我有一个清单:
[(14, 2), (14, 2), (16, 2), (14, 2), (15, 2), (15, 2), (21, 2), (15, 2), (18, 2), (15, 2), (19, 2), (25, 2), (22, 2), (17, 2), (31, 2), (26, 2), (21, 2), (25, 2), (29, 2), (33, 2), (25, 2), (23, 2), (25, 2), (19, 2), (12, 2), (29, 2), (18, 2), (21, 2), (13, 2), (13, 2), (18, 2), (11, 2), (12, 2), (20, 2), (23, 2), (17, 2), (14, 2), (17, 2), (12, 2), (13, 2), (15, 2), (21, 2), (15, 2), (19, 2), (22, 2), (16, 2), (16, 2), (13, 2), (17, 2), (18, 2), (20, 2), (18, 2), (13, 2), (13, 2), (18, 2), (14, 2), (13, 2), (22, 2), (14, 2), (25, 2), (22, 2), (9, 2), (18, 2), (22, 2), (19, 2), (13, 2), (14, 2), (15, 2), (13, 2), (17, 2), (21, 2), (18, 2), (21, 2), (18, 2), (15, 2), (16, 2), (13, 2), (16, 2), (16, 2), (15, 2), (11, 2), (24, 2), (15, 2), (12, 2), (20, 2), (21, 2), (21, 2), (14, 2), (11, 2), (26, 2), (17, 2), (21, 2), (16, 2), (13, 2), (15, 2), (13, 2), (12, 2), (22, 2), (16, 2), (13, 2), (13, 2), (22, 2), (12, 2), (16, 2), (16, 2), (21, 2), (19, 2), (15, 2), (16, 2), (16, 2), (13, 2), (14, 2), (14, 2), (20, 2), (14, 2), (20, 2), (13, 2), (19, 2), (20, 2), (17, 2), (17, 2), (25, 2), (22, 2), (22, 2), (22, 2), (14, 2), (19, 2), (20, 2), (16, 2), (13, 2), (19, 2), (16, 2), (12, 2), (18, 2), (20, 2), (19, 2), (18, 2), (15, 2), (22, 2), (18, 2), (20, 2), (14, 2), (19, 2), (16, 2), (18, 2), (28, 2), (14, 2), (17, 2), (17, 2), (23, 2), (18, 2), (24, 2), (17, 2), (18, 2), (18, 2), (22, 2)]
我希望我的输出是一个列表列表,其中每个子列表是可以添加在一起的元素数,而不超过某个阈值,并且只存储为(每个元素的)和:
例如,如果阈值为50(含50):
[[16, 16, 18], [16, 17, 17], [23, 17], [20, 17], [21, 27], [24, 19], [33], [28], [23, 27], [31], [35], [27], [25], [27, 21] ...]
元组的第二个值可能不同。作为列表优先。你知道吗
编辑:
根据要求,我要清理/优化的原始代码:
padding = len("Packages () ") + math.floor(math.log10(len(apps))+1)
line_length = columns - (padding * 2) - 2
spacings = 2
element_length = [item for sublist in [list(a) for a in zip([len(i) for i in apps],[i for i in itertools.repeat(spacings, len([len(i) for i in apps]))])] for item in sublist]
limits = []
outer_limit = 0
while line_length <= sum(element_length):
while line_length >= sum(element_length[0:outer_limit]):
outer_limit += 1
limits.append(outer_limit - 1)
element_length = element_length[outer_limit - 1:]
outer_limit = 0
message = ""
a = 0
b = 0
for amount in limits:
b += math.ceil(amount / 2)
message += (" " * spacings).join(apps[a:b]) + ("\n" + " " * padding)
a = b
print("Packages ({}) {}".format(len(apps), message))
下面是一个非常简单的方法,只需一个for循环:
这可以通过functools的reduce函数完成:
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