如何确定哪些点在多边形内而哪些不在多边形内(大量点)? [英] How to determine which points are inside of a polygon and which are not (large number of points)?

问题描述

我有一大组数据点(100,000+)存储在二维numpy数组中(第一列:x坐标，第二列:y坐标).我还有几个一维数组，用于存储每个数据点的其他信息.我现在想从这些一维数组的子集中创建图，这些图仅包含给定多边形中的点.

I've got a large set of data points (100,000+) stored in a 2-dimensional numpy array (1st column: x coordinates, 2nd column: y coordinates). I've also got several 1-dimensional arrays storing additional information for each data point. I'd now like to create plots from subsets of these 1D arrays which include only the points which are in a given polygon.

我想出了以下既不优雅也不快捷的解决方案:

I came up with the following solution which is neither elegant nor fast:

#XY is the 2D array.
#A is one of the 1D arrays.
#poly is a matplotlib.patches.Polygon

mask = np.array([bool(poly.get_path().contains_point(i)) for i in XY])

matplotlib.pylab.hist(A[mask], 100)
matplotlib.pylab.show()

能否请您帮助我改进此代码?我尝试使用np.vectorize而不是列表理解，但是无法使其正常工作.

Could you please help me to improve this code? I tried playing around with np.vectorize instead of the list comprehension but could not manage to get it to work.

推荐答案

使用

Use matplotlib.nxutils.points_inside_poly, which implements a very efficient test.

在 更新:请注意，自1.2.0版的matplotlib开始不推荐使用points_inside_poly.请改用 matplotlib.path.Path.contains_points

Update: Note that points_inside_poly is deprecated since version 1.2.0 of matplotlib. Use matplotlib.path.Path.contains_points instead.

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