问题
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]))
回答1:
You'd have to implement it yourself. Here is the basic idea:
def remove_sublist(lst, sub):
i = 0
out = []
while i < len(lst):
if lst[i:i+len(sub)] == sub:
i += len(sub)
else:
out.append(lst[i])
i += 1
return out
This steps along every element of the original list and adds it to an output list if it isn't a member of the subset. This version is not very efficient, but it works like the string example you provided, in the sense that it creates a new list not containing your subset. It also works for arbitrary element types as long as they support ==. Removing [1,1,1] from [1,1,1,1] will correctly result in [1], as for a string.
Here is an IDEOne link showing off the result of
>>> remove_sublist([1, 'a', int, 3, float, 'a', int, 5], ['a', int])
[1, 3, <class 'float'>, 5]
回答2:
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]
回答3:
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]
回答4:
A improved version to check whether lst[i:i+len(sub)] < len(lst)
def remove_sublist(lst, sub):
i = 0
out = []
sub_len = len(sub)
lst_len = len(lst)
while i < lst_len:
if (i+sub_len) < lst_len:
if lst[i: i+sub_len] == sub:
i += sub_len
else:
out.append(lst[i])
i += 1
else:
out.append(lst[i])
i += 1
return out
回答5:
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.
回答6:
Use itertools.zip_longest to create n element tuples (where n is length of sub_list) and then filter the current element and next n-1 elements when one of the element matched the sub_list
>>> from itertools import zip_longest, islice
>>> itr = zip_longest(*(big_list[i:] for i in range(len(sub_list))))
>>> [sl[0] for sl in itr if not (sl == tuple(sub_list) and next(islice(itr, len(sub_list)-2, len(sub_list)-1)))]
[2, 3, 4]
To improve the efficiency, you can calculate tuple(sub_list) and len(sub_list) before hand you start filtering
>>> l = len(sub_list)-1
>>> tup = tuple(sub_list)
>>> [sl[0] for sl in itr if not (sl == tup and next(islice(itr, l-1, l)))]
[2, 3, 4]
回答7:
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).
回答8:
More readable than any above and with no additional memory footprint:
def remove_sublist(sublist, mainlist):
cursor = 0
for b in mainlist:
if cursor == len(sublist):
cursor = 0
if b == sublist[cursor]:
cursor += 1
else:
cursor = 0
yield b
for i in range(0, cursor):
yield sublist[i]
This is for onliner if you wanted a function from library, let it be this
[x for x in remove_sublist([1, 2], [2, 1, 2, 3, 1, 2, 4])]
回答9:
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.
回答10:
(For final approach, see last code snippet)
I'd have thought a simple string conversion would be sufficient:
big_list = [2, 1, 2, 3, 1, 2, 4]
sub_list = [1, 2]
new_list = list(map(int, list((''.join(map(str, big_list))).replace((''.join(map(str, sub_list))), ''))))
I'm essentially doing a find/replace with the string equivalents of the lists. I'm mapping them to integers afterwards so that the original types of the variables are retained. This will work for any size of the big and sub lists.
However, it's likely that this won't work if you're calling it on arbitrary objects if they don't have a textual representation. Moreover, this method results in only the textual version of the objects being retained; this is a problem if the original data types need to be maintained.
For this, I've composed a solution with a different approach:
new_list = []
i = 0
while new_list != big_list:
if big_list[i:i+len(sub_list)] == sub_list:
del big_list[i:i+len(sub_list)]
else:
new_list.append(big_list[i])
i += 1
Essentially, I'm removing every duplicate of the sub_list when I find them and am appending to the new_list when I find an element which isn't part of a duplicate. When the new_list and big_list are equal, all of the duplicates have been found, which is when I stop. I haven't used a try-except as I don't think there should be any indexing errors.
This is similar to @MadPhysicist's answer and is of roughly the same efficiency, but mine consumes less memory.
This second approach will work for any type of object with any size of lists and therefore is much more flexible than the first approach. However, the first approach is quicker if your lists are just integers.
However, I'm not done yet! I've concocted a one-liner list comprehension which has the same functionality as the second approach!
import itertools
new_list = [big_list[j] for j in range(len(big_list)) if j not in list(itertools.chain.from_iterable([ list(range(i, i+len(sub_list))) for i in [i for i, x in enumerate(big_list) if x == sub_list[0]] if big_list[i:i+len(sub_list)] == sub_list ]))]
Initially, this seems daunting, but I assure you it's quite simple! First, i create a list of the indices where the first element of the sublist has occured. Next, for each of these indices, I check if the following elements form the sublist. If they do, the range of indices which form the duplicate of the sublist is added to another list. Afterwards, I use a function from itertools to flatten the resulting list of lists. Every element in this flattened list is an index which is in a duplicate of the sublist. Finally, I create a new_list which consists of every element of the big_list which has an index not found in the flattened list.
I don't think that this method is in any of the other answers. I like it the most as it's quite neat once you realise how it works and is very efficient (due to the nature of list comprehensions).
回答11:
You can use recursion with a generator:
def remove(d, sub_list):
if d[:len(sub_list)] == sub_list and len(sub_list) <= len(d[:len(sub_list)]):
yield from [[], remove(d[len(sub_list):], sub_list)][bool(d[len(sub_list):])]
else:
yield d[0]
yield from [[], remove(d[1:], sub_list)][bool(d[1:])]
tests = [[[2, 1, 2, 3, 1, 2, 4], [1, 2]], [[1, 2, 1, 2], [1, 2]], [[1, 'a', int, 3, float, 'a', int, 5], ['a', int]], [[1, 1, 1, 1, 1], [1,1,1]]]
for a, b in tests:
print(list(remove(a, b)))
Output:
[2, 3, 4]
[]
[1, 3, <class 'float'>, 5]
[1, 1]
回答12:
Just for fun, here is the closest approximation to a one-liner:
from functools import reduce
big_list = [2, 1, 2, 3, 1, 2, 4]
sub_list = [1, 2]
result = reduce(lambda r, x: r[:1]+([1]+r[2:-r[1]],[min(len(r[0]),r[1]+1)]+r[2:])[r[-r[1]:]!=r[0]]+[x], big_list+[0], [sub_list, 1])[2:-1]
Don't trust that it works? Check it on IDEone!
Of course it's far from efficient and is disgustfully cryptic, however it should help to convince the OP to accept @Mad Physicist's answer.
回答13:
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]
来源：https://stackoverflow.com/questions/51518601/how-to-remove-every-occurrence-of-sub-list-from-list