python3解压缩元组列表

Question

在python2.7中,下面的代码获取字典fd(在此示例中表示单词及其计数的频率分布),并将其分为两个列表的列表:[[keys],[values]]:

sortedDKandVs = [zip(*sorted(fd.items(), key=itemgetter(1), reverse=True))] #[word,word,...],[count,count]

我可以做,例如:

keys = sortedDKandVs[0]
values = sortedDKandVs[1]

这不再适用于Python3,我想知道如何转换代码.

这里没有答案如何将元组列表解压缩到单个列表中？因为在Python3中,zip对象返回迭代器而不是列表,但是我不知道如何转换答案.

Answer 1

Python 2:

Python 2.7.6 (default, Apr  9 2014, 11:48:52) 
[GCC 4.2.1 Compatible Apple LLVM 5.1 (clang-503.0.38)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> di={'word1':22, 'word2':45, 'word3':66}
>>> zip(*sorted(di.items(), key=itemgetter(1), reverse=True))
[('word3', 'word2', 'word1'), (66, 45, 22)]
>>> k,v=zip(*sorted(di.items(), key=itemgetter(1), reverse=True))
>>> k
('word3', 'word2', 'word1')
>>> v
(66, 45, 22)

Python 3:

Python 3.4.1 (default, May 19 2014, 13:10:29) 
[GCC 4.2.1 Compatible Apple LLVM 5.1 (clang-503.0.40)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> di={'word1':22, 'word2':45, 'word3':66}
>>> k,v=zip(*sorted(di.items(), key=itemgetter(1), reverse=True))
>>> k
('word3', 'word2', 'word1')
>>> v
(66, 45, 22)

对于Python 2和Python 3,它完全相同

如果你想要列表与元组(Python 3和Python 2):

>>> k,v=map(list, zip(*sorted(di.items(), key=itemgetter(1), reverse=True)))
>>> k
['word3', 'word2', 'word1']
>>> v
[66, 45, 22]