使列表中所有可能的值組合大小不同

Question

我有一個 Python字典，其中包含不同長度的列表作為鍵值。

{
    'basis1': ['key11', 'key12', 'key13', 'key14'],
    'basis2': ['key21'],
    'basis3': ['key31', 'key32', 'key33'],
    'basis4': ['key41', 'key42'],
    ...'basisX': ['keyX1', ..., 'keyXX']
}

如何獲得所有dict鍵值的所有可能組合（每個鍵的一個值到另一個鍵的一個值）？ 我的意思不僅是“basis1 + basic2 +... basicX”值，還包括“basis2 + basic1 + basic3”和“basis3 + basicX”值。

我曾經使用“itertools”中的“產品”函數進行迭代，以通過預定公式生成關鍵字。 但是由於公式和這些公式中的列表數量存在限制。 但我需要使它不依賴於將饋送到 function 輸入的列表的數量，以及以不同的順序混合列表中的值。

from itertools import product
...
...
...
# Each [keysX] is a list of values

    formula1 = [keys0] + [keys1]
    formula2 = [keys0] + [keys2]
    formula3 = [keys1] + [keys2]
    formula4 = [keys0] + [keys1] + [keys2]
    all_keywords = []

    for combo in product(*formula1):
        all_keywords.append(" ".join(combo))

    for combo in product(*formula2):
        all_keywords.append(" ".join(combo))

    for combo in product(*formula3):
        all_keywords.append(" ".join(combo))

    for combo in product(*formula4):
        all_keywords.append(" ".join(combo))

Answer 1

注意：考慮到它是一個冪集的所有元素的排列，生成的迭代將是巨大的。 所以建議不要存放在memory中，除非必要。 出於這個原因，我使用了generators 。

您可以使用以下內容：

from itertools import chain, combinations, permutations

def powerset(iterable):
    '''
    >>> list(powerset([1, 2, 3]))
    [(1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)]
    '''
    iterable = list(iterable)
    return chain.from_iterable(
        combinations(iterable, r)
        for r, _ in enumerate(iterable, start=1)
    )

def perm_powerset(iterable):
    '''
    Given a powerset, returns a generator consisting
    all possible permutations of each element in the powerset.
    '''
    for each_set in powerset(iterable):
        for elem in permutations(each_set):
            yield elem

d = {'k1': [1, 2], 'k2': [3], 'k4': [4]}

for elem in perm_powerset(chain.from_iterable(d.values())):
    print(elem)

Output：

(1,)
(2,)
(3,)
(4,)
(1, 2)
(2, 1)
(1, 3)
(3, 1)
(1, 4)
(4, 1)
(2, 3)
(3, 2)
(2, 4)
(4, 2)
(3, 4)
(4, 3)
(1, 2, 3)
(1, 3, 2)
(2, 1, 3)
(2, 3, 1)
(3, 1, 2)
(3, 2, 1)
(1, 2, 4)
(1, 4, 2)
(2, 1, 4)
(2, 4, 1)
(4, 1, 2)
(4, 2, 1)
(1, 3, 4)
(1, 4, 3)
(3, 1, 4)
(3, 4, 1)
(4, 1, 3)
(4, 3, 1)
(2, 3, 4)
(2, 4, 3)
(3, 2, 4)
(3, 4, 2)
(4, 2, 3)
(4, 3, 2)
(1, 2, 3, 4)
(1, 2, 4, 3)
(1, 3, 2, 4)
(1, 3, 4, 2)
(1, 4, 2, 3)
(1, 4, 3, 2)
(2, 1, 3, 4)
(2, 1, 4, 3)
(2, 3, 1, 4)
(2, 3, 4, 1)
(2, 4, 1, 3)
(2, 4, 3, 1)
(3, 1, 2, 4)
(3, 1, 4, 2)
(3, 2, 1, 4)
(3, 2, 4, 1)
(3, 4, 1, 2)
(3, 4, 2, 1)
(4, 1, 2, 3)
(4, 1, 3, 2)
(4, 2, 1, 3)
(4, 2, 3, 1)
(4, 3, 1, 2)
(4, 3, 2, 1)

