到/来自列表字典的字典列表，但具有所有可能的排列

Question

Coming from question number 5558418, suppose I have a dictionary where values are lists:

DL = {'a': [0, 1], 'b': [2, 3]}

And I want to convert it to a list of dictionaries, but with all possible permutations . That is, instead of

LD = [{'a': 0, 'b': 2}, {'a': 1, 'b': 3}]

I now want:

LD = [{'a': 0, 'b': 2}, {'a': 0, 'b': 3}, {'a': 1, 'b': 2}, {'a': 1, 'b': 3}]

How can I do that? I think itertools' product() may be handy but I can't quite figure it out.

Bonus: In my actual problem, some values are not lists. How can I do the same as above, but preserving all values that are not lists? eg:

D = {'a': [0, 1], 'b': [2, 3], 'c':7, 'd':9}

Becomes:

L = [{'a': 0, 'b': 2, 'c':7, 'd':9}, {'a': 0, 'b': 3, 'c':7, 'd':9}}, {'a': 1, 'b': 2, 'c':7, 'd':9}}, {'a': 1, 'b': 3, 'c':7, 'd':9}}]

Thanks!

Answer 1

Make sure every value is a list and then use itertools.product :

from itertools import product

# DL = {'a': [0, 1], 'b': [2, 3]}
D = {'a': [0, 1], 'b': [2, 3], 'c': 7, 'd': 9}

res = [dict(zip(D, p)) for p in product(*[v if isinstance(v, list) else [v] for v in D.values()])]
print(res)

Output

[{'a': 0, 'b': 2, 'c': 7, 'd': 9}, {'a': 0, 'b': 3, 'c': 7, 'd': 9}, {'a': 1, 'b': 2, 'c': 7, 'd': 9}, {'a': 1, 'b': 3, 'c': 7, 'd': 9}]