我有一个工作订单管理系统，其中包含Python 2.7。
Line Vertex 1: 676561.00, 4860927.00
Line Vertex 2: 676557.00, 4860939.00
Point 100: 676551.00, 4860935.00
Point 200: 676558.00, 4860922.00
我希望您能将点“snap”到最近的线段上。
捕捉可以定义为:
将一个点移动，使其恰好与一条线的最近部分重合。
似乎有两种不同的数学情景：
A. 直线上最接近的部分是点垂直于直线的位置。
B. 或者，直线上最接近的部分只是最接近的顶点。
是否可能使用Python 2.7（和标准2.7库）将一个点捕捉到最近的线段上？
这是一个非常有用的资源。
import collections
import math
Line = collections.namedtuple('Line', 'x1 y1 x2 y2')
Point = collections.namedtuple('Point', 'x y')
def lineLength(line):
dist = math.sqrt((line.x2 - line.x1)**2 + (line.y2 - line.y1)**2)
return dist
## See http://paulbourke.net/geometry/pointlineplane/
## for helpful formulas
## Inputs
line = Line(0.0, 0.0, 100.0, 0.0)
point = Point(50.0, 1500)
## Calculations
print('Inputs:')
print('Line defined by: ({}, {}) and ({}, {})'.format(line.x1, line.y1, line.x2, line.y2))
print('Point "P": ({}, {})'.format(point.x, point.y))
len = lineLength(line)
if (len == 0):
raise Exception('The points on input line must not be identical')
print('\nResults:')
print('Length of line (calculated): {}'.format(len))
u = ((point.x - line.x1) * (line.x2 - line.x1) + (point.y - line.y1) * (line.y2 - line.y1)) / (
len**2)
# restrict to line boundary
if u > 1:
u = 1
elif u < 0:
u = 0
nearestPointOnLine = Point(line.x1 + u * (line.x2 - line.x1), line.y1 + u * (line.y2 - line.y1))
shortestLine = Line(nearestPointOnLine.x, nearestPointOnLine.y, point.x, point.y)
print('Nearest point "N" on line: ({}, {})'.format(nearestPointOnLine.x, nearestPointOnLine.y))
print('Length from "P" to "N": {}'.format(lineLength(shortestLine)))
import collections
import math
Point = collections.namedtuple('Point', "x, y")
def distance(pt, a, b, c):
# line eq: ax + by + c = 0
return math.fabs(a * pt.x + b * pt.y + c) / math.sqrt(a**2 + b**2)
l1 = Point(676561.00, 4860927.00)
l2 = Point(676557.00, 4860939.00)
# line equation
a = l2.y - l1.y
b = l1.x - l2.x
c = l2.x * l1.y - l2.y * l1.x
assert a * l1.x + b * l1.y + c == 0
assert a * l2.x + b * l2.y + c == 0
p100 = Point(676551.00, 4860935.00)
p200 = Point(676558.00, 4860922.00)
print(distance(p100, a, b, c))
print(distance(p200, a, b, c))
你得到：
6.957010852370434
4.427188724235731
编辑1：计算正交投影
你想要的是点p100和p200在直线(l1, l2)上的正交投影的坐标。
你可以按照以下方式进行计算：
import collections
import math
Point = collections.namedtuple('Point', "x, y")
def snap(pt, pt1, pt2):
# type: (Point, Point, Point) -> Point
v = Point(pt2.x - pt1.x, pt2.y - pt1.y)
dv = math.sqrt(v.x ** 2 + v.y ** 2)
bh = ((pt.x - pt1.x) * v.x + (pt.y - pt1.y) * pt2.y) / dv
h = Point(
pt1.x + bh * v.x / dv,
pt1.y + bh * v.y / dv
)
return h
l1 = Point(676561.00, 4860927.00)
l2 = Point(676557.00, 4860939.00)
p100 = Point(676551.00, 4860935.00)
p200 = Point(676558.00, 4860922.00)
s100 = snap(p100, l1, l2)
s200 = snap(p200, l1, l2)
print(s100)
print(p100)
你得到：
Point(x=-295627.7999999998, y=7777493.4)
Point(x=676551.0, y=4860935.0)
# line equation
a = l2.y - l1.y
b = l1.x - l2.x
c = l2.x * l1.y - l2.y * l1.x
assert math.fabs(a * s100.x + b * s100.y + c) < 1e-6
assert math.fabs(a * s200.x + b * s200.y + c) < 1e-6
编辑2：对齐到线段
如果您想对齐到一条线段，您需要检查正交投影是否在线段内部。
您可以按照以下方式执行：
def distance_pts(pt1, pt2):
v = Point(pt2.x - pt1.x, pt2.y - pt1.y)
dv = math.sqrt(v.x ** 2 + v.y ** 2)
return dv
def snap(pt, pt1, pt2):
# type: (Point, Point, Point) -> Point
v = Point(pt2.x - pt1.x, pt2.y - pt1.y)
dv = distance_pts(pt1, pt2)
bh = ((pt.x - pt1.x) * v.x + (pt.y - pt1.y) * pt2.y) / dv
h = Point(pt1.x + bh * v.x / dv, pt1.y + bh * v.y / dv)
if 0 <= (pt1.x - h.x) / (pt2.x - h.y) < 1:
# in the line segment
return h
elif distance_pts(h, pt1) < distance_pts(h, pt2):
# near pt1
return pt1
else:
# near pt2
return pt2
问题100和问题200的解决方案如下：
Point(x=676557.0, y=4860939.0)
Point(x=676551.0, y=4860935.0)
另一个选项：
# snap a point to a 2d line
# parameters:
# A,B: the endpoints of the line
# C: the point we want to snap to the line AB
# all parameters must be a tuple/list of float numbers
def snap_to_line(A,B,C):
Ax,Ay = A
Bx,By = B
Cx,Cy = C
# special case: A,B are the same point: just return A
eps = 0.0000001
if abs(Ax-Bx) < eps and abs(Ay-By) < eps: return [Ax,Ay]
# any point X on the line can be represented by the equation
# X = A + t * (B-A)
# so for point C we compute its perpendicular D on the line
# and the parameter t for D
# if t is between 0..1 then D is on the line so the snap point is D
# if t < 0 then the snap point is A
# if t > 1 then the snap point is B
#
# explanation of the formula for distance from point to line:
# http://paulbourke.net/geometry/pointlineplane/
#
dx = Bx-Ax
dy = By-Ay
d2 = dx*dx + dy*dy
t = ((Cx-Ax)*dx + (Cy-Ay)*dy) / d2
if t <= 0: return A
if t >= 1: return B
return [dx*t + Ax, dy*t + Ay]
if __name__=="__main__":
A,B = [676561.00, 4860927.00],[676557.00, 4860939.00]
C = [676551.00, 4860935.00]
print(snap_to_line(A,B,C))
C = [676558.00, 4860922.00]
print(snap_to_line(A,B,C))
PIL或Opencv。 - jizhihaoSAMA