python – Get the cartesian product of a series of lists?

The Question :

355 people think this question is useful

How can I get the Cartesian product (every possible combination of values) from a group of lists?

Input:

somelists = [
   [1, 2, 3],
   ['a', 'b'],
   [4, 5]
]

Desired output:

[(1, 'a', 4), (1, 'a', 5), (1, 'b', 4), (1, 'b', 5), (2, 'a', 4), (2, 'a', 5) ...]

The Question Comments :
  • be aware that ‘every possible combination’ is not quite the same as ‘Cartesian product’, since in Cartesian products, duplicates are allowed.
  • Is there a non duplicate version of cartesian product?
  • @KJW Yes, set(cartesian product)
  • There should be no duplicates in a Cartesian product, unless the input lists contain duplicates themselves. If you want no duplicates in the Cartesian product, use set(inputlist) over all your input lists. Not on the result.
  • @Triptych what? The standard definition of a Cartesian product is a set. Why do so many people upvote?

The Answer 1

427 people think this answer is useful

itertools.product

Available from Python 2.6.

import itertools

somelists = [
   [1, 2, 3],
   ['a', 'b'],
   [4, 5]
]
for element in itertools.product(*somelists):
    print(element)

Which is the same as,

for element in itertools.product([1, 2, 3], ['a', 'b'], [4, 5]):
    print(element)

The Answer 2

88 people think this answer is useful
import itertools
>>> for i in itertools.product([1,2,3],['a','b'],[4,5]):
...         print i
...
(1, 'a', 4)
(1, 'a', 5)
(1, 'b', 4)
(1, 'b', 5)
(2, 'a', 4)
(2, 'a', 5)
(2, 'b', 4)
(2, 'b', 5)
(3, 'a', 4)
(3, 'a', 5)
(3, 'b', 4)
(3, 'b', 5)
>>>

The Answer 3

41 people think this answer is useful

For Python 2.5 and older:

>>> [(a, b, c) for a in [1,2,3] for b in ['a','b'] for c in [4,5]]
[(1, 'a', 4), (1, 'a', 5), (1, 'b', 4), (1, 'b', 5), (2, 'a', 4), 
 (2, 'a', 5), (2, 'b', 4), (2, 'b', 5), (3, 'a', 4), (3, 'a', 5), 
 (3, 'b', 4), (3, 'b', 5)]

Here’s a recursive version of product() (just an illustration):

def product(*args):
    if not args:
        return iter(((),)) # yield tuple()
    return (items + (item,) 
            for items in product(*args[:-1]) for item in args[-1])

Example:

>>> list(product([1,2,3], ['a','b'], [4,5])) 
[(1, 'a', 4), (1, 'a', 5), (1, 'b', 4), (1, 'b', 5), (2, 'a', 4), 
 (2, 'a', 5), (2, 'b', 4), (2, 'b', 5), (3, 'a', 4), (3, 'a', 5), 
 (3, 'b', 4), (3, 'b', 5)]
>>> list(product([1,2,3]))
[(1,), (2,), (3,)]
>>> list(product([]))
[]
>>> list(product())
[()]

The Answer 4

25 people think this answer is useful

I would use list comprehension :

somelists = [
   [1, 2, 3],
   ['a', 'b'],
   [4, 5]
]

cart_prod = [(a,b,c) for a in somelists[0] for b in somelists[1] for c in somelists[2]]

The Answer 5

22 people think this answer is useful

with itertools.product:

import itertools
result = list(itertools.product(*somelists))

The Answer 6

14 people think this answer is useful

Here is a recursive generator, which doesn’t store any temporary lists

def product(ar_list):
    if not ar_list:
        yield ()
    else:
        for a in ar_list[0]:
            for prod in product(ar_list[1:]):
                yield (a,)+prod

print list(product([[1,2],[3,4],[5,6]]))

Output:

[(1, 3, 5), (1, 3, 6), (1, 4, 5), (1, 4, 6), (2, 3, 5), (2, 3, 6), (2, 4, 5), (2, 4, 6)]

The Answer 7

11 people think this answer is useful

In Python 2.6 and above you can use ‘itertools.product`. In older versions of Python you can use the following (almost — see documentation) equivalent code from the documentation, at least as a starting point:

def product(*args, **kwds):
    # product('ABCD', 'xy') --> Ax Ay Bx By Cx Cy Dx Dy
    # product(range(2), repeat=3) --> 000 001 010 011 100 101 110 111
    pools = map(tuple, args) * kwds.get('repeat', 1)
    result = [[]]
    for pool in pools:
        result = [x+[y] for x in result for y in pool]
    for prod in result:
        yield tuple(prod)

The result of both is an iterator, so if you really need a list for furthert processing, use list(result).

