Python 当列表总和低于阈值时，将列表分组为列表

我有一份清单：

[(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，并且仅存储为（每个元素的）总和：

例如，如果阈值为50（含50）：

元组的第二个值可能会有所不同。最好是列表理解

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))

padding=len（“Packages（）”）+math.floor（math.log10（len（apps））+1）
行长度=列-（填充*2）-2
间距=2
element_length=[子列表中的项目[列表（a）中的项目用于zip中的项目（[len（i）用于应用程序中的i]，[i用于itertools中的i。重复（间隔，len（[len（i）用于应用程序中的i]））]]子列表中的项目]
限制=[]
外极限=0
而行长度=和（元素长度[0:外部限制]）：
外极限+=1
限制。追加（外部限制-1）
元素长度=元素长度[外部限制-1:]
外极限=0
message=“”
a=0
b=0
对于限额中的金额：
b+=数学单元（金额/2）
消息+=（“”*间距）.join（应用程序[a:b]）+（“\n”+“”*填充）
a=b
打印（“包（{}）{}”。格式（len（应用程序），消息））

这里有一个非常简单的方法，只需一个for循环即可：

tups = [(14, 2), (14, 2), (16, 2), (14, 2)]
threshold = 50
result = [[]]
for tup in tups:
    tupSum = sum(tup)
    # Start a new sublist if adding this tuple's sum would exceed the threshold
    if tupSum + sum(result[-1]) > threshold:
        result.append([])
    result[-1].append(tupSum)

---

这可以通过functools的reduce函数完成：

a = [(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)]

from functools import reduce
b = list(reduce(lambda a,b:a+[[b]] if sum(a[-1])+b>50 else a[:-1]+[a[-1]+[b]],map(sum,a),[[]]))
print(b) 

# [[16, 16, 18], [16, 17, 17], [23, 17], [20, 17], [21, 27], [24, 19], [33], [28], [23, 27], [31], [35], [27], [25], [27, 21], [14, 31], [20, 23], [15, 15, 20], [13, 14, 22], [25, 19], [16, 19, 14], [15, 17], [23, 17], [21, 24], [18, 18], [15, 19], [20, 22], [20, 15, 15], [20, 16], [15, 24], [16, 27], [24, 11], [20, 24], [21, 15], [16, 17, 15], [19, 23], [20, 23], [20, 17], [18, 15], [18, 18], [17, 13], [26, 17], [14, 22], [23, 23], [16, 13], [28, 19], [23, 18], [15, 17, 15], [14, 24], [18, 15, 15], [24, 14], [18, 18], [23, 21], [17, 18], [18, 15, 16], [16, 22], [16, 22], [15, 21], [22, 19], [19, 27], [24, 24], [24, 16], [21, 22], [18, 15], [21, 18], [14, 20], [22, 21], [20, 17], [24, 20], [22, 16], [21, 18], [20, 30], [16, 19], [19, 25], [20, 26], [19, 20], [20, 24]]

---

是否有任何理由要列出一份清单？那样的话，它的可读性可能会降低。你试过什么吗？看起来你只是想让我们为你做这件事。我怀疑有一个更优雅的解决方案是基于itertools.groupby的，但我不知道这会是什么样子，所以我提出了这个问题。我有一个基于列表理解+2 while循环的解决方案，但我不喜欢这个解决方案的复杂性。我没有提到它，因为它不是（直接）我问题的一部分。如果元组本身超过50，例如

（50，2）

，该怎么办？或者这是保证不会发生的？你能发布这个解决方案吗？我们可以看看它是否可以简化

a = [(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)]

from functools import reduce
b = list(reduce(lambda a,b:a+[[b]] if sum(a[-1])+b>50 else a[:-1]+[a[-1]+[b]],map(sum,a),[[]]))
print(b) 

# [[16, 16, 18], [16, 17, 17], [23, 17], [20, 17], [21, 27], [24, 19], [33], [28], [23, 27], [31], [35], [27], [25], [27, 21], [14, 31], [20, 23], [15, 15, 20], [13, 14, 22], [25, 19], [16, 19, 14], [15, 17], [23, 17], [21, 24], [18, 18], [15, 19], [20, 22], [20, 15, 15], [20, 16], [15, 24], [16, 27], [24, 11], [20, 24], [21, 15], [16, 17, 15], [19, 23], [20, 23], [20, 17], [18, 15], [18, 18], [17, 13], [26, 17], [14, 22], [23, 23], [16, 13], [28, 19], [23, 18], [15, 17, 15], [14, 24], [18, 15, 15], [24, 14], [18, 18], [23, 21], [17, 18], [18, 15, 16], [16, 22], [16, 22], [15, 21], [22, 19], [19, 27], [24, 24], [24, 16], [21, 22], [18, 15], [21, 18], [14, 20], [22, 21], [20, 17], [24, 20], [22, 16], [21, 18], [20, 30], [16, 19], [19, 25], [20, 26], [19, 20], [20, 24]]