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    copied!<p>You can use scipy.spatial.cKDTree to speedup this kind of calculation.</p> <p>Here is the code:</p> <pre><code>import time import numpy as np #### create some sample data #### np.random.seed(1) V_NUM = 6000 P_NUM = 8000 size = 0.1 vertices = np.random.rand(V_NUM, 3) points = np.random.rand(P_NUM, 3) numpoints = np.zeros(V_NUM, np.int32) #### brute force #### start = time.clock() for i in xrange(len(vertices)): cube=((vertices[i,0]&lt;= points[:,0]) &amp; (points[:,0]&lt;(vertices[i,0]+size)) &amp; (vertices[i,1]&lt;= points[:,1]) &amp; (points[:,1] &lt; (vertices[i,1]+size)) &amp; (vertices[i,2]&lt;= points[:,2]) &amp; (points[:,2] &lt; (vertices[i,2]+size)) ) numpoints[i]=len(points[cube]) print time.clock() - start #### KDTree #### from scipy.spatial import cKDTree center_vertices = vertices + [[size/2, size/2, size/2]] start = time.clock() tree_points = cKDTree(points) _, result = tree_points.query(center_vertices, k=100, p = np.inf, distance_upper_bound=size/2) numpoints2 = np.zeros(V_NUM, np.int32) for i, neighbors in enumerate(result): numpoints2[i] = np.sum(neighbors!=P_NUM) print time.clock() - start print np.all(numpoints == numpoints2) </code></pre> <ul> <li>Change the cube corner position to center position first. </li> </ul> <p><code>center_vertices = vertices + [[size/2, size/2, size/2]]</code></p> <ul> <li>Create a cKDTree from points</li> </ul> <p><code>tree_points = cKDTree(points)</code></p> <ul> <li>Do query, k is the the number of nearest neighbors to return, p=np.inf means maximum-coordinate-difference distance, distance_upper_bound is the max distance.</li> </ul> <p><code>_, result = tree_points.query(center_vertices, k=100, p = np.inf, distance_upper_bound=size/2)</code></p> <p>The output is:</p> <pre><code>2.04113164434 0.11087783696 True </code></pre> <p>If there are more then 100 points in a cube, you can check this by <code>neighbors[-1] == P_NUM</code> in the for loop, and do a k=1000 query for these vertices.</p>
 

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