python: sorting key:value pair based on key

Question

I have a dictionary as

    i={106:0.33,107:0.21,98:0.56}

I want to sort the key:value pairs based on keys. I am expecting output to look like following:

    {98:0.56, 106:0.33, 107:0.21}

When i use this:

    od = collections.OrderedDict(sorted(i.items()))
    print(od)

I am getting wrong output as:

    [(98,0.56),(106,0.33),(107,0.21)]

Can anyone please help me with correct code for this?

Answer 1

An OrderedDict remembers the order that you added items to the dictionary, not the sort order.

But you can easily create a sorted list from a regular dictionary. For example:

>>> list1={106:0.33,107:0.21,98:0.56}
>>> sorted(list1.items())
[(98, 0.56), (106, 0.33), (107, 0.21)]

Or if you want it in an OrderedDict, just apply the OrderedDict after the sort:

>>> from collections import OrderedDict
>>> 
>>> list1={106:0.33,107:0.21,98:0.56}
>>> OrderedDict(sorted(list1.items()))
OrderedDict([(98, 0.56), (106, 0.33), (107, 0.21)])

But a regular dict has no concept of sort order, so you can't sort the dictionary itself.

However, if you really want that display, you can create your own class:

>>> from collections import OrderedDict
>>> class MyDict(OrderedDict):
...     def __repr__(self):
...         return '{%s}' % ', '.join(str(x) for x in self.items())
... 
>>> list1={106:0.33,107:0.21,98:0.56}
>>> MyDict(sorted(list1.items()))
{(98, 0.56), (106, 0.33), (107, 0.21)}