view python/tests/moving.txt @ 372:349eb1e09f45

Cleaned the methods/functions indicating if a point is in a polygon In general, shapely should be used, especially for lots of points: from shapely.geometry import Polygon, Point poly = Polygon(array([[0,0],[0,1],[1,1],[1,0]])) p = Point(0.5,0.5) poly.contains(p) -> returns True poly.contains(Point(-1,-1)) -> returns False You can convert a moving.Point to a shapely point: p = moving.Point(1,2) p.asShapely() returns the equivalent shapely point If you have several points to test, use moving.pointsInPolygon(points, polygon) where points are moving.Point and polygon is a shapely polygon.
author Nicolas Saunier <nicolas.saunier@polymtl.ca>
date Tue, 16 Jul 2013 17:00:17 -0400
parents 027e254f0b53
children cd342a774806
line wrap: on
line source

>>> from moving import *
>>> import numpy as np

>>> Interval().empty()
True
>>> Interval(0,1).empty()
False
>>> Interval(0,1)
[0, 1]
>>> Interval(0,1).length()
1.0
>>> Interval(23.2,24.9).length()
1.6999999999999993
>>> Interval(10,8).length()
0.0

>>> TimeInterval(0,1).length()
2.0
>>> TimeInterval(10,8).length()
0.0

>>> [i for i in TimeInterval(9,13)]
[9, 10, 11, 12, 13]

>>> TimeInterval(2,5).equal(TimeInterval(2,5))
True
>>> TimeInterval(2,5).equal(TimeInterval(2,4))
False
>>> TimeInterval(2,5).equal(TimeInterval(5,2))
False

>>> TimeInterval(3,6).distance(TimeInterval(4,6))
0
>>> TimeInterval(3,6).distance(TimeInterval(6,10))
0
>>> TimeInterval(3,6).distance(TimeInterval(8,10))
2
>>> TimeInterval(20,30).distance(TimeInterval(3,15))
5
>>> unionIntervals([TimeInterval(3,6), TimeInterval(8,10),TimeInterval(11,15)])
[3, 15]

>>> Point(3,4)-Point(1,7)
(2.000000,-3.000000)

>>> Point(3,2).norm2Squared()
13

>>> Point.distanceNorm2(Point(3,4),Point(1,7))
3.605551275463989

>>> Point(3,2).inPolygonNoShapely(np.array([[0,0],[1,0],[1,1],[0,1]]))
False
>>> Point(3,2).inPolygonNoShapely(np.array([[0,0],[4,0],[4,3],[0,3]]))
True

>>> predictPositionNoLimit(10, Point(0,0), Point(1,1)) # doctest:+ELLIPSIS
((1.0...,1.0...), (10.0...,10.0...))

>>> segmentIntersection(Point(0,0),Point(1,1), Point(0,1), Point(1,2))
>>> segmentIntersection(Point(0,1),Point(1,0), Point(0,2), Point(2,1))
>>> segmentIntersection(Point(0,0),Point(2,0), Point(1,-1),Point(1,1))
(1.000000,0.000000)
>>> segmentIntersection(Point(0,1),Point(2,0),Point(1,1),Point(1,2))

>>> Trajectory().length()
0
>>> t1 = Trajectory([[0.5,1.5,2.5],[0.5,3.5,6.5]])
>>> t1.length() == 3.
True
>>> t1[1]
(1.500000,3.500000)
>>> t1.getTrajectoryInPolygonNoShapely(np.array([[0,0],[4,0],[4,3],[0,3]]))
(0.500000,0.500000)
>>> t1.getTrajectoryInPolygonNoShapely(np.array([[10,10],[14,10],[14,13],[10,13]])).length()
0

>>> from utils import LCSS
>>> lcss = LCSS(lambda x,y: Point.distanceNorm2(x,y) <= 0.1)
>>> Trajectory.lcss(t1, t1, lcss)
3
>>> lcss = LCSS(lambda p1, p2: (p1-p2).normMax() <= 0.1)
>>> Trajectory.lcss(t1, t1, lcss)
3

>>> o1 = MovingObject(positions = Trajectory([[0]*3,[2]*3]), velocities = Trajectory([[0]*3,[1]*3]))
>>> o1.classifyUserTypeSpeed(0.5, np.median)
>>> userTypeNames[o1.getUserType()]
'car'
>>> o1.classifyUserTypeSpeed(0.5, np.mean)
>>> userTypeNames[o1.getUserType()]
'car'
>>> o1.classifyUserTypeSpeed(1.5, np.median)
>>> userTypeNames[o1.getUserType()]
'pedestrian'