设定值排列字典的Python递归

Question

I'm trying to get all combinations of a dictionary recursively but I can't wrap my head around how to get the output properly. Essentially a depth-first search which saves the input in a (key, value) tuple or something similar. Any help appreciated. Thanks /Fred

input:

d = {"item1": {1, 2},
     "item2": {3, 4},
     "item3": {5, 6}}

output:

"item1" 1
"item2" 3
"item3" 5
"item3" 6
"item2" 4
"item3" 5
"item3" 6
"item1" 2
"item2" 3
"item3" 5
"item3" 6
"item2" 4
"item3" 5
"item3" 6

edit: The perms needs to be drawn recursively. Maybe an illustration clarifies a bit:

A tree structure worked but was not generalised enough for my purposes, and cumbersome to edit.

Update: Currently I'm hardcoding these like so:

d = {"item1": {1, 2},
     "item2": {3, 4},
     "item3": {i for i in range(1, 5)}}

for k in d["item1"]:
    print ("item1", k)
    for j in d["item2"]:
        print ("item2", j)
        for i in d["item3"]:
            print("item3", i)

It seems obvious where the recursion happens but I'm still having troubles with it. Thank you all for all suggestions so far! Also it's in python3 if that makes any difference.

Answer 1

This will return a list of permutations.

from itertools import product

perms = [[perm for perm in product([key], d[key]) for key in d]]

An update in case you're looking for a possible combinations of key values pairs which would be 18 in total.

[print(prod) for prod in product(d, itertools.chain.from_iterable(d.values()))]

outputs:

('item3', 5)
('item3', 6)
('item3', 1)
('item3', 2)
('item3', 3)
('item3', 4)
('item1', 5)
('item1', 6)
('item1', 1)
('item1', 2)
('item1', 3)
('item1', 4)
('item2', 5)
('item2', 6)
('item2', 1)
('item2', 2)
('item2', 3)
('item2', 4)

Answer 2

How about this:

>>> for pair in itertools.chain(*([(k, i) for i in v] for k, v in d.items())):
    print(pair)

('item2', 3)
('item2', 4)
('item3', 5)
('item3', 6)
('item1', 1)
('item1', 2)

Or a bit more legible:

def key_value_permutations(d):
    for k, v in d:
       for p in itertools.product([k], v):
           yield p

Then:

>>> for p in key_value_permutations(d):
    print p

('item2', 3)
('item2', 4)
('item3', 5)
('item3', 6)
('item1', 1)
('item1', 2)

Both solutions use generators because they assume your dicts and sets will be bigger than in the sample you provided and you will just iterate over it so you don't really need to create a huge list in memory. If the size is negligible, list comprehensions may be faster.

You can't count on consistent ordering because both dicts and sets have no guaranteed ordering.

Answer 3

Is this what you're looking for?

for first_level in d:
     item_list = d[first_level]
     for item in item_list:
             print((first_level, item))

Output

('item2', 3)
('item2', 4)
('item3', 5)
('item3', 6)
('item1', 1)
('item1', 2)