如何从列表生成有序字典？

Question

问题是：

有一个名字的列表，还有一个列表的列表，如何创建一个新的列表，其中每个项目都是一个有序字典，字典的键是名字，值是来自列表的列表的项目？下面的代码可能会更清楚：

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']

我想得到的是：

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

我知道这个问题已经很久了，但我想推荐一个用namedtuple的解决方案，作为OrderedDict的替代方案，这在这种情况下会很好用：

from collections import namedtuple

Bar = namedtuple('Bar', ['date', 'open', 'high', 'low', 'close'])

bars = [Bar(date, o, h, l, c) for date, o, h, l, c in list_of_lists]

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

>>> bars[2].date
'20010105'

>>> bars[2].close
'0.9575'

更棒的是，可以用字典推导式，把日期作为键来使用：

Bar = namedtuple('Bar', ['open', 'high', 'low', 'close'])

bars = {date: Bar(o, h, l, c) for date, o, h, l, c in list_of_lists}

>>> bars
{'20010103': Bar(open='0.9507', high='0.9569', low='0.9262', close='0.9271'),
 '20010104': Bar(open='0.9271', high='0.9515', low='0.9269', close='0.9507'),
 '20010105': Bar(open='0.9507', high='0.9591', low='0.9464', close='0.9575')}

>>> bars['20010105']
Bar(open='0.9507', high='0.9591', low='0.9464', close='0.9575')

>>> bars['20010105'].close
'0.9575'

Answer 2

使用 zip() 函数可以把名字和对应的值组合在一起。你可以用一种叫做列表推导式的方法来实现：

from collections import OrderedDict

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

这样就能得到：

>>> 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')])]