All possible permutations of a set of lists in Python
Question:
In Python I have a list of n lists, each with a variable number of elements. How can I create a single list containing all the possible permutations:
For example
[ [ a, b, c], [d], [e, f] ]
I want
[ [a, d, e] , [a, d, f], [b, d, e], [b, d, f], [c, d, e], [c, d, f] ]
Note I don’t know n in advance. I thought itertools.product would be the right approach but it requires me to know the number of arguments in advance
Answers:
You can do it with a multi-level list comprehension:
>>> L1=['a','b','c']
>>> L2=['d']
>>> L3=['e','f']
>>> [[i,j,k] for i in L1 for j in L2 for k in L3]
[['a', 'd', 'e'], ['a', 'd', 'f'], ['b', 'd', 'e'], ['b', 'd', 'f'], ['c', 'd', 'e'], ['c', 'd', 'f']]
You don’t need to know n in advance to use itertools.product
>>> import itertools
>>> s=[ [ 'a', 'b', 'c'], ['d'], ['e', 'f'] ]
>>> list(itertools.product(*s))
[('a', 'd', 'e'), ('a', 'd', 'f'), ('b', 'd', 'e'), ('b', 'd', 'f'), ('c', 'd', 'e'), ('c', 'd', 'f')]
itertools.product works for me.
>>> l=[ [ 1, 2, 3], [4], [5, 6] ]
>>> list(itertools.product(*l))
[(1, 4, 5), (1, 4, 6), (2, 4, 5), (2, 4, 6), (3, 4, 5), (3, 4, 6)]
>>> l=[ [ 1, 2, 3], [4], [5, 6],[7,8] ]
>>> list(itertools.product(*l))
[(1, 4, 5, 7), (1, 4, 5, 8), (1, 4, 6, 7), (1, 4, 6, 8), (2, 4, 5, 7), (2, 4, 5, 8), (2, 4, 6, 7), (2, 4, 6, 8), (3, 4, 5, 7), (3, 4, 5, 8), (3, 4, 6,
7), (3, 4, 6, 8)]
>>>
In Python I have a list of n lists, each with a variable number of elements. How can I create a single list containing all the possible permutations:
For example
[ [ a, b, c], [d], [e, f] ]
I want
[ [a, d, e] , [a, d, f], [b, d, e], [b, d, f], [c, d, e], [c, d, f] ]
Note I don’t know n in advance. I thought itertools.product would be the right approach but it requires me to know the number of arguments in advance
You can do it with a multi-level list comprehension:
>>> L1=['a','b','c']
>>> L2=['d']
>>> L3=['e','f']
>>> [[i,j,k] for i in L1 for j in L2 for k in L3]
[['a', 'd', 'e'], ['a', 'd', 'f'], ['b', 'd', 'e'], ['b', 'd', 'f'], ['c', 'd', 'e'], ['c', 'd', 'f']]
You don’t need to know n in advance to use itertools.product
>>> import itertools
>>> s=[ [ 'a', 'b', 'c'], ['d'], ['e', 'f'] ]
>>> list(itertools.product(*s))
[('a', 'd', 'e'), ('a', 'd', 'f'), ('b', 'd', 'e'), ('b', 'd', 'f'), ('c', 'd', 'e'), ('c', 'd', 'f')]
itertools.product works for me.
>>> l=[ [ 1, 2, 3], [4], [5, 6] ]
>>> list(itertools.product(*l))
[(1, 4, 5), (1, 4, 6), (2, 4, 5), (2, 4, 6), (3, 4, 5), (3, 4, 6)]
>>> l=[ [ 1, 2, 3], [4], [5, 6],[7,8] ]
>>> list(itertools.product(*l))
[(1, 4, 5, 7), (1, 4, 5, 8), (1, 4, 6, 7), (1, 4, 6, 8), (2, 4, 5, 7), (2, 4, 5, 8), (2, 4, 6, 7), (2, 4, 6, 8), (3, 4, 5, 7), (3, 4, 5, 8), (3, 4, 6,
7), (3, 4, 6, 8)]
>>>