sorting python list based on key [duplicate]

Question

I have some Entrys in a python list.Each Entry has a creation date and creation time.The values are stored as python datetime.date and datetime.time (as two separate fields).I need to get the list of Entrys sorted sothat previously created Entry comes before the others.

I know there is a list.sort() function that accepts a key function.In this case ,do I have to use the date and time to create a datetime and use that as key to sort()? There is a datetime.datetime.combine(date,time) for this. But how do I specify this inside the sort function?

I tried key = datetime.datetim.combine(created_date,created_time)

but the interpreter complains that the name created_date is not defined

class Entry: 
   created_date = #datetime.date
   created_time  = #datetime.time
   ...

my_entries_list=[Entry1,Entry2...Entry10]
my_entries_list.sort(key = datetime.datetim.combine(created_date,created_time))

This can never work. You need define your own method or lambda method for performing the calculation needed here. — user2665694
– user2665694, Commented Jan 7, 2013 at 14:58

user4815162342 · Accepted Answer · 2013-01-07 15:17:04Z

9

You probably want something like:

my_entries_list.sort(key=lambda v:
                           datetime.datetime.combine(v.created_date, v.created_time))

Passing datetime.datetime.combine(created_date, created_time) tries to call combine immediately and breaks since created_date and created_time are not available as local variables. The lambda provides delayed evaluation: instead of executing the code immediately, it creates a function that will, when called, execute the specified code and return the result. The function also provides the parameter that will be used to access the created_date and created_time attributes.

edited Jan 7, 2013 at 15:17

answered Jan 7, 2013 at 14:58

user4815162342

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Fred Foo · Accepted Answer · 2013-01-07 14:58:52Z

3

Use lambda:

sorted(my_entries_list,
       key=lambda e: datetime.combine(e.created_date, e.created_time))

answered Jan 7, 2013 at 14:58

Fred Foo

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Useless · Accepted Answer · 2013-01-07 15:18:36Z

1

From the documentation

key specifies a function of one argument that is used to extract a comparison key from each list element

eg.

def combined_date_key(elem):
    return datetime.datetime.combine(v.created_date, v.created_time)

entries.sort(key=combined_date_key)

(this is identical to the lambda already posted, just broken out for clarity).

Your other option is to use the cmp argument (it's not required here, but for a sufficiently expensive key function which can be short-circuited, this could be faster)

edited Jan 7, 2013 at 15:18

answered Jan 7, 2013 at 15:02

Useless

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xorsyst · Accepted Answer · 2013-01-07 15:26:15Z

0

It may be slightly faster to use a tuple like this:

sorted(my_entries_list,
       key = lambda e: (e.created_date, e.created_time))

Then things will be sorted by date, and only if the dates match will the time be considered.

answered Jan 7, 2013 at 15:26

xorsyst

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