四边形坐标顺时针排序

1. 做图像检测的时候处理数据经常遇到给出矩形的四个坐标点，要求找出左上角坐标并对乱序的坐标按顺时针或者逆时针进行排序。Its extremely important to maintain a consistent ordering of points !

from scipy.spatial import distance as distimport numpy as npimport mathdef cos_dist(a, b):    if len(a) != len(b):        return None    part_up = 0.0    a_sq = 0.0    b_sq = 0.0    print a, b    print zip(a, b)    for a1, b1 in zip(a, b):        part_up += a1*b1        a_sq += a1**2        b_sq += b1**2    part_down = math.sqrt(a_sq*b_sq)    if part_down == 0.0:        return None    else:        return part_up / part_down# this function is confined to rectangledef order_points(pts):    # sort the points based on their x-coordinates    xSorted = pts[np.argsort(pts[:, 0]), :]    # grab the left-most and right-most points from the sorted    # x-roodinate points    leftMost = xSorted[:2, :]    rightMost = xSorted[2:, :]    # now, sort the left-most coordinates according to their    # y-coordinates so we can grab the top-left and bottom-left    # points, respectively    leftMost = leftMost[np.argsort(leftMost[:, 1]), :]    (tl, bl) = leftMost    # now that we have the top-left coordinate, use it as an    # anchor to calculate the Euclidean distance between the    # top-left and right-most points; by the Pythagorean    # theorem, the point with the largest distance will be    # our bottom-right point    D = dist.cdist(tl[np.newaxis], rightMost, "euclidean")[0]    (br, tr) = rightMost[np.argsort(D)[::-1], :]    # return the coordinates in top-left, top-right,    # bottom-right, and bottom-left order    return np.array([tl, tr, br, bl], dtype="float32")def order_points_quadrangle(pts):    # sort the points based on their x-coordinates    xSorted = pts[np.argsort(pts[:, 0]), :]    # grab the left-most and right-most points from the sorted    # x-roodinate points    leftMost = xSorted[:2, :]    rightMost = xSorted[2:, :]    # now, sort the left-most coordinates according to their    # y-coordinates so we can grab the top-left and bottom-left    # points, respectively    leftMost = leftMost[np.argsort(leftMost[:, 1]), :]    (tl, bl) = leftMost    # now that we have the top-left and bottom-left coordinate, use it as an    # base vector to calculate the angles between the other two vectors    vector_0 = np.array(bl-tl)    vector_1 = np.array(rightMost[0]-tl)    vector_2 = np.array(rightMost[1]-tl)    angle = [np.arccos(cos_dist(vector_0, vector_1)), np.arccos(cos_dist(vector_0, vector_2))]    (br, tr) = rightMost[np.argsort(angle), :]    # return the coordinates in top-left, top-right,    # bottom-right, and bottom-left order    return np.array([tl, tr, br, bl], dtype="float32")

参考：

http://www.pyimagesearch.com/2016/03/21/ordering-coordinates-clockwise-with-python-and-opencv/

https://www.coder4.com/archives/3826