Is there any `strip`-liked method for a list?

Question

The buildin strip method in python can strip padding substring that meet a custom condition easily. eg

"000011110001111000".strip("0")

will trim the padding zero on both side of the string, and return 11110001111.

I would like to find a similar function for a list. eg, for a given list

input = ["0", "0", "1", "1", "0", "0", "1", "0", "1", "0", "0", "0"]

the expect output will be

output = ["1", "1", "0", "0", "1", "0", "1"]

The items in the example input are over simplified, they might be any other python objects.

list comprehension will remove all the items, instead of the padding ones.

[i for i in input if i != "0"]

Answer 1

Use itertools.dropwhile from both ends:

from itertools import dropwhile

input_data = ["0", "0", "1", "1", "0", "0", "1", "0", "1", "0", "0", "0"]

def predicate(x):
    return x == '0'

result = list(dropwhile(predicate, list(dropwhile(predicate, input_data))[::-1]))[::-1]
result

Output:

['1', '1', '0', '0', '1', '0', '1']

Answer 2

No list method, but it's not hard to implement such a function: Scan for the desired indexes and then slice to them.

def strip_seq(predicate, xs):
    def scan(xs):
        return next((i for i, x in enumerate(xs) if not predicate(x)), 0)
    return xs[scan(xs) : -scan(reversed(xs)) or None]

xs = ["0", "0", "a", "1", "0", "0", "1", "0", "b", "0", "0", "0"]
print(strip_seq(lambda x: x=='0', xs))  # ['a', '1', '0', '0', '1', '0', 'b']

This should work on any of the sliceable sequence types, including strings and tuples.