I think the issue was by ths following , Using itertools.product: code :
>>> import itertools
>>> a = [0, 0, [0, 1], [0, 1, 2]]
>>> a2 = [x if isinstance(x, list) else [x] for x in a]
>>> # = [[0], [0], [0, 1], [0, 1, 2]]
>>> list(itertools.product(*a2))
[(0, 0, 0, 0), (0, 0, 0, 1), (0, 0, 0, 2), (0, 0, 1, 0), (0, 0, 1, 1), (0, 0, 1, 2)]
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permutations of two lists in python
By : user1641709
Date : March 29 2020, 07:55 AM
may help you . I have two lists like: code :
>>> import itertools
>>> map(''.join, itertools.chain(itertools.product(list1, list2), itertools.product(list2, list1)))
['squarered', 'squaregreen', 'circlered',
'circlegreen', 'trianglered', 'trianglegreen',
'redsquare', 'redcircle', 'redtriangle', 'greensquare',
'greencircle', 'greentriangle']

permutations of lists python
By : Jose Perez
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , I think you want to use the itertools module, which is part of the standard Python distribution. For example: code :
import itertools
a = ["1"]
b = ["0"]
c = ["a","b","c"]
d = ["d","e","f"]
for item in itertools.product(a, b, c, d):
print(item)

permutations of a lists python
By : qoda
Date : March 29 2020, 07:55 AM
wish of those help So I posted this question here. , Unpacking everything from a list using *: code :
>>> import itertools
>>> a = ["1"]
>>> b = ["0"]
>>> c = ["a","b","c"]
>>> d = ["d","e","f"]
>>> lists = [a,b,c,d]
>>> for item in itertools.product(*lists):
print item
('1', '0', 'a', 'd')
('1', '0', 'a', 'e')
('1', '0', 'a', 'f')
('1', '0', 'b', 'd')
('1', '0', 'b', 'e')
('1', '0', 'b', 'f')
('1', '0', 'c', 'd')
('1', '0', 'c', 'e')
('1', '0', 'c', 'f')

Permutations on 2D lists (Python)
By : William Menant
Date : March 29 2020, 07:55 AM
Does that help I might be wrong, but from what I understand, you want to do the cartesian product of several ensembles (whose element are themselves ensembles) : [[(1, 2)], [(1, 2),(1, 3)]] X [[(2, 3)]] code :
from itertools import product, permutations, chain
L = [ [[(1, 2)], [(1, 2),(1, 3)]], [[(2, 3)]] ]
for element in product(*L):
for permutation in permutations(element):
print(list(chain(*permutation)))
[(1, 2), (2, 3)]
[(2, 3), (1, 2)]
[(1, 2), (1, 3), (2, 3)]
[(2, 3), (1, 2), (1, 3)]
result = [list(chain(*permutation)) for element in product(*L) for permutation in permutations(element)]
[[(1, 2), (2, 3)],
[(2, 3), (1, 2)],
[(1, 2), (1, 3), (2, 3)],
[(2, 3), (1, 2), (1, 3)]]

2 lists permutations in python
By : NiTiN
Date : March 29 2020, 07:55 AM
will help you Here is the code with comments as explanation line by line: Use isPermutation as a helper method in your is_list_permutation function. This makes the code easier to read and cleaner. code :
L1 = [1, 'b', 1, 'c', 'c', 1]
L2 = ['c', 1, 'b', 1, 1, 'c']
def isPermutation(list1,list2):
if len(list1) != len(list2):
return False; #two list does not have same length so impossible being permutation of each other
for i in range(0, len(list1)):
if list1.count(list1[i]) != list2.count(list1[i]):
return False
def is_list_permutation(list1,list2):
if (isPermutation(list1,list2) == False): #use the above method isPermutation to check if they are permutation of each other
return False #if not return false
elif not list1:
return (None, None, None)
else:
mostOccurItem = max(set(list1), key=list1.count)
numberOfTimes = list1.count(mostOccurItem)
theType = type(mostOccurItem)
return (mostOccurItem, numberOfTimes, theType)
print(is_list_permutation(L1,L2))

