如何在字典理解中创建某些值的元组?

时间:2019-05-02 07:49:38

标签: python list dictionary list-comprehension

我有很多这样的字典:

data_sample = {'query_result': {'imdbid': 50083, 'file name': '12.Angry.Men.1957.Criterion.Collection.720p.BluRay.x264-WiKi.fre.srt', 'IDSubtitleFile': '1952985556'}, 
                'movie directory': 'C:\\...\\Movies\\12 Angry Men 1957 1080p BluRay x264 AAC - Ozlem'}

基本上:

data_collection = [data_sample] * 10

我想要的输出是一个元组,该元组包含:

(query_result["IDSubtitleFile"], movie_directory)

我将如何使用列表理解来创建所需的输出?我知道如何使用for循环来实现它,但是我正在努力使用列表理解来管理它。我唯一的想法是,但这会引发错误,并且实际上没有任何意义:

[[(value["IDSubtitleFile"], value) for value in data_sample.values()] for data_sample in data_collection]

写完这些之后,我想对于这种情况完全不使用列表推导可能更明智。

1 个答案:

答案 0 :(得分:2)

这种理解会做到:

results = [
    (d["query_result"]["IDSubtitleFile"], d["movie directory"]) 
    for d in data_collection
]

结果:

>>> pprint([(d["query_result"]["IDSubtitleFile"], d["movie directory"]) for d in data_collection])
[('1952985556',
  'C:\\...\\Movies\\12 Angry Men 1957 1080p BluRay x264 AAC - Ozlem'),
 ('1952985556',
  'C:\\...\\Movies\\12 Angry Men 1957 1080p BluRay x264 AAC - Ozlem'),
 ('1952985556',
  'C:\\...\\Movies\\12 Angry Men 1957 1080p BluRay x264 AAC - Ozlem'),
 ('1952985556',
  'C:\\...\\Movies\\12 Angry Men 1957 1080p BluRay x264 AAC - Ozlem'),
 ('1952985556',
  'C:\\...\\Movies\\12 Angry Men 1957 1080p BluRay x264 AAC - Ozlem'),
 ('1952985556',
  'C:\\...\\Movies\\12 Angry Men 1957 1080p BluRay x264 AAC - Ozlem'),
 ('1952985556',
  'C:\\...\\Movies\\12 Angry Men 1957 1080p BluRay x264 AAC - Ozlem'),
 ('1952985556',
  'C:\\...\\Movies\\12 Angry Men 1957 1080p BluRay x264 AAC - Ozlem'),
 ('1952985556',
  'C:\\...\\Movies\\12 Angry Men 1957 1080p BluRay x264 AAC - Ozlem'),
 ('1952985556',
  'C:\\...\\Movies\\12 Angry Men 1957 1080p BluRay x264 AAC - Ozlem')]