基于Z值~Python的分裂矢量三维阵列

Question

我正在制作C4D插件并且想知道：如何使用groupby根据包含的值拆分数组？

我将此代码发送到splinebuilder函数，但如果我按原样发送样条列表，我会在行之间建立连接，这是不想要的。

所以我需要发送单独的列表，以便 splinebuilder 创建这些行的集合，避免互连。

#initial contains XYZ vectors

splinelist = [(-200, 0, -200), (0, 0, -200), (200, 0, -200), (-200, 0, 0), (0, 0, 0), (200, 0, 0), (-200, 0, 200), (0, 0, 200), (200, 0, 200)]


#desired handcoded for example separated based on Z value

la = [(-200, 0, -200), (0, 0, -200), (200, 0, -200)]
lb = [(-200, 0, 0), (0, 0, 0), (200, 0, 0)]
lc = [(-200, 0, 200), (0, 0, 200), (200, 0, 200)]


#desired list of lists based on Z value

desiredlist = [[(-200, 0, -200), (0, 0, -200), (200, 0, -200)],[(-200, 0, 0), (0, 0, 0), (200, 0, 0)],[(-200, 0, 200), (0, 0, 200), (200, 0, 200)]]

Answer 1

我认为这可以满足您的需求：

import itertools
splinelist = [(-200, 0, -200), (0, 0, -200), (200, 0, -200), (-200, 0, 0), (0, 0, 0), (200, 0, 0), (-200, 0, 200), (0, 0, 200), (200, 0, 200)]
grouped = itertools.groupby(splinelist, lambda x : x[2])
desiredlist  = [list(group) for key, group in grouped]
print(desiredlist)

输出：

[[(-200, 0, -200), (0, 0, -200), (200, 0, -200)], [(-200, 0, 0), (0, 0, 0), (200, 0, 0)], [(-200, 0, 200), (0, 0, 200), (200, 0, 200)]]