擅长:python、mysql、java
<p>你可以用<a href="https://docs.python.org/2/library/collections.html#collections.Counter" rel="nofollow noreferrer">^{<cd1>}</a>来做</p>
<pre><code>In [42]: from collections import Counter
In [43]: lst = [('ABC',123,10),('ABC',123,10),('DEF',123,5)]
In [44]: [(i[0],i[1],i[2]*j) for i,j in Counter(lst).items()]
Out[44]: [('DEF', 123, 5), ('ABC', 123, 20)]
</code></pre>
<p>根据OP的建议,如果它有不同的值,请使用<code>groupby</code></p>
<pre><code>In [26]: lst = [('ABC',123,10),('ABC',123,10),('ABC',123,25),('DEF',123,5)]
In [27]: [tuple(list(n)+[sum([i[2] for i in g])]) for n,g in groupby(sorted(lst,key = lambda x:x[:2]), key = lambda x:x[:2])]
Out[27]: [('ABC', 123, 45), ('DEF', 123, 5)]
</code></pre>