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 !
注意，这个方法仅适用于凸多边形。
from scipy.spatial import distance as dist
import numpy as np
import math
def 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 rectangle
def 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