如何从Python列表字典中的值生成所有组合

Question

I would like to generate all combinations of values which are in lists indexed in a dict:

{'A':['D','E'],'B':['F','G','H'],'C':['I','J']}

Each time, one item of each dict entry would be picked and combined to items from other keys, so I have:

['D','F','I']
['D','F','J']
['D','G','I']
['D','G','J']
['D','H','I']
...
['E','H','J']

I know there is a something to generate combinations of items in list in itertools but I don't think I can use it here since I have different "pools" of values.

Is there any existing solution to do this, or how should I proceed to do it myself, I am quite stuck with this nested structure.

Answer 1

import itertools as it

my_dict={'A':['D','E'],'B':['F','G','H'],'C':['I','J']}
allNames = sorted(my_dict)
combinations = it.product(*(my_dict[Name] for Name in allNames))
print(list(combinations))

Which prints:

[('D', 'F', 'I'), ('D', 'F', 'J'), ('D', 'G', 'I'), ('D', 'G', 'J'), ('D', 'H', 'I'), ('D', 'H', 'J'), ('E', 'F', 'I'), ('E', 'F', 'J'), ('E', 'G', 'I'), ('E', 'G', 'J'), ('E', 'H', 'I'), ('E', 'H', 'J')]

Answer 2

If you want to keep the key:value in the permutations you can use:

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

this will provide you a list of dicts with the permutations:

print(permutations_dicts)
[{'A':'D', 'B':'F', 'C':'I'}, 
 {'A':'D', 'B':'F', 'C':'J'},
 ...
 ]

disclaimer not exactly what the OP was asking, but google send me here looking for that.

Answer 3

How about using ParameterGrid from scikit-learn? It creates a generator over which you can iterate in a normal for loop. In each iteration, you will have a dictionary containing the current parameter combination.

from sklearn.model_selection import ParameterGrid

params = {'A':['D','E'],'B':['F','G','H'],'C':['I','J']}
param_grid = ParameterGrid(params)
for dict_ in param_grid:
    # Do something with the current parameter combination in ``dict_``
    print(dict_["A"])
    print(dict_["B"])
    print(dict_["C"])

Answer 4

As a complement, here is code that does it Python so that you get the idea, but itertools is indeed more efficient.

res = [[]]
for _, vals in my_dict.items():
    res = [x+[y] for x in res for y in vals]
print(res)

Answer 5

from itertools import combinations

a=['I1','I2','I3','I4','I5']

list(combinations(a,2))

The output will be:

[('I1', 'I2'),
 ('I1', 'I3'),
 ('I1', 'I4'),
 ('I1', 'I5'),
 ('I2', 'I3'),
 ('I2', 'I4'),
 ('I2', 'I5'),
 ('I3', 'I4'),
 ('I3', 'I5'),
 ('I4', 'I5')]

Answer 6

除了使用Maryam提供的笛卡尔积之外，您还可以使用朴素的嵌套循环来获得您想要的效果。