Python - collections.OrderedDict() 未正确排序字典

Question

I have a list of dictionary as follows:

[{'17': 1}, {'17': 1, '19': 1}, {'8': 1, '9': 2, '12': 3}, {'23': 3}]

I want to merge the dictionaries in the list which I do by:

    from collections import Counter

    c = Counter()
    for d in hourofDayData:
        c.update(d)
    temp = dict(c)  

for which I get the following output:

{'17': 2, '19': 1, '8': 1, '9': 2, '12': 3, '23': 3}  

which is what I want, except that it is not ordered. I want the above dictionary to be like:

{'8': 1, '9': 2, '12': 3,'17': 2, '19': 1,  '23': 3}  

I tried to use collections.OrderedDict like this:

OrderedDict([('12', 3), ('17', 2), ('19', 1), ('23', 3), ('8', 1), ('9', 2)])  

Again, which is not ordered. How do I make the dictionary ordered?

Answer 1

Two points to note:

1. OrderedDict is insertion ordered, not ordered by size.
2. Your keys are strings. For ordering by integer size, you need to convert them to integers.

Considering these aspects, you can use sorted with a custom key defined to construct a list of tuples. Then feed this ordered collection to OrderedDict :

d = {'17': 2, '19': 1, '8': 1, '9': 2, '12': 3, '23': 3}

from collections import OrderedDict

res = OrderedDict(sorted(d.items(), key=lambda x: int(x[0])))

Result:

OrderedDict([('8', 1), ('9', 2), ('12', 3), ('17', 2), ('19', 1), ('23', 3)])

It is worth noting in Python 3.7 dictionaries are insertion ordered and this fact can be relied upon. In this case, OrderedDict can be replaced by dict provided the additional methods of OrderedDict are not required.

Answer 2

You should sort the dictionary first then you should convert it to ordereddict. Check the following code.

import collections
from collections import Counter

hourofDayData = [{'17': 1}, {'17': 1, '19': 1}, {'8': 1, '9': 2, '12': 3}, {'23': 3}]

c = Counter()
for d in hourofDayData:
    c.update(d)
temp = dict(c)
def get_key(key):
    try:
        return int(key)
    except ValueError:
        return key
print(collections.OrderedDict(sorted(temp.items(), key=lambda t: get_key(t[0]))))

Prints:

OrderedDict([('8', 1), ('9', 2), ('12', 3), ('17', 2), ('19', 1), ('23', 3)])

Answer 3

Given:

times=[{'17': 1}, {'17': 1, '19': 1}, {'8': 1, '9': 2, '12': 3}, {'23': 3}]

You can do:

from collections import OrderedDict

cnt={}
for d in times:
    for k, v in d.items():
        cnt[k]=cnt.get(k, 0)+v

>>> OrderedDict(sorted([(k,v) for k,v in cnt.items()], key=lambda t:int(t[0])))
OrderedDict([('8', 1), ('9', 2), ('12', 3), ('17', 2), ('19', 1), ('23', 3)])

With Python 3.6+, you do not need to use OrderedDict since the keys of a regular dict use insertion order.

Python 3.7:

cnt={}
for d in times:
    for k, v in d.items():
        cnt[k]=cnt.get(k, 0)+v

>>> dict(sorted([(k,v) for k,v in cnt.items()], key=lambda t:int(t[0])))  
{'8': 1, '9': 2, '12': 3, '17': 2, '19': 1, '23': 3}