Re: Sort list of dictionaries

From	Peter Otten <__peter__@web.de>
Subject	Re: Sort list of dictionaries
Date	2015-03-03 18:44 +0100
Organization	None
References	(1 earlier) <mailman.56.1425321135.13471.python-list@python.org> <946797be-10e6-433b-9411-2db0d5697ac8@googlegroups.com> <8ccd65d8-3f1f-41ac-9092-9b0832d2fc49@googlegroups.com> <mailman.61.1425324201.13471.python-list@python.org> <432a0d61-a8ba-46ca-bc53-318e234fe168@googlegroups.com>
Newsgroups	comp.lang.python
Message-ID	<mailman.23.1425404702.21433.python-list@python.org> (permalink)

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Charles Heizer wrote:

> On Monday, March 2, 2015 at 11:23:37 AM UTC-8, Peter Otten wrote:
>> Charles Heizer wrote:
>> 
>> > Never mind, the light bulb finally went off. :-\
>> > 
>> > sortedlist = sorted(mylist , key=lambda elem: "%s %s" % ( elem['name'],
>> > (".".join([i.zfill(5) for i in elem['version'].split(".")])) ),
>> > reverse=True)
>> 
>> This lightbulb will break with version numbers > 99999 ;)
>> 
>> Here are two alternatives:
>> 
>> result = sorted(
>>     mylist,
>>     key=lambda elem: (elem['name'], LooseVersion(elem['version'])),
>>     reverse=True)
>> 
>> result = sorted(
>>     mylist,
>>     key=lambda e: (e["name"], tuple(map(int, e["version"].split(".")))),
>>     reverse=True)
>> 
>> 
>> Personally, I prefer to not use a lambda:
>> 
>> def name_version(elem):
>>     return elem['name'], LooseVersion(elem['version'])
>>     
>> result = sorted(mylist, key=name_version, reverse=True)
> 
> Peter, thank you. Me being new to Python why don't you prefer to use a
> lambda?

I find

def name_version(elem):
    return elem['name'], LooseVersion(elem['version'])

more readable than

lambda elem: (elem['name'], LooseVersion(elem['version']))

and I can understand what

>> result = sorted(mylist, key=name_version, reverse=True)

is supposed to do without grokking the implementation of name_version() 
first. I can spot a potential bug -- are the names really supposed to occur 
in reverse order, not just the versions? -- again without a look at the 
implementation.

If I intend to use the script more than once or if I want to rely on the 
result I can write unit tests for name_version() to increase confidence that 
it does what I expect.

Finally I can add a docstring to make it more discoverable in the 
interactive interpreter.

Thread

Sort list of dictionaries Charles Heizer <ceh329@gmail.com> - 2015-03-02 10:17 -0800
  Re: Sort list of dictionaries Emile van Sebille <emile@fenx.com> - 2015-03-02 10:31 -0800
    Re: Sort list of dictionaries Charles Heizer <ceh329@gmail.com> - 2015-03-02 10:38 -0800
      Re: Sort list of dictionaries Ian Kelly <ian.g.kelly@gmail.com> - 2015-03-02 11:55 -0700
      Re: Sort list of dictionaries Charles Heizer <ceh329@gmail.com> - 2015-03-02 10:58 -0800
        Re: Sort list of dictionaries Peter Otten <__peter__@web.de> - 2015-03-02 20:23 +0100
          Re: Sort list of dictionaries Charles Heizer <ceh329@gmail.com> - 2015-03-03 07:56 -0800
            Re: Sort list of dictionaries Chris Angelico <rosuav@gmail.com> - 2015-03-04 03:09 +1100
              Re: Sort list of dictionaries Paul  Moore <p.f.moore@gmail.com> - 2015-03-03 08:48 -0800
            Re: Sort list of dictionaries Peter Otten <__peter__@web.de> - 2015-03-03 18:44 +0100
      Re: Sort list of dictionaries Dave Angel <davea@davea.name> - 2015-03-02 13:59 -0500
      Re: Sort list of dictionaries Jason Friedman <jsf80238@gmail.com> - 2015-03-02 22:33 -0700
      Re: Sort list of dictionaries Chris Angelico <rosuav@gmail.com> - 2015-03-03 18:07 +1100
      Re: Sort list of dictionaries Jason Friedman <jsf80238@gmail.com> - 2015-03-03 07:45 -0700
      Re: Sort list of dictionaries Chris Angelico <rosuav@gmail.com> - 2015-03-04 01:50 +1100
  Re: Sort list of dictionaries Steven D'Aprano <steve+comp.lang.python@pearwood.info> - 2015-03-03 18:55 +1100

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