def testEnds(self):
     cc = Chaincode([0, 0], "2460") # Can I test more than one path? How?
     end = [0, 0]
     self.assertTrue(cc.matchend(end))
Exemple #2
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 def testEndsFalse(self):
     cc = Chaincode([0, 0], "24460")
     end = [0, 0]
     self.failIf(cc.matchend(end))
Exemple #3
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 def testSubBoundingBoxY(self):
     cc = Chaincode([0, 0], "44464660012075602223")
     self.failUnless(cc.subBoundingBox(yedge=[-1, 0.5]) == [0, 3])
 def testScaleSecond(self):
     cc = Chaincode([0, 0], "0723", xdelta=0.5, ydelta=0.5)
     second = [-1, 0.5]
     self.failUnless(cc.matchany(second, 2))
Exemple #5
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 def testnparray(self):
     # Let's test that the shape of the array matches the expected
     # To do so we need to use np.array, instead of lists.
     cc = Chaincode([0, 0], "2460")
     shape = (2, 5)
     self.failUnless(cc.coordinates.shape == shape)
Exemple #6
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 def testBoundingBoxFalse(self):
     cc = Chaincode([0, 0], "002")
     boundingbox = [[-1, 0], [-1, 0]]
     self.failIf(np.all(cc.BoundingBox() != np.array(boundingbox)))
 def testSecondCoordinate(self):
     cc = Chaincode([0, 0], "0023")
     second = [-2, 0]
     self.assertTrue(cc.matchany(second, 2))
Exemple #8
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 def testScaleSecond(self):
     cc = Chaincode([0, 0], "0723", xdelta=0.5, ydelta=0.5)
     second = [-1, 0.5]
     self.failUnless(cc.matchany(second, 2))
Exemple #9
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 def testEnds(self):
     cc = Chaincode([0, 0], "2460")  # Can I test more than one path? How?
     end = [0, 0]
     self.assertTrue(cc.matchend(end))
Exemple #10
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 def testEndsFalse(self):
     cc = Chaincode([0, 0], "24460")
     end = [0, 0]
     self.assertFalse(cc.matchend(end))
Exemple #11
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 def testScaleSecond(self):
     cc = Chaincode([0, 0], "0723", xdelta=0.5, ydelta=0.5)
     second = [-1, 0.5]
     self.assertTrue(cc.matchany(second, 2))
Exemple #12
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 def testSecondCoordinate(self):
     cc = Chaincode([0, 0], "0023")
     second = [-2, 0]
     self.assertTrue(cc.matchany(second, 2))
 def testEnds(self):
     cc = Chaincode([0, 0], "2460") # Can I test more than one path? How?
     end = [0, 0]
     self.failUnless(cc.matchend(end))
Exemple #14
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 def testSecondCoordinate(self):
     cc = Chaincode([0, 0], "0023")
     second = [-2, 0]
     self.failUnless(cc.matchany(second, 2))
Exemple #15
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 def testSecondCoordinateFails(self):
     cc = Chaincode([1, 0], "0023")
     second = [-2, 0]
     self.assertFalse(cc.matchany(second, 2))
Exemple #16
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 def testSecondCoordinateFails(self):
     cc = Chaincode([1, 0], "0023")
     second = [-2, 0]
     self.failIf(cc.matchany(second, 2))
Exemple #17
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 def testScaleSecond(self):
     cc = Chaincode([0, 0], "0723", xdelta=0.5, ydelta=0.5)
     second = [-1, 0.5]
     self.assertTrue(cc.matchany(second, 2))
Exemple #18
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 def testScaleEnd(self):
     cc = Chaincode([1.2, 3], "0723", xdelta=2.629, ydelta=2.629)
     end = [-1.429, 0.371]
     self.failUnless(cc.matchany(end, -1))
Exemple #19
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 def testScaleEnd(self):
     cc = Chaincode([1.2, 3],"0723", xdelta=2.629, ydelta=2.629)
     end = [-1.429, 0.371]
     self.assertTrue(cc.matchany(end, -1))
Exemple #20
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 def testBoundingBox(self):  #needs of np.array... I think
     cc = Chaincode([0, 0], "00033344")
     boundingbox = [[-3, 2], [-3, 0]]  # [[x0,x1],[y0,y1]] (like cc)
     self.failUnless(np.all(cc.BoundingBox() == np.array(boundingbox)))
Exemple #21
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 def testSubBoundingBoxX(self):
     cc = Chaincode([0, 0], "44464660012075602223")
     self.assertTrue(cc.subBoundingBox(xedge=[0.1, 2]) == [0, 3])
Exemple #22
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 def testSubBoundingBoxX(self):
     cc = Chaincode([0, 0], "44464660012075602223")
     self.failUnless(cc.subBoundingBox(xedge=[0.1, 2]) == [0, 3])
Exemple #23
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 def testSubBoundingBoxY(self):
     cc = Chaincode([0, 0], "44464660012075602223")
     self.assertTrue(cc.subBoundingBox(yedge=[-1, 0.5]) == [0, 3])
Exemple #24
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 def testEnds(self):
     cc = Chaincode([0, 0], "2460")  # Can I test more than one path? How?
     end = [0, 0]
     self.failUnless(cc.matchend(end))
 def testSecondCoordinate(self):
     cc = Chaincode([0, 0], "0023")
     second = [-2, 0]
     self.failUnless(cc.matchany(second, 2))