列表编辑的高效算法

本文介绍了列表编辑的高效算法的处理方法，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3D点云，保存在两个列表中.具有5个点(x，y，z)的示例:(3,3,3)，(1、1,1)，(4、4、4)，(2、2、2)，(5、5、5 )->我的列表如下:

I have a 3D-point cloud, saved in two lists. Example with 5 points (x,y,z): (3,3,3), (1,1,1), (4,4,4), (2,2,2), (5,5,5) -> My lists looks like this:

z = [3, 1, 4, 2, 5] # the z values pts = [(3,3), (1,1), (4,4), (2,2), (5,5)] # the x and y values

现在我要消除z值大于3的所有值.

Now I want to eliminate all values where the z-value is higher than 3:

# what I want to receive: z = [3, 1, 2] pts = [(3,3), (1,1), (2,2)]

我的算法在这里:

k = -1 for i in range(len(z)): k += 1 if z[k] > h: z.pop(k) pts.pop(k) k -= 1

这完全返回了我想要的-但是它非常慢(对于> 100,000个值). 我考虑过先通过z.sort()对列表进行排序，然后对z = z [:index]进行排序-但是当我对z-list执行此操作时，我的pts-list仍未排序.即使我可以对两个列表进行排序，我也不必经过很长的loop来找到我的条件为true的索引吗? 有谁知道更有效的解决方案?

This returns me exactly what I want - but it's very slow (for >100,000 values). I thought about first sorting my list via z.sort() and then do z = z[:index] - but when I do this for my z-list then my pts-list is still unsorted. And even if I could get both lists sorted, doesn't I also have to go through a long loop to find the index where my condition is true? Does anyone knows a more efficient solution?

推荐答案

z, pts = zip(*[(z, pt) for z, pt in zip(z, pts) if z <= 3]) print z, pts

输出

(3, 1, 2) ((3, 3), (1, 1), (2, 2))