将嵌套列表简化为笛卡尔积

Question

我在 Python 中有一个嵌套列表：

lst = ['alpha', ['beta', 'gamma'], ['delta', 'peta', 'lambda']]

我需要一个函数来 return 一个包含它们的笛卡尔积的列表。好吧，我认为笛卡尔积不是一个正确的词，但仍然以合乎逻辑的方式结果看起来像：

final_lst = your_magical_function(lst)
print final_lst

'''
[['alpha','beta','delta'],
['alpha','beta','peta'],
['alpha','beta','lambda'],
['alpha','gamma','delta'],
['alpha','gamma','peta']
['alpha','gamma','lambda']]
'''

无论有没有递归的函数，都欢迎。

Answer 1

使用itertools.product，这将需要您稍微修改输入（'alpha' 到 ['alpha']）：

from itertools import product

lst = [['alpha'],['beta','gamma'],['delta','peta','lambda']]

for res in product(*lst):
    print(res)

>> ('alpha', 'beta', 'delta')

('alpha', 'beta', 'peta')
('alpha', 'beta', 'lambda')
('alpha', 'gamma', 'delta')
('alpha', 'gamma', 'peta')
('alpha', 'gamma', 'lambda')

Answer 2

将所有项目转换为列表后，您可以使用 itertools.product：

>>> from itertools import product
>>> lst = ['alpha',['beta','gamma'],['delta','peta','lambda']]
>>> list(product(*(x if isinstance(x, list) else [x] for x in lst)))
[('alpha', 'beta', 'delta'), ('alpha', 'beta', 'peta'), ('alpha', 'beta', 'lambda'), ('alpha', 'gamma', 'delta'), ('alpha', 'gamma', 'peta'), ('alpha', 'gamma', 'lambda')]

Answer 3

使用 Python 中的库很棒！但是，如果您正在寻找纯 Python 实现：

def CartesianProduct(list_entry):
# Save Sizes of Everything
size_dictionary = {}

# Get Size of Entire Entry List
size_dictionary["full_size"] = len(list_entry)

# Get Sizes of All Sub Entries
for i in range(len(list_entry)):
    if not (isinstance(list_entry[i],list)):
        list_entry[i] = [list_entry[i]]
    size_dictionary[i] = len(list_entry[i])

# Now lets create the cartesian product
# Lets Create a Dictionary to hold all of the results
cartesian_result = {}

# Lets get the size of the final result
final_result_amount = 1
for i in range(size_dictionary["full_size"]):
    final_result_amount = final_result_amount * size_dictionary[i]

# And create the final results
for i in range(final_result_amount):
    cartesian_result[i] = []

    for j in range(size_dictionary["full_size"]):
        cartesian_result[i].append(list_entry[j][i % size_dictionary[j]])

    print(cartesian_result[i])

def main():
    lst = ['alpha',['beta','gamma'],['delta','peta','lambda']]
    CartesianProduct(lst)

main()

它不像使用 itertools 那样漂亮和简单，但是实现库偶尔使用的逻辑仍然很有趣。

将嵌套列表简化为笛卡尔积

Simplifying nested-lists as a Cartesian product

python

recursion

nested-lists