Answer 2

這段代碼會生成您需要的內容。 首先，您需要擁有所有值組合，然后為每個組合生成關鍵字的所有產品。

from itertools import product, chain, combinations

my_dict = {
    'basis1': ['key11', 'key12', 'key13', 'key14'],
    'basis2': ['key21'],
    'basis3': ['key31', 'key32', 'key33'],
    'basis4': ['key41', 'key42']}

list_of_values = list(my_dict.values())

all_combs_of_lists = [combinations(list_of_values, r) for r in range(1, len(list_of_values) + 1)]
list_combinations = list(chain.from_iterable(all_combs_of_lists))

all_keywords = [" ".join(combo) for item in list_combinations for combo in product(*item)]
        
print(all_keywords)

Output：

['key11', 'key12', 'key13', 'key14', 'key21', 'key31', 'key32', 'key33', 'key41', 'key42', 'key11 key21', 'key12 key21', 'key13 key21', 'key14 key21', 'key11 key31', 'key11 key32', 'key11 key33', 'key12 key31', 'key12 key32', 'key12 key33', 'key13 key31', 'key13 key32', 'key13 key33', 'key14 key31', 'key14 key32', 'key14 key33', 'key11 key41', 'key11 key42', 'key12 key41', 'key12 key42', 'key13 key41', 'key13 key42', 'key14 key41', 'key14 key42', 'key21 key31', 'key21 key32', 'key21 key33', 'key21 key41', 'key21 key42', 'key31 key41', 'key31 key42', 'key32 key41', 'key32 key42', 'key33 key41', 'key33 key42', 'key11 key21 key31', 'key11 key21 key32', 'key11 key21 key33', 'key12 key21 key31', 'key12 key21 key32', 'key12 key21 key33', 'key13 key21 key31', 'key13 key21 key32', 'key13 key21 key33', 'key14 key21 key31', 'key14 key21 key32', 'key14 key21 key33', 'key11 key21 key41', 'key11 key21 key42', 'key12 key21 key41', 'key12 key21 key42', 'key13 key21 key41', 'key13 key21 key42', 'key14 key21 key41', 'key14 key21 key42', 'key11 key31 key41', 'key11 key31 key42', 'key11 key32 key41', 'key11 key32 key42', 'key11 key33 key41', 'key11 key33 key42', 'key12 key31 key41', 'key12 key31 key42', 'key12 key32 key41', 'key12 key32 key42', 'key12 key33 key41', 'key12 key33 key42', 'key13 key31 key41', 'key13 key31 key42', 'key13 key32 key41', 'key13 key32 key42', 'key13 key33 key41', 'key13 key33 key42', 'key14 key31 key41', 'key14 key31 key42', 'key14 key32 key41', 'key14 key32 key42', 'key14 key33 key41', 'key14 key33 key42', 'key21 key31 key41', 'key21 key31 key42', 'key21 key32 key41', 'key21 key32 key42', 'key21 key33 key41', 'key21 key33 key42', 'key11 key21 key31 key41', 'key11 key21 key31 key42', 'key11 key21 key32 key41', 'key11 key21 key32 key42', 'key11 key21 key33 key41', 'key11 key21 key33 key42', 'key12 key21 key31 key41', 'key12 key21 key31 key42', 'key12 key21 key32 key41', 'key12 key21 key32 key42', 'key12 key21 key33 key41', 'key12 key21 key33 key42', 'key13 key21 key31 key41', 'key13 key21 key31 key42', 'key13 key21 key32 key41', 'key13 key21 key32 key42', 'key13 key21 key33 key41', 'key13 key21 key33 key42', 'key14 key21 key31 key41', 'key14 key21 key31 key42', 'key14 key21 key32 key41', 'key14 key21 key32 key42', 'key14 key21 key33 key41', 'key14 key21 key33 key42']