How to remove every occurrence of sub-list from li

Question

I have two lists:

big_list = [2, 1, 2, 3, 1, 2, 4]
sub_list = [1, 2]

I want to remove all sub_list occurrences in big_list.

result should be [2, 3, 4]

For strings you could use this:

'2123124'.replace('12', '')

But AFAIK this does not work for lists.

This is not a duplicate of Removing a sublist from a list since I want to remove all sub-lists from the big-list. In the other question the result should be [5,6,7,1,2,3,4].

Update: For simplicity I took integers in this example. But list items could be arbitrary objects.

Update2:

if big_list = [1, 2, 1, 2, 1] and sub_list = [1, 2, 1], I want the result to be [2, 1] (like '12121'.replace('121', ''))

Update3:

I don't like copy+pasting source code from StackOverflow into my code. That's why I created second question at software-recommendations: https://softwarerecs.stackexchange.com/questions/51273/library-to-remove-every-occurrence-of-sub-list-from-list-python

Update4: if you know a library to make this one method call, please write it as answer, since this is my preferred solution.

The test should pass this test:

def test_remove_sub_list(self):
    self.assertEqual([1, 2, 3], remove_sub_list([1, 2, 3], []))
    self.assertEqual([1, 2, 3], remove_sub_list([1, 2, 3], [4]))
    self.assertEqual([1, 3], remove_sub_list([1, 2, 3], [2]))
    self.assertEqual([1, 2], remove_sub_list([1, 1, 2, 2], [1, 2]))
    self.assertEquals([2, 1], remove_sub_list([1, 2, 1, 2, 1], [1, 2, 1]))
    self.assertEqual([], remove_sub_list([1, 2, 1, 2, 1, 2], [1, 2]))

Answer 1

Try del and slicing. The worst time complexity is O(N^2).

sub_list=['a', int]
big_list=[1, 'a', int, 3, float, 'a', int, 5]
i=0
while i < len(big_list):
    if big_list[i:i+len(sub_list)]==sub_list:
        del big_list[i:i+len(sub_list)]
    else:
        i+=1

print(big_list)

result:

[1, 3, <class 'float'>, 5]

Answer 2

How about this:

def remove_sublist(lst, sub):
    max_ind_sub = len(sub) - 1
    out = []
    i = 0
    tmp = []

    for x in lst:
        if x == sub[i]:
            tmp.append(x)
            if i < max_ind_sub: # partial match 
                i += 1
            else:  # found complete match
                i = 0
                tmp = []
        else:
            if tmp:  # failed partial match 
                i = 0
                out += tmp
            if x == sub[0]:  # partial match
                i += 1
                tmp = [x]
            else:
                out.append(x)

    return out

Performance:

lst = [2, 1, 2, 3, 1, 2, 4]
sub = [1, 2]
%timeit remove_sublist(lst, sub)  # solution of Mad Physicist
%timeit remove_sublist_new(lst, sub)
>>> 2.63 µs ± 112 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
>>> 1.77 µs ± 13.7 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)

Update

My first solution had a bug. Was able to fix it (updated my code above) but the method looks way more complicated now. In terms of performance it still does better than the solution from Mad Physicist on my local machine.

Answer 3

Kinda different approach in Python 2.x!

from more_itertools import locate, windowed
big_list = [1, 2, 1, 2, 1]
sub_list = [1, 2, 1]

"""
Fetching all starting point of indexes (of sub_list in big_list)
to be removed from big_list. 
"""

i = list(locate(windowed(big_list, len(sub_list)), pred=lambda x: x==tuple(sub_list)))

""" 
Here i comes out to be [0, 2] in above case. But index from 2 which 
includes 1, 2, 1 has last 1 from the 1st half of 1, 2, 1 so further code is
to handle this case.
PS: this won't come for-
big_list = [2, 1, 2, 3, 1, 2, 4]
sub_list = [1, 2]
as here i comes out to be [1, 4]
"""

# The further code.
to_pop = []
for ele in i:
    if to_pop:
        if ele == to_pop[-1]:
            continue
    to_pop.extend(range(ele, ele+len(sub_list)))

# Voila! to_pop consists of all the indexes to be removed from big_list.

# Wiping out the elements!
for index in sorted(to_pop, reverse=True):
    del big_list[index]

Note that you need to delete them in reverse order so that you don't throw off the subsequent indexes.

In Python3, signature of locate() will differ.

