python - 如何对列表列表进行分组

Question

我有一个如下所示的列表:

list=[
 ('2013-01-04', u'crid2557171372', 1),
 ('2013-01-04', u'crid9904536154', 719677),
 ('2013-01-04', u'crid7990924609', 577352),
 ('2013-01-04', u'crid7990924609', 399058),
 ('2013-01-04', u'crid9904536154', 385260),
 ('2013-01-04', u'crid2557171372', 78873)
]

问题是第二个具有重复 ID 但计数不同的列。我需要一个列表来汇总计数，因此列表看起来像这样。 python中有group by cluase吗？

list=[
     ('2013-01-04', u'crid9904536154', 1104937),
     ('2013-01-04', u'crid7990924609', 976410),
     ('2013-01-04', u'crid2557171372', 78874)
    ]

Answer 1

让我们将您的列表命名为 a 而不是 list(list 是 Python 中非常有用的函数，我们不想屏蔽它):

import itertools as it

a = [('2013-01-04', u'crid2557171372', 1),
     ('2013-01-04', u'crid9904536154', 719677),
     ('2013-01-04', u'crid7990924609', 577352),
     ('2013-01-04', u'crid7990924609', 399058),
     ('2013-01-04', u'crid9904536154', 385260),
     ('2013-01-04', u'crid2557171372', 78873)]

b = []
for k,v in it.groupby(sorted(a, key=lambda x: x[:2]), key=lambda x: x[:2]):
    b.append(k + (sum(x[2] for x in v),))

b 现在是:

[('2013-01-04', u'crid2557171372', 78874),
 ('2013-01-04', u'crid7990924609', 976410),
 ('2013-01-04', u'crid9904536154', 1104937)]