python - 元组列表的列表，按第一个元素分组并添加第二个元素

Question

假设我有以下元组列表:

tuples = [
             [ 
                 ('2017-04-11', '2000000.00'), 
                 ('2017-04-12', '1000000.00'), 
                 ('2017-04-13', '3000000.00')
             ],
             [
                 ('2017-04-12', '472943.00'), 
                 ('2017-04-13', '1000000.00')
             ]
             # ...
         ]

我将如何根据第一个元素(日期)对它们进行分组并添加另一个元素。

例如我想要这样的东西:

tuples = [('2017-04-11', '2000000.00'), ('2017-04-12', '1472943.00'), ('2017-04-13', '4000000.00')],

Answer 1

使用itertools.chain.from_iterable、itertools.groupby和sum函数的解决方案:

import itertools, operator

tuples = [
         [('2017-04-11', '2000000.00'), ('2017-04-12', '1000000.00'), ('2017-04-13', '3000000.00')],
         [('2017-04-12', '472943.00'), ('2017-04-13', '1000000.00')]
         ]

result = [(k, "%.2f" % sum(float(t[1]) for t in g)) 
          for k,g in itertools.groupby(sorted(itertools.chain.from_iterable(tuples)), operator.itemgetter(0))]

print(result)

输出:

[('2017-04-11', '2000000.00'), ('2017-04-12', '1472943.00'), ('2017-04-13', '4000000.00')]