Cartesian products of lists without duplicates

Question

Given an array a=['a','b','c'], how would you go about returning the Cartesian product of the array without duplicates. Example:

[['a', 'a' , 'a' ,'a']
['a', 'a' , 'a' ,'b']
['a', 'a' , 'a' ,'c']
['a', 'a' , 'b' ,'b']
['a', 'a' , 'b' ,'c']
['a', 'a' , 'c' ,'c']
...etc..]

Following How to generate all permutations of a list in Python, I tried :

print list(itertools.permutations(['a', 'b' , 'c'], 4))
[]

print list(itertools.product(['a', 'b' , 'c'], repeat=4)

But I get the Cartesian product with duplicates. For example the list will contain both ['a','a','b','b'] and ['a','b','b','a'] which are clearly the equal.

Note: my 'a','b','c' are variables which store numbers say 1,2,3. So after getting the list of combinations of the letters, I would need to: say,

['a','b','c','c'] ----> a*b*c*c = 1*2*3*3 = 18

What is the fastest way of doing this in python? Would it be possible/faster to do it with numpy?? Thanks!

Answer 1

Maybe you actually want combinations_with_replacement?

>>> from itertools import combinations_with_replacement
>>> a = ['a', 'b', 'c']
>>> c = combinations_with_replacement(a, 4)
>>> for x in c:
...     print x
...     
('a', 'a', 'a', 'a')
('a', 'a', 'a', 'b')
('a', 'a', 'a', 'c')
('a', 'a', 'b', 'b')
('a', 'a', 'b', 'c')
('a', 'a', 'c', 'c')
('a', 'b', 'b', 'b')
('a', 'b', 'b', 'c')
('a', 'b', 'c', 'c')
('a', 'c', 'c', 'c')
('b', 'b', 'b', 'b')
('b', 'b', 'b', 'c')
('b', 'b', 'c', 'c')
('b', 'c', 'c', 'c')
('c', 'c', 'c', 'c')

Without more information about how you're mapping strings to numbers I can't comment on your second question, but writing your own product function or using numpy's isn't too difficult.