Question

我有一点思想缠绕，给出这样的数据：

data = [('topic1', (['apples', 'oranges'], 0.14975108213820515)),
       ('topic2', (['oranges', 'raisins'], 0.14975108213820515)),
       ('topic3', (['grapes', 'raisins'], 0.14975108213820515)),
       ('topic4', (['trees', 'flowers'], 0.14975108213820515))]

我想根据数组中的至少一个文本（在元组的第二个元素的第一个元素中）是否共同来连接主题。所以在上面的例子中：

topic1 is connected to topic2 
topic2 is connected to topic1 and topic3
topic3 is connected to topic2
topic4 is unconnected

理想情况下，我的输出看起来像：

output = [(topic1,topic2), 
         (topic1,topic2, topic3),
         (topic3, topic2),
         (topic4)]

所以，给定像data这样的输入我怎样才能获得像output这样的输出。我认为itertools可能会以某种方式参与其中，但我现在真的陷入困境。

Answer 1

有效的方法是使用set s。

>>> set1= set(['apples', 'oranges'])
>>> set2= set(['oranges', 'raisins'])
>>> print len(set1.intersection(set2))
1

所以，基本上：

为每个主题的文字创建一个集合
对于每个主题，迭代彼此主题并检查其文本集交集的len

topic_text_sets= {topic:set(text) for topic,(text,_) in data}
topic_related= {}
for topic1, text1 in topic_text_sets.iteritems():
    related= [topic2 for topic2, text2 in topic_text_sets.iteritems() if topic1!=topic2 and len(text1.intersection(text2))>0]
    print related

topic1 ['topic2']
topic3 ['topic2']
topic2 ['topic1', 'topic3']
topic4 []

Answer 2

您创建一个包含列表的字典以捕获连接：

connections = {}
for topic, (conns, some_number) in data:
    for conn in conns:
        connections.setdefault(conn, set()).add(topic)

这会将连接值映射到主题集。

现在你可以查找反向连接;如果订单不重要，只需获取所有连接值集的并集：

output = [tuple(set().union(*(connections[c] for c in conns))) 
          for topic, (conns, some_number) in data]

演示：

>>> data = [('topic1', (['apples', 'oranges'], 0.14975108213820515)),
...        ('topic2', (['oranges', 'raisins'], 0.14975108213820515)),
...        ('topic3', (['grapes', 'raisins'], 0.14975108213820515)),
...        ('topic4', (['trees', 'flowers'], 0.14975108213820515))]
>>> connections = {}
>>> for topic, (conns, some_number) in data:
...     for conn in conns:
...         connections.setdefault(conn, set()).add(topic)
... 
>>> [tuple(set().union(*(connections[c] for c in conns))) 
...               for topic, (conns, some_number) in data]
[('topic1', 'topic2'), ('topic1', 'topic3', 'topic2'), ('topic3', 'topic2'), ('topic4',)]
>>> from pprint import pprint
>>> pprint(_)
[('topic1', 'topic2'),
 ('topic1', 'topic3', 'topic2'),
 ('topic3', 'topic2'),
 ('topic4',)]

否则先将topic从第一组中删除，将output = [(topic,) + tuple(set().union(*(connections[c] for c in conns)) - {topic}) for topic, (conns, some_number) in data] >>> [(topic,) + tuple(set().union(*(connections[c] for c in conns)) - {topic}) ... for topic, (conns, some_number) in data] [('topic1', 'topic2'), ('topic2', 'topic1', 'topic3'), ('topic3', 'topic2'), ('topic4',)] >>> pprint(_) [('topic1', 'topic2'), ('topic2', 'topic1', 'topic3'), ('topic3', 'topic2'), ('topic4',)]移到前面：

{{1}}

Answer 3

带有两个for循环的简单：

>>> for i in range(len(data)):
...     x = set(data[i][1][0])
...     for j in range(len(data)):
...         if len(x & set(data[j][1][0]))>=1:
...             print data[j][0],             # for python 3 use print()
...     print
... 
topic1 topic2
topic1 topic2 topic3
topic2 topic3
topic4

Answer 4

将其分解为子问题。

首先，您需要获取所有不同的文本，可能使用列表理解（或设置理解以避免重复）。然后你需要遍历它，并为每个文本找到data中包含它作为其一部分的每个部分。你不应该使用itertools - 这可能会使它复杂化。

Python中的高级分组

4 个答案: