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@@ -18,30 +18,30 @@ class LicensePlate:
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self.height = int(properties['height'])
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self.read_xml()
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-
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+
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def are_corners_sorted(corners):
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- ''' Check if points are sorted clockwise, starting in the left-top
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- corner.'''
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+ '''Check if points are sorted clockwise, starting in the left-top
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+ corner.'''
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x0, y0 = corners[0].to_tuple()
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x1, y1 = corners[1].to_tuple()
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x2, y2 = corners[2].to_tuple()
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x3, y3 = corners[3].to_tuple()
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-
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+
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return x0 < x1 and y1 <= y2 and x2 >= x3 and y3 > y0
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-
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+
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def sort_corners(corners):
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- '''Sort the corners clockwise, starting in the left-top corner. '''
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+ '''Sort the corners clockwise, starting in the left-top corner.'''
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tuples = []
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output = []
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-
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+
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for point in corners:
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tuples.append(point.to_tuple())
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-
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+
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bot1 = (0, 0)
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bot2 = (0, 0)
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top1 = None
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top2 = None
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-
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+
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# Get bottom points (where the y value is the largest). The top points
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# are the points that are not a bottom point.
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for tup in tuples:
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@@ -50,7 +50,7 @@ class LicensePlate:
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top1 = bot2
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else:
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top2 = bot2
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-
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+
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if tup[1] > bot1[1]:
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bot2 = bot1
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bot1 = tup
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@@ -61,7 +61,7 @@ class LicensePlate:
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top1 = tup
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else:
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top2 = tup
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-
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+
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# First point is the smallest x-value top point, second is the other
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# top point
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if top1[0] < top2[0]:
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@@ -70,7 +70,7 @@ class LicensePlate:
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else:
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output.append(Point(top2[0], top2[1]))
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output.append(Point(top1[0], top1[1]))
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-
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+
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# Third point is the bottom point with the largest x-value, fourth is
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# the other bottom point
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if bot1[0] > bot2[0]:
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@@ -79,9 +79,9 @@ class LicensePlate:
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else:
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output.append(Point(bot2[0], bot2[1]))
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output.append(Point(bot1[0], bot1[1]))
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-
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+
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return output
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-
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+
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# sets the entire license plate of an image
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def retrieve_data(self, corners):
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x0, y0 = corners[0].to_tuple()
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@@ -111,16 +111,16 @@ class LicensePlate:
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or_coor = dot(P, ([[i],[j],[1]]))
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or_coor_h = (or_coor[1][0] / or_coor[2][0],
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or_coor[0][0] / or_coor[2][0])
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-
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+
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data[j][i] = self.pV(or_coor_h[0], or_coor_h[1])
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return data
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def get_transformation_matrix(self, matrix):
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# Get the vector p and the values that are in there by taking the SVD.
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- # Since D is diagonal with the eigenvalues sorted from large to small on
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- # the diagonal, the optimal q in min ||Dq|| is q = [[0]..[1]]. Therefore,
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- # p = Vq means p is the last column in V.
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+ # Since D is diagonal with the eigenvalues sorted from large to small
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+ # on the diagonal, the optimal q in min ||Dq|| is q = [[0]..[1]].
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+ # Therefore, p = Vq means p is the last column in V.
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U, D, V = svd(matrix)
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p = V[8][:]
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Taddeus Kroes