I have a list of numbers:
lst = [1, 2, 3, 1,4]
def permutation(lst):
# If lst is empty then there are no permutations
if len(lst) == 0:
return []
# If there is only one element in lst then, only
# one permuatation is possible
if len(lst) == 1:
return [lst]
# Find the permutations for lst if there are
# more than 1 characters
l = [] # empty list that will store current permutation
# Iterate the input(lst) and calculate the permutation
for i in range(len(lst)):
m = lst[i]
# Extract lst[i] or m from the list. remLst is
# remaining list
remLst = lst[:i] + lst[i + 1:]
# Generating all permutations where m is first
# element
for p in permutation(remLst):
l.append([m] + p)
return l
if __name__ == "__main__":
lst = [1, 2, 3, 1,4]
v_out = permutation(lst)
print(v_out)
I am only getting permutations of 4 length, I want permutatins of all lengths, and only distinct permutations. But within each permutation, repetition is allowed.
This should work... Using the permutations function from itertools and making a set out of everything to prevent duplicates from being added to the overall result
In [20]: from itertools import permutations
In [21]: a = [1, 1, 2, 3]
In [22]: all_results = set()
In [23]: for i in range(1, len(a)):
...: all_results.update(set(permutations(a, i)))
...:
In [24]: all_results
Out[24]:
{(1,),
(1, 1),
(1, 1, 2),
(1, 1, 3),
(1, 2),
(1, 2, 1),
(1, 2, 3),
(1, 3),
(1, 3, 1),
(1, 3, 2),
(2,),
(2, 1),
(2, 1, 1),
(2, 1, 3),
(2, 3),
(2, 3, 1),
(3,),
(3, 1),
(3, 1, 1),
(3, 1, 2),
(3, 2),
(3, 2, 1)}
In [25]: