删除列表中连续整数的最快，更有效的方法是什么？

Question

If I have a list like: infs = [0, 19, 20, 21, 24] I'd like to remove consecutive values but leave the first only from this group, so here I expect a result: infs = [0, 19, 24]

My attempts:

 for k,(i,j) in enumerate(zip(infs, infs[1:])):
        print(k,i,j)
        if j-i == 1:
            del infs[k+1]

It leaves '21' because it was deleted, so this is bad idea to remove it in loop.

Answer 1

You can pair adjacent items in the list by zipping the list with itself but with a padding of its first item, so that you can use a list comprehension that filters out adjacent pairs that differ by just 1:

[b for a, b in zip(infs[:1] + infs, infs) if b - a != 1]

This returns:

[0, 19, 24]

Answer 2

You can use itertools.groupby over the enumeration of the given list, with a key function that returns the difference between the number and its index:

from itertools import groupby
[next(g)[1] for _, g in groupby(enumerate(infs), lambda t: t[1] - t[0])]

This returns:

[0, 19, 24]

Answer 3

You simply can do:

infs=[0, 19, 20, 21, 24]
[v for (i, v) in enumerate(infs) if i==0 or v - infs[i-1] != 1]

[0, 19, 24]