擅长:python、mysql、java
<p>让我们把你的列表命名为<code>a</code>,而不是<code>list</code>(<code>list</code>在Python中是一个非常有用的函数,我们不想屏蔽它):</p>
<pre><code>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),))
</code></pre>
<p><code>b</code>现在是:</p>
<pre><code>[('2013-01-04', u'crid2557171372', 78874),
('2013-01-04', u'crid7990924609', 976410),
('2013-01-04', u'crid9904536154', 1104937)]
</code></pre>