The Answer 8

10 people think this answer is useful

Although there are many answers already, I would like to share some of my thoughts:

Iterative approach

def cartesian_iterative(pools):
  result = [[]]
  for pool in pools:
    result = [x+[y] for x in result for y in pool]
  return result

Recursive Approach

def cartesian_recursive(pools):
  if len(pools) > 2:
    pools[0] = product(pools[0], pools[1])
    del pools[1]
    return cartesian_recursive(pools)
  else:
    pools[0] = product(pools[0], pools[1])
    del pools[1]
    return pools
def product(x, y):
  return [xx + [yy] if isinstance(xx, list) else [xx] + [yy] for xx in x for yy in y]

Lambda Approach

def cartesian_reduct(pools):
  return reduce(lambda x,y: product(x,y) , pools)

The Answer 9

7 people think this answer is useful

Recursive Approach:

def rec_cart(start, array, partial, results):
  if len(partial) == len(array):
    results.append(partial)
    return 

  for element in array[start]:
    rec_cart(start+1, array, partial+[element], results)

rec_res = []
some_lists = [[1, 2, 3], ['a', 'b'], [4, 5]]  
rec_cart(0, some_lists, [], rec_res)
print(rec_res)

Iterative Approach:

def itr_cart(array):
  results = [[]]
  for i in range(len(array)):
    temp = []
    for res in results:
      for element in array[i]:
        temp.append(res+[element])
    results = temp

  return results

some_lists = [[1, 2, 3], ['a', 'b'], [4, 5]]  
itr_res = itr_cart(some_lists)
print(itr_res)

The Answer 10

3 people think this answer is useful

A minor modification to the above recursive generator solution in variadic flavor:

def product_args(*args):
    if args:
        for a in args[0]:
            for prod in product_args(*args[1:]) if args[1:] else ((),):
                yield (a,) + prod

And of course a wrapper which makes it work exactly the same as that solution:

def product2(ar_list):
    """
    >>> list(product(()))
    [()]
    >>> list(product2(()))
    []
    """
    return product_args(*ar_list)

with one trade-off: it checks if recursion should break upon each outer loop, and one gain: no yield upon empty call, e.g.product(()), which I suppose would be semantically more correct (see the doctest).

Regarding list comprehension: the mathematical definition applies to an arbitrary number of arguments, while list comprehension could only deal with a known number of them.

The Answer 11

1 people think this answer is useful

Just to add a bit to what has already been said: if you use sympy, you can use symbols rather than strings which makes them mathematically useful.

import itertools
import sympy

x, y = sympy.symbols('x y')

somelist = [[x,y], [1,2,3], [4,5]]
somelist2 = [[1,2], [1,2,3], [4,5]]

for element in itertools.product(*somelist):
  print element

About sympy.

The Answer 12

1 people think this answer is useful

I believe this works:

def cartesian_product(L):  
   if L:
       return {(a,) + b for a in L[0] 
                        for b in cartesian_product(L[1:])}
   else:
       return {()}

The Answer 13

-1 people think this answer is useful

Stonehenge approach:

def giveAllLists(a, t):
    if (t + 1 == len(a)):
        x = []
        for i in a[t]:
            p = [i]
            x.append(p)
        return x
    x = []

    out = giveAllLists(a, t + 1)
    for i in a[t]:

        for j in range(len(out)):
            p = [i]
            for oz in out[j]:
                p.append(oz)
            x.append(p)
    return x

xx= [[1,2,3],[22,34,'se'],['k']]
print(giveAllLists(xx, 0))



output:

[[1, 22, 'k'], [1, 34, 'k'], [1, 'se', 'k'], [2, 22, 'k'], [2, 34, 'k'], [2, 'se', 'k'], [3, 22, 'k'], [3, 34, 'k'], [3, 'se', 'k']]

The Answer 14

-1 people think this answer is useful

You can use itertools.product in the standard library to get the cartesian product. Other cool, related utilities in itertools include permutations, combinations, and combinations_with_replacement. Here is a link to a python codepen for the snippet below:

from itertools import product

somelists = [
   [1, 2, 3],
   ['a', 'b'],
   [4, 5]
]

result = list(product(*somelists))
print(result)

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