72 lines
2.7 KiB
Python
72 lines
2.7 KiB
Python
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# author: Adrian Rosebrock
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# website: http://www.pyimagesearch.com
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# import the necessary packages
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from scipy.spatial import distance as dist
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import numpy as np
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import cv2
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def order_points(pts):
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# sort the points based on their x-coordinates
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xSorted = pts[np.argsort(pts[:, 0]), :]
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# grab the left-most and right-most points from the sorted
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# x-roodinate points
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leftMost = xSorted[:2, :]
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rightMost = xSorted[2:, :]
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# now, sort the left-most coordinates according to their
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# y-coordinates so we can grab the top-left and bottom-left
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# points, respectively
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leftMost = leftMost[np.argsort(leftMost[:, 1]), :]
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(tl, bl) = leftMost
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# now that we have the top-left coordinate, use it as an
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# anchor to calculate the Euclidean distance between the
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# top-left and right-most points; by the Pythagorean
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# theorem, the point with the largest distance will be
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# our bottom-right point
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D = dist.cdist(tl[np.newaxis], rightMost, "euclidean")[0]
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(br, tr) = rightMost[np.argsort(D)[::-1], :]
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# return the coordinates in top-left, top-right,
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# bottom-right, and bottom-left order
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return np.array([tl, tr, br, bl], dtype="float32")
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def four_point_transform(image, pts):
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# obtain a consistent order of the points and unpack them
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# individually
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rect = order_points(pts)
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(tl, tr, br, bl) = rect
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# compute the width of the new image, which will be the
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# maximum distance between bottom-right and bottom-left
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# x-coordiates or the top-right and top-left x-coordinates
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widthA = np.sqrt(((br[0] - bl[0]) ** 2) + ((br[1] - bl[1]) ** 2))
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widthB = np.sqrt(((tr[0] - tl[0]) ** 2) + ((tr[1] - tl[1]) ** 2))
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maxWidth = max(int(widthA), int(widthB))
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# compute the height of the new image, which will be the
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# maximum distance between the top-right and bottom-right
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# y-coordinates or the top-left and bottom-left y-coordinates
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heightA = np.sqrt(((tr[0] - br[0]) ** 2) + ((tr[1] - br[1]) ** 2))
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heightB = np.sqrt(((tl[0] - bl[0]) ** 2) + ((tl[1] - bl[1]) ** 2))
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maxHeight = max(int(heightA), int(heightB))
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# now that we have the dimensions of the new image, construct
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# the set of destination points to obtain a "birds eye view",
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# (i.e. top-down view) of the image, again specifying points
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# in the top-left, top-right, bottom-right, and bottom-left
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# order
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dst = np.array([
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[0, 0],
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[maxWidth - 1, 0],
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[maxWidth - 1, maxHeight - 1],
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[0, maxHeight - 1]], dtype="float32")
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# compute the perspective transform matrix and then apply it
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M = cv2.getPerspectiveTransform(rect, dst)
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warped = cv2.warpPerspective(image, M, (maxWidth, maxHeight))
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# return the warped image
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return warped
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