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我有两个列表，其中 1 存储几个名人的信息，另一个列表存储与这些演员有关的奖项信息。

我试图解决的问题是将这两个列表组合成一个，其中奖励信息成为一个属性，本质上是一个奖励列表。是的，这很容易实现。

for actor in actor_info:
    for award in award_list:
        if actor['personid'] == award['personid']:
            if not actor.get('awards', False):
                actor.update({'awards':[]})
            actor['awards'].append(award)

但是，如果您观察上面的代码，它会重复len(actor_info) * len(award_list)多次，这不是一个优雅的解决方案。这个问题是否有其他观点，执行周期要少得多。

笔记：

为了更清楚地解释这个问题，我在下面描述了正在使用的数据结构。actor_info 和award_info 列表中的每个元素本质上都是一个字典。

actor_info = []

d = {}
d['personid'] = 1210
d['firstname'] = 'Robert , Jr'
d['lastname'] = 'Downey'
d['birthplace'] = 'manhattan, NY'

d1 = {}
d1['personid'] = 2842
d1['firstname'] = 'Brad'
d1['lastname'] = 'Pitt'
d1['birthplace'] = 'Shawnee, OK'

d2 = {}
d2['personid'] = 361
d2['fname'] = 'Cate'
d2['lname'] = 'Blanchett'
d2['birthplace'] = 'Melbournce, Victoria'

d3 = {}
d3['personid'] = 261
d3['fname'] = 'Meg'
d3['lname'] = 'Ryan'
d3['birthplace'] = 'Melbournce, Victoria'

actor_info.append(d)
actor_info.append(d1)
actor_info.append(d2)
actor_info.append(d3)

获奖信息：

k = {} 
k['year'] = '1992'
k['won'] = 'NO'
k['category'] = 'Best Actor'
k['name'] = 'Academy Award'
k['movie'] = 'Chaplin'
k['personid'] = 1210

k1 = {}
k1['year'] = '2008'
k1['won'] = 'NO'
k1['category'] = 'Best Actor'
k1['name'] = 'Academy Award'
k1['movie'] = 'Tropic thunder'
k1['personid'] = 1210

k2 = {}
k2['year'] = '2008'
k2['won'] = 'NO'
k2['category'] = 'Best Actor'
k2['name'] = 'Academy Award'
k2['movie'] = 'The Curious Case of Benjamin Button'
k2['personid'] = 2842

k3 = {}
k3['year'] = '1989'
k3['won'] = 'yes'
k3['category'] = 'Best supporting Actress'
k3['name'] = 'Academy award'
k2['movie'] = 'Aviator'
k3['personid'] = 361

award_list = []
award_list.append(k)
award_list.append(k1)
award_list.append(k2)
award_list.append(k3)
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2 回答 2

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首先，您应该从字典切换到collections.namedtuple，它允许您以普通属性的形式访问数据。

无论如何，您可以通过事先制作查找表来避免二次迭代。

idToActor = {a['personid']:a for a in actor_info}

for award in award_list:
    actor = idToActor[award['personid']]
    actor.setdefault('awards',[]).append(award)
于 2013-05-13T00:08:27.440 回答
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首先，我会将您存储 actor_info 的方式重组为如下所示：

actor_info['personid'] = {'fname':'Cate',...,"awards":list()}

使用您当前的代码，这可以通过以下代码实现：

actor_info = {}
for actor in [d,d1,d2,d3]:
    actor_info[actor['personid']] = dict([(k,v) for (k,v) in actor.items() if k != 'personid'])
    actor_info[actor['personid']]['awards'] = list()

现在给每个演员他们的奖项

for award in award_list:
    if actor_info.has_key(award['personid']): #if an actor exists with the same personID
        actor_info[award['personid']]['awards'].append(award) #append the award to their award list

这将在大约 O(N) 时间内运行。

于 2013-05-13T00:23:46.750 回答