CODEWITHSUNDEEP
pythoncombinationscombinatoricspython-itertools
David Pekker
David Pekker

Reputation: 353

getting all possible combinations of a list in a form of sublists

I wonder if someone can help with the following task: What is the way to get all combinations a list can be split into sublists, when order does not matter?

Let's say I have a list of 4 items:

import itertools as it

a = [1, 2, 3, 4]
print(list(it.combinations(a, 2)))

That will give me a list of 6 possible pairs:

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

How to make (out of it? or any other way) a set of lists that contain original [1, 2, 3, 4] sequence in any order? So for this example it will contain three sublists:

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

UPDATE: A small clarification: In other words, I need to get all sets of n-tuples such that their members contain all the population of original list, when the order within an n-tuple does not matter. Thus [(1, 2), (3, 4)] is ok, but [(2, 1), (3, 4)] is not needed because it is the same as the first set, if we ignore the order.

UPDATE2: So for the list of length 6, and for chunks of size 2 this fun function should work as follows:

import itertools as it
a = [1, 2, 3, 4, 5, 6,]
r = 2

# fun(a,r):
# OUT:
# [
#    (1, 2), (3, 4), (5, 6)
#    (1, 3), (2, 4), (5, 6),
#    (1, 4), (2, 3), (5, 6),
#    (1, 5), (2, 3), (4, 6),
#    (1, 6), (2, 3), (4, 5),
#  ]

Upvotes: 6

Views: 2040

Answers (3)

Raviteja Ainampudi
Raviteja Ainampudi

Reputation: 282

a = [1, 2, 3, 4]
choices = []
for i in range(len(a)):
  for j in range(i+1, len(a)):
      sub_a1 = (a[i], a[j])
      sub_a2 = tuple(j for j in a if j not in sub_a1)
      sub_a = sorted([sub_a1, sub_a2])
      if sub_a not in choices:
         choices.append(sub_a)
for k in choices:
  print(k)

May be this code can help your requirements.

Upvotes: 0

aydow
aydow

Reputation: 3801

As you require the "remainder" of what is in a, you can use a set difference

In [901]: [(c, tuple(set(a) - set(c))) for c in it.combinations(a, 2)]
Out[901]:
[((1, 2), (3, 4)),
 ((1, 3), (2, 4)),
 ((1, 4), (2, 3)),
 ((2, 3), (1, 4)),
 ((2, 4), (1, 3)),
 ((3, 4), (1, 2))]

However, this gives you twice as many results as you require so to eliminate these duplicates we need to sort and setify the result set which will give us a set of lists which can't be turned into sets. You should then cast the lists to tuples.

In [900]: {tuple(sorted((c, tuple(set(a) - set(c))))) for c in it.combinations(a, 2)}
Out[900]: {((1, 2), (3, 4)), ((1, 3), (2, 4)), ((1, 4), (2, 3))}

If you require a list of list of tuple then you can use the following

In [902]: s = {tuple(sorted((c, tuple(set_a - set(c))))) for c in it.combinations(a, 2)}

In [903]: [list(l) for l in s]
Out[903]: [[(1, 3), (2, 4)], [(1, 4), (2, 3)], [(1, 2), (3, 4)]]

Upvotes: 0

Sunitha
Sunitha

Reputation: 12025

Just zip the combinations, with its reverse and take only the first half of the resulting list

>>> import itertools as it
>>> lst = [1, 2, 3, 4]
>>> r = len(lst)//2
>>> combs = list(it.combinations(lst, r))
>>> list(it.islice(zip(combs, reversed(combs)), len(combs)//2))
[((1, 2), (3, 4)), ((1, 3), (2, 4)), ((1, 4), (2, 3))]

Upvotes: 2

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