擅长:python、mysql、java
<p>我希望python代码足够快:</p>
<pre><code>movielist = [
('imdbid22', 'genreid1'),
('imdbid22', 'genreid2'),
('imdbid44, 'genreid1'),
('imdbid55', 'genreid8'),
]
dict = {}
for items in movielist:
if dict[items[0]] not in dict:
dict[items[0]] = items[1]
else:
dict[items[0]] = dict[items[0]].append(items[1])
print dict
</code></pre>
<p>输出:</p>
^{pr2}$
<p>如果你只需要电影名,数数:
在原始查询中更改此项,您将得到不需要python代码的答案</p>
<p><code>SELECT "movie"."imdb_id", count("moviegenre"."genre_id")</code></p>
<p><code>group by "movie"."imdb_id"</code></p>