Cartesian product of a dictionary of lists

Question

I'm trying to write some code to test out the Cartesian product of a bunch of input parameters.

I've looked at itertools, but its product function is not exactly what I want. Is there a simple obvious way to take a dictionary with an arbitrary number of keys and an arbitrary number of elements in each value, and then yield a dictionary with the next permutation?

Input:

options = {"number": [1,2,3], "color": ["orange","blue"] }
print list( my_product(options) )

Example output:

[ {"number": 1, "color": "orange"},
  {"number": 1, "color": "blue"},
  {"number": 2, "color": "orange"},
  {"number": 2, "color": "blue"},
  {"number": 3, "color": "orange"},
  {"number": 3, "color": "blue"}
]

Answer 1

Ok, thanks @dfan for telling me I was looking in the wrong place. I've got it now:

from itertools import product
def my_product(inp):
    return (dict(zip(inp.keys(), values)) for values in product(*inp.values())

EDIT: after years more Python experience, I think a better solution is to accept kwargs rather than a dictionary of inputs; the call style is more analogous to that of the original itertools.product. Also I think writing a generator function, rather than a function that returns a generator expression, makes the code clearer. So:

def product_dict(**kwargs):
    keys = kwargs.keys()
    vals = kwargs.values()
    for instance in itertools.product(*vals):
        yield dict(zip(keys, instance))

and if you need to pass in a dict, list(product_dict(**mydict)). The one notable change using kwargs rather than an arbitrary input class is that it prevents the keys/values from being ordered, at least until Python 3.6.