;
// calculate the difference of your points y value from the reference value calculated from lines definition
// in ideal cases this would equal 0 but we are dealing with floating point values so we need some threshold value not to lose results
// this is up to you to fine tune
$diff = abs($pt[1]-$y);
$thr = 0.000001;
return $diff<=$thr;
}