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The problem is: Having a list of names, and a list of lists, how to create a list, in which each item is an ordered dictionary with names as keys, and items from list of lists as values? It might be more clear from code below:
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']
I would like to get:
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')]),
]
809
To convert a list to a dictionary using the same values, you can use the dict. fromkeys() method. To convert two lists into one dictionary, you can use the Python zip() function. The dictionary comprehension lets you create a new dictionary based on the values of a list.
OrderedDict is part of python collections module. We can create an empty OrderedDict and add items to it. If we create an OrderedDict by passing a dict argument, then the ordering may be lost because dict doesn't maintain the insertion order. If an item is overwritten in the OrderedDict, it's position is maintained.
We can convert a nested list to a dictionary by using dictionary comprehension. It will iterate through the list. It will take the item at index 0 as key and index 1 as value.
Use zip() to combine the names and the values. With a list comprehension:
from collections import OrderedDict
ordered_dictionary = [OrderedDict(zip(names, subl)) for subl in list_of_lists]
which gives:
>>> 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')])]
I know this question is very old, but I thought I'd suggest a namedtuple solution as an alternative to OrderedDict that would work well in this situation:
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'
Even better, one could use dictionary comprehension with the date as the key:
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'
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