列表字典的笛卡尔积

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.

Answer 2

Python 3 version of Seth's answer .

import itertools

def dict_product(dicts):
    """
    >>> list(dict_product(dict(number=[1,2], character='ab')))
    [{'character': 'a', 'number': 1},
     {'character': 'a', 'number': 2},
     {'character': 'b', 'number': 1},
     {'character': 'b', 'number': 2}]
    """
    return (dict(zip(dicts, x)) for x in itertools.product(*dicts.values()))

Answer 3

By the way, this is not a permutation. A permutation is a rearrangement of a list. This is an enumeration of possible selections from lists.

Edit: after remembering that it was called a Cartesian product, I came up with this:

import itertools
options = {"number": [1,2,3], "color": ["orange","blue"] }
product = [x for x in apply(itertools.product, options.values())]
print [dict(zip(options.keys(), p)) for p in product]

Answer 4

# I would like to do
keys,values = options.keys(), options.values()
# but I am not sure that the keys and values would always
# be returned in the same relative order. Comments?
keys = []
values = []
for k,v in options.iteritems():
    keys.append(k)
    values.append(v)

import itertools
opts = [dict(zip(keys,items)) for items in itertools.product(*values)]

results in

opts = [
    {'color': 'orange', 'number': 1},
    {'color': 'orange', 'number': 2},
    {'color': 'orange', 'number': 3},
    {'color': 'blue', 'number': 1},
    {'color': 'blue', 'number': 2},
    {'color': 'blue', 'number': 3}
]