Kinh Nghiệm Hướng dẫn Generate all sublists of a list Python Chi Tiết

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Find all possible sublists of a list

Let’s say I have the following list

[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18]

I want to find all possible sublists of a certain lenght where they don’t contain one certain number and without losing the order of the numbers.

For example all possible sublists with length 6 without the 12 are:

[1,2,3,4,5,6]
[2,3,4,5,6,7]
[3,4,5,6,7,8]
[4,5,6,7,8,9]
[5,6,7,8,9,10]
[6,7,8,9,10,11]
[13,14,15,16,17,18]

The problem is that I want to do it in a very big list and I want the most quick way.

Update with my method:

oldlist = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18]
newlist = []
length = 6
exclude = 12
for i in oldlist:
if length+i>len(oldlist):
break
else:
mylist.append(oldlist[i:(i+length)]
for i in newlist:
if exclude in i:
newlist.remove(i)

I know it’s not the best method, that’s why I need a better one.

A straightforward, non-optimized solution would be

result = [sublist for sublist in
(lst[x:x+size] for x in range(len(lst) – size + 1))
if item not in sublist
]

An optimized version:

result = []
start = 0
while start < len(lst):
try:
end = lst.index(item, start + 1)
except ValueError:
end = len(lst)
result.extend(lst[x+start:x+start+size] for x in range(end – start – size + 1))
start = end + 1

Use itertools.combinations:

import itertools
mylist = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18]
def contains_sublist(lst, sublst):
n = len(sublst)
return any((sublst == lst[i:i+n]) for i in xrange(len(lst)-n+1))
print [i for i in itertools.combinations(mylist,6) if 12 not in i and contains_sublist(mylist, list(i))]

Prints:

[(1, 2, 3, 4, 5, 6), (2, 3, 4, 5, 6, 7), (3, 4, 5, 6, 7, 8), (4, 5, 6, 7, 8, 9), (5, 6, 7, 8, 9, 10), (6, 7, 8, 9, 10, 11), (13, 14, 15, 16, 17, 18)]

I like to build solutions out of small composable parts. A few years of writing Haskell does that to you. So I’d do it like this…

First, this will return an iterator over all sublists, in ascending order of length, starting with the empty list:

from itertools import chain, combinations
def all_sublists(l):
return chain(*(combinations(l, i) for i in range(len(l) + 1)))

Generally we’re discouraged from using single-letter variable names, but I think that in short bursts of highly abstract code it’s a perfectly reasonable thing to do.

(BTW to omit the empty list, use range(1, len(l) + 1) instead.)

Then we can solve your problem in general by adding your criteria:

def filtered_sublists(input_list, length, exclude):
return (
l for l in all_sublists(input_list)
if len(l) == length and exclude not in l
)

So for example:

oldlist = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18]
length = 6
exclude = 12
newlist = filtered_sublists(old_list, length, exclude)

The simplest way I can think of would be to remove the excluded number from the list and then use itertools.combinations() to generate the desired sublists, This has the added advantage that it will produce the sublists iteratively.

from itertools import combinations
def combos_with_exclusion(lst, exclude, length):
for combo in combinations((e for e in lst if e != exclude), length):
yield list(combo)
mylist = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18]
for sublist in combos_with_exclusion(mylist, 12, 6):
print sublist

Output:

[1, 2, 3, 4, 5, 6]
[1, 2, 3, 4, 5, 7]
[1, 2, 3, 4, 5, 8]
[1, 2, 3, 4, 5, 9]
[1, 2, 3, 4, 5, 10]
[1, 2, 3, 4, 5, 11]
[1, 2, 3, 4, 5, 13]

[11, 14, 15, 16, 17, 18]
[13, 14, 15, 16, 17, 18]

I have an answer but i think it is not the best:

It will get the result but cause it will have much same sublist so it need a lot memory and a lot time. I think it is not a good way.

My attempt at recursively creating all possible list of lists. The depth parameter just takes the number of items to remove from each list. This is not a sliding window.

Code:

def sublists(input, depth):
output= []
if depth > 0:
for i in range(0, len(input)):
sub= input[0:i] + input[i+1:]
output += [sub]
output.extend(sublists(sub, depth-1))
return output

Examples (typed interactively into python3):

sublists([1,2,3,4],1)

[[2, 3, 4], [1, 3, 4], [1, 2, 4], [1, 2, 3]]

sublists([1,2,3,4],2)

[[2, 3, 4], [3, 4], [2, 4], [2, 3], [1, 3, 4], [3, 4], [1, 4], [1, 3], [1, 2, 4], [2, 4], [1, 4], [1, 2], [1, 2, 3], [2, 3], [1, 3], [1, 2]]

sublists([1,2,3,4],3)

[[2, 3, 4], [3, 4], [4], [3], [2, 4], [4], [2], [2, 3], [3], [2], [1, 3, 4], [3, 4], [4], [3], [1, 4], [4], [1], [1, 3], [3], [1], [1, 2, 4], [2, 4], [4], [2], [1, 4], [4], [1], [1, 2], [2], [1], [1, 2, 3], [2, 3], [3], [2], [1, 3], [3], [1], [1, 2], [2], [1]]

Some edge cases:

sublists([1,2,3,4],100)

[[2, 3, 4], [3, 4], [4], [3], [2, 4], [4], [2], [2, 3], [3], [2], [1, 3, 4], [3, 4], [4], [3], [1, 4], [4], [1], [1, 3], [3], [1], [1, 2, 4], [2, 4], [4], [2], [1, 4], [4], [1], [1, 2], [2], [1], [1, 2, 3], [2, 3], [3], [2], [1, 3], [3], [1], [1, 2], [2], [1]]

sublists([], 1)

[]

NOTE: the output list of lists includes duplicates.

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