（4,1,2）numpy的数组排序顺时针

本文介绍了（4,1,2）numpy的数组排序顺时针的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我有一个numpy的数组如下：

I have a numpy array as the following:

my_array = np.float32([[[ 323. , 143.]], [[ 237. , 143.]], [[ 227. , 230.]], [[ 318. , 233.]]])

这4个点重新present摆在一个图像上的矩形的顶点，我需要顺时针重新排序，并将其保存到一个新的NP阵列（顶部左>右上 - >下 - 右键 - >底部 - 左）。在我的例子将是：

This 4 points represent the vertices of a rectangle that lies on a image, I need to reorder them clockwise and save it to a new np array, (top-left-> top-right -> bottom - right -> bottom - left). In my example it will be:

[237, 143] -> [323, 143] -> [318, 233] -> [227, 230]

我已阅读这但我对numpy的技能是不是好实现它...

I have read this but my skills on numpy aren't as good to implement it...

谢谢！

推荐答案

您可以做这样的事情 -

You could do something like this -

import numpy as np from scipy.spatial import distance def sortpts_clockwise(A): # Sort A based on Y(col-2) coordinates sortedAc2 = A[np.argsort(A[:,1]),:] # Get top two and bottom two points top2 = sortedAc2[0:2,:] bottom2 = sortedAc2[2:,:] # Sort top2 points to have the first row as the top-left one sortedtop2c1 = top2[np.argsort(top2[:,0]),:] top_left = sortedtop2c1[0,:] # Use top left point as pivot & calculate sq-euclidean dist against # bottom2 points & thus get bottom-right, bottom-left sequentially sqdists = distance.cdist(top_left[None], bottom2, 'sqeuclidean') rest2 = bottom2[np.argsort(np.max(sqdists,0))[::-1],:] # Concatenate all these points for the final output return np.concatenate((sortedtop2c1,rest2),axis =0)

样的输入，输出 -

Sample input, output -

In [85]: A Out[85]: array([[ 281., 147.], [ 213., 170.], [ 239., 242.], [ 307., 219.]], dtype=float32) In [86]: sortpts_clockwise(A) Out[86]: array([[ 213., 170.], [ 281., 147.], [ 307., 219.], [ 239., 242.]], dtype=float32)