Sorting dictionary by datetime value
Frank Millman
frank at chagford.com
Sat Feb 8 03:03:42 EST 2014
"Chris Angelico" <rosuav at gmail.com> wrote in message
news:CAPTjJmqDusdFC1eLbU6LF5-up__LAE-63ii0UUvAGGNem9U4+w at mail.gmail.com...
> On Sat, Feb 8, 2014 at 6:06 PM, Igor Korot <ikorot01 at gmail.com> wrote:
>>>>> sorted(a.items(), key=a.get)
>> [('1', datetime.datetime(2012, 12, 28, 12, 15, 30, 100)), ('3',
>> datetime.datetim
>> e(2012, 12, 28, 12, 16, 44, 100)), ('2', datetime.datetime(2012, 12, 28,
>> 12, 17,
>> 29, 100))]
That seemed like a neat trick, so I thought I would try to understand it a
bit better in case I could use it some day.
I am using python3. I don't know if that makes a difference, but I cannot
get it to work.
>>> d = {1: 'abc', 2: 'xyz', 3: 'pqr'}
>>> sorted(d.items(), key=d.get)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: unorderable types: NoneType() < NoneType()
>>>
I know that python3 is stricter regarding ordering of non-comparable types,
but I don't see where None is coming from.
I have python 2.7.3 on another machine. Here are the results -
>>> d = {1: 'abc', 2: 'xyz', 3: 'pqr'}
>>> sorted(d.items(), key=d.get)
[(1, 'abc'), (2, 'xyz'), (3, 'pqr')]
It did not crash, but it did not sort.
Then I changed the keys to strings, to match Igor's example -
>>> d = {'1': 'abc', '2': 'xyz', '3': 'pqr'}
>>> sorted(d.items(), key=d.get)
[('1', 'abc'), ('3', 'pqr'), ('2', 'xyz')]
It works - now I am even more confused.
Any hints will be appreciated.
Frank Millman
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