Python 使用itertools.groupby汇总报告

有人能帮我按函数第一列、第二列和第三列分组吗

from itertools import groupby
from operator import itemgetter

things = [('2009-09-02','j', 12),
          ('2009-09-02','j', 3),
          ('2009-09-03','k',10),
          ('2009-09-03','k',4),
          ('2009-09-03','u', 22),
          ('2009-09-06','m',33)]

for k, items in groupby(things, itemgetter(1)):    
    for subitem in items:
        print(subitem)

得到这个结果：

('2009-09-02', 'j', 12) ('2009-09-02', 'j', 3) ('2009-09-03', 'k', 10) ('2009-09-03', 'k', 4) ('2009-09-03', 'u', 22) ('2009-09-06', 'm', 33) 

 ('2009-09-02', 'j', 15) ('2009-09-03', 'k', 14) ('2009-09-03', 'u', 22) ('2009-09-06', 'm', 33)

期待这一结果：

('2009-09-02', 'j', 12) ('2009-09-02', 'j', 3) ('2009-09-03', 'k', 10) ('2009-09-03', 'k', 4) ('2009-09-03', 'u', 22) ('2009-09-06', 'm', 33) 

 ('2009-09-02', 'j', 15) ('2009-09-03', 'k', 14) ('2009-09-03', 'u', 22) ('2009-09-06', 'm', 33)

========================================================================

   sales = [('Scotland', 'Edinburgh', 20000),
         ('Scotland', 'Glasgow', 12500),
         ('Wales', 'Cardiff', 29700),
         ('Wales', 'Bangor', 12800),
         ('England', 'London', 90000),
         ('England', 'Manchester', 45600),
         ('England', 'London', 29700)]

您不需要groupby作为一种更有效的方法，您可以将dictionary与dict.setdefault方法结合使用：

如果您想使用groupby作为提示，您可能会注意到需要将索引传递给itemgetter函数：

itemgetter(0, 1)

如果你想求和，你必须求和，简单地打印它不会神奇地为你求和

另外，根据您的示例，您似乎应该基于第一列和第二列进行分组。范例-

for k,items in groupby(things, itemgetter(0, 1)):    
    print(k + (sum(x[2] for x in items),)

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谢谢你，莎伦。我在上面的销售示例中又添加了一个。然而，出于某种原因，它不会总结英格兰和伦敦？