python - 如何从列表列表中制作有序字典？

Question

问题是:有一个名称列表和一个列表列表，如何创建一个列表，其中每个项目都是一个有序的字典，名称作为键，列表列表中的项目作为值？从下面的代码中可能会更清楚:

from collections import OrderedDict

list_of_lists = [
                ['20010103', '0.9507', '0.9569', '0.9262', '0.9271'],
                ['20010104', '0.9271', '0.9515', '0.9269', '0.9507'],
                ['20010105', '0.9507', '0.9591', '0.9464', '0.9575'],
                ]

names = ['date', 'open', 'high', 'low', 'close']

我想得到:

ordered_dictionary = [
                     OrderedDict([('date', '20010103'), ('open', '0.9507'), ('high', '0.9569'), ('low', '0.9262'), ('close', '0.9271')]),
                     OrderedDict([('date', '20010104'), ('open', '0.9271'), ('high', '0.9515'), ('low', '0.9269'), ('close', '0.9507')]),
                     OrderedDict([('date', '20010105'), ('open', '0.9507'), ('high', '0.9591'), ('low', '0.9464'), ('close', '0.9575')]),
                     ]

Answer 1

使用zip()组合名称和值。通过列表理解:

from collections import OrderedDict

ordered_dictionary = [OrderedDict(zip(names, subl)) for subl in list_of_lists]

给出:

>>> from pprint import pprint
>>> pprint([OrderedDict(zip(names, subl)) for subl in list_of_lists])
[OrderedDict([('date', '20010103'), ('open', '0.9507'), ('high', '0.9569'), ('low', '0.9262'), ('close', '0.9271')]),
 OrderedDict([('date', '20010104'), ('open', '0.9271'), ('high', '0.9515'), ('low', '0.9269'), ('close', '0.9507')]),
 OrderedDict([('date', '20010105'), ('open', '0.9507'), ('high', '0.9591'), ('low', '0.9464'), ('close', '0.9575')])]