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我需要根据 n 个输入列表获取所有可能的组合并对它们做一些事情。

当前代码示例：

import itertools

# example inputs
list_small = [1, 2, 3]
list_medium = [444, 666, 242]
list_huge = [1680, 7559, 5573, 43658, 530, 11772, 284, 50078, 783, 37809, 6740, 37765, 74492, 50078, 783, 37809, 6740, 37765, 74492]

# out of the input list, I need to generate all numbers from 0 to the current list element
# e.g. if I have 6, I need to get [0, 1, 2, 3, 4, 5, 6]
# if I get a list [1, 2, 3], the output will be [[0, 1], [0, 1, 2], [0, 1, 2, 3]]
# I achieved this by doing it with xrange: [x for x in xrange(0, current_list_element + 1)]
# after that, I need to generate all possible combinations using the generated lists
# I managed to do this by using itertools.product()

# print this to get all possible combinations
# print list(itertools.product(*[[x for x in xrange(0, current_list_element + 1)] for current_list_element in list_medium]))

cumulative_sum = 0
for current_combination in itertools.product(*[[x for x in xrange(0, current_list_element + 1)] for current_list_element in list_medium]):
    # now I need to do some calculations to the current combination
    # e.g. get sum of all combinations, this is just an example
    cumulative_sum += sum(current_combination)

    # another example
    # get XOR sum of current combination, more at https://en.wikipedia.org/wiki/Exclusive_or
    print reduce(operator.xor, current_combination, 0)

# runs fast for list_small, then takes some time for list_medium and then takes ages for list_huge
print cumulative_sum

这适用于较小的列表，但对于较大的列表需要无穷大/或引发运行时错误。有没有更好的方法来做到这一点？获得所有组合的更好方法？还是我以某种错误的方式使用 xrange？

我用 Python 2.7 和 Pypy 2 试过这个。

编辑：感谢@famagusta，我摆脱了xrange，但问题仍然存在

import itertools

# example inputs
list_small = [1, 2, 3]
list_medium = [444, 666, 242]
list_huge = [1680, 7559, 5573, 43658, 530, 11772, 284, 50078, 783, 37809, 6740, 37765, 74492, 50078, 783, 37809, 6740, 37765, 74492]

max_element = max(get_input_stones)
combo_list = range(0, max_element + 1)

cumulative_sum = 0
for current_combination in itertools.product(*combo_list):
    # now I need to do some calculations to the current combination
    # e.g. get sum of all combinations, this is just an example
    cumulative_sum += sum(current_combination)

    # another example
    # get XOR sum of current combination, more at https://en.wikipedia.org/wiki/Exclusive_or
    print reduce(operator.xor, current_combination, 0)

# runs fast for list_small, then takes some time for list_medium and then takes ages for list_huge
print cumulative_sum
4

1 回答 1

1

生成这样的嵌套列表可能会给您带来内存限制的麻烦。您可以只使用从列表中最大数生成的一个超级列表，而不是重复生成子列表。只需将索引存储在较小元素会停止的位置。

例如，[1, 6, 10] - [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [1, 6, 10]

第二个列表告诉您在第一个列表中的何处停止以提取感兴趣的子列表进行计算

这应该可以为您节省一些空间。

list_small = [1, 2, 3]
list_medium = [444, 666, 242]
list_huge = [1680, 7559, 5573, 43658, 530, 11772, 284, 50078, 783, 37809, 6740, 37765, 74492, 50078, 783, 37809, 6740, 37765, 74492]

max_element = max(list_huge)   # being lazy here - write a max function
combo_list = range(0, max_element + 1)  # xrange does not support slicing

cumulative_sum = 0
for element in list_huge:
    cumulative_sum += sum(combo_list[:element])

print(cumulative_sum)
于 2016-08-25T09:45:56.903 回答