我有一个元组列表，('Day of Week', n)如下所示：
[('Wed', 1), ('Wed', 1), ('Thu', 1), ('Thu', 0), ('Tue', 0), ('Mon', 0), ('Sun', 0), ('Sat', 0),
('Fri', 0)]
我想产生以下输出，一个元组列表('Day of Week', sum_n)，其中sum_n是n初始列表中所有相应值的总和。请注意，元素应按工作日顺序排列。
[('Wed', 2), ('Thu', 1), ('Tue', 0), ('Mon', 0), ('Sun', 0), ('Sat', 0), ('Fri', 0)]
如何做到这一点？
具有列表理解：
>>> l = [('Wed', 1), ('Wed', 1), ('Thu', 1), ('Thu', 0), ('Tue', 0), ('Mon', 0), ('Sun', 0), ('Sat', 0), ('Fri', 0)]
>>> [(x, sum(y[1] for y in l if y[0] == x)) for x in set(z[0] for z in l)]
[('Sun', 0),
('Tue', 0),
('Mon', 0),
('Wed', 2),
('Sat', 0),
('Thu', 1),
('Fri', 0)]
要使第二个元素降序排序：
>>> res = [(x, sum(y[1] for y in l if y[0] == x)) for x in set(z[0] for z in l)]
>>> sorted(res, key=lambda x: x[1], reverse=True)
[('Wed', 2),
('Thu', 1),
('Sun', 0),
('Tue', 0),
('Mon', 0),
('Sat', 0),
('Fri', 0)]
按星期几排序（更多答案在此）：
>>> sorted_days = ["Fri", "Sat", "Sun", "Mon", "Tue", "Wed", "Thu"]
>>> sorted(res, key=lambda x: sorted_days.index(x[0]))
[('Fri', 0),
('Sat', 0),
('Sun', 0),
('Mon', 0),
('Tue', 0),
('Wed', 2),
('Thu', 1)]
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