如何根据相邻元素之间的差异将列表拆分为列表列表？

Question

Eg, if I've got

MAX_ALLOWED_DIFF = 3
nums=[1, 2, 4, 10, 13, 2, 5, 5, 5]

the output should be

groups = [[1, 2, 4], [10, 13], [2, 5, 5, 5]]

The context: I had a pandas.Series object nums and I used

nums = nums.diff().gt(DETECTION_MAX_DIFF_NS).cumsum()).apply(list).tolist()

to subsample in the same fashion but I noticed that there're a lot of duplicates in my Series nums and after I use .unique() method I don't have Series object anymore, I've got numpy.ndarray (1D) instead.

I believe I may use sth like pandas.Series(nums.unique) but I don't like this hack.

Answer 1

Here's one approach -

>>> import numpy as np
>>> idx = np.r_[0,np.flatnonzero(np.abs(np.diff(nums))>MAX_ALLOWED_DIFF)+1,len(nums)]
>>> [nums[i:j] for (i,j) in zip(idx[:-1],idx[1:])]
[[1, 2, 4], [10, 13], [2, 5, 5, 5]]

Answer 2

So we using drop_duplicates , keep nums stay in pd.Series

nums=nums.drop_duplicates()
nums.groupby(nums.diff().abs().gt(MAX_ALLOWED_DIFF).cumsum()).apply(list).tolist()
Out[447]: [[1, 2, 4], [10, 13], [5]]

Answer 3

Given that you've tagged with numpy too, here's one way to do it:

thr = 3
ix = np.flatnonzero(np.concatenate([[False], np.abs(np.diff(nums))>thr]))
np.split(nums, ix)

[array([1, 2, 4]), array([10, 13]), array([2, 5, 5, 5])]