Answer 4

Update: The more_itertools library has released more_itertool.replace, a tool that solves this particular problem (see Option 3).

First, here are some other options that work on generic iterables (lists, strings, iterators, etc.):

Code

Option 1 - without libraries:

def remove(iterable, subsequence):
    """Yield non-subsequence items; sans libraries."""
    seq = tuple(iterable)
    subsequence = tuple(subsequence)
    n = len(subsequence)
    skip = 0

    for i, x in enumerate(seq):
        slice_ = seq[i:i+n]
        if not skip and (slice_ == subsequence):
            skip = n
        if skip:
            skip -= 1
            continue
        yield x   

Option 2 - with more_itertools

import more_itertools as mit


def remove(iterable, subsequence):
    """Yield non-subsequence items."""
    iterable = tuple(iterable)
    subsequence = tuple(subsequence)
    n = len(subsequence)
    indices = set(mit.locate(mit.windowed(iterable, n), pred=lambda x: x == subsequence))

    it_ = enumerate(iterable)
    for i, x in it_:
        if i in indices:
            mit.consume(it_, n-1)
        else:
            yield x

Demo

list(remove(big_list, sub_list))
# [2, 3, 4]

list(remove([1, 2, 1, 2], sub_list))
# []

list(remove([1, "a", int, 3, float, "a", int, 5], ["a", int]))
# [1, 3, float, 5]

list(remove("11111", "111"))
# ['1', '1']

list(remove(iter("11111"), iter("111")))
# ['1', '1']

---

Option 3 - with more_itertools.replace:

Demo

pred = lambda *args: args == tuple(sub_list)
list(mit.replace(big_list, pred=pred, substitutes=[], window_size=2))
# [2, 3, 4]

pred=lambda *args: args == tuple(sub_list)
list(mit.replace([1, 2, 1, 2], pred=pred, substitutes=[], window_size=2))
# []

pred=lambda *args: args == tuple(["a", int])
list(mit.replace([1, "a", int, 3, float, "a", int, 5], pred=pred, substitutes=[], window_size=2))
# [1, 3, float, 5]

pred=lambda *args: args == tuple("111")
list(mit.replace("11111", pred=pred, substitutes=[], window_size=3))
# ['1', '1']

pred=lambda *args: args == tuple(iter("111"))
list(mit.replace(iter("11111"), pred=pred, substitutes=[], window_size=3))
# ['1', '1']

---

Details

In all of these examples, we are scanning the main sequence with smaller window slices. We yield whatever is not found in the slice and skip whatever is in the slice.

Option 1 - without libraries

Iterate an enumerated sequence and evaluate slices of size n (the length of the sub-sequence). If the upcoming slice equals the sub-sequence, reset skip and yield the item. Otherwise, iterate past it. skip tracks how many times to advance the loop, e.g. sublist is of size n=2, so it skips twice per match.

Note, you can convert this option to work with sequences alone by removing the first two tuple assignments and replacing the iterable parameter with seq, e.g. def remove(seq, subsequence):.

Option 2 - with more_itertools

Indices are located for every matching sub-sequence in an iterable. While iterating an enumerated iterator, if an index is found in indices, the remaining sub-sequence is skipped by consuming the next n-1 elements from the iterator. Otherwise, an item is yielded.

Install this library via > pip install more_itertools.

Option 3 - with more_itertools.replace:

This tool replaces a sub-sequence of items defined in a predicate with substitute values. To remove items, we substitute an empty container, e.g. substitutes=[]. The length of replaced items is specified by the window_size parameter (this value is equal to the length of the sub-sequence).

Answer 5

What you are trying to achieve can be done by converting it into list of strings and after replacing again convert it to integer type.

In a single line you can do it like this

map(int,list(("".join(map(str, big_list))).replace("".join(map(str, sub_list)),'').replace(''.join((map(str, sub_list))[::-1]),'')))

Input

big_list = [1, 2, 1, 2, 1]
sub_list = [1, 2, 1]

Output

[2, 1]

Input

big_list = [2, 1, 2, 3, 1, 2, 4]
sub_list = [1, 2]

Ouput

[2, 3, 4]

Answer 6

A recursive approach:

def remove(lst, sub):
    if not lst:
        return []
    if lst[:len(sub)] == sub:
        return remove(lst[len(sub):], sub)
    return lst[:1] + remove(lst[1:], sub)
print(remove(big_list, sub_list))

This outputs:

[2, 3, 4]