views:

151

answers:

5
[{'date': '2010-04-01', 'people': 1047, 'hits': 4522}, {'date': '2010-04-03', 'people': 617, 'hits': 2582}, {'date': '2010-04-02', 'people': 736, 'hits': 3277}]

Suppose I have this list. How do I sort by "date", which is an item in the dictionary. But, "date" is a string...

+5  A: 
.sort(key=lambda x: datetime.datetime.strptime(x['date'], '%Y-%m-%d'))
Ignacio Vazquez-Abrams
this looks cool but i got "None" for an answer when i tried it!
controlfreak123
Most `list` methods work in-place, instead of returning the list.
Ignacio Vazquez-Abrams
+1  A: 
records = [
     {'date': '2010-04-01', 'people': 1047, 'hits': 4522}, 
     {'date': '2010-04-03', 'people': 617, 'hits': 2582}, 
     {'date': '2010-04-02', 'people': 736, 'hits': 3277}
     ]
records.sort(key=lambda x: x['date'].split('-'))
Satoru.Logic
+6  A: 

Fortunately, ISO format dates, which seems to be what you have here, sort perfectly well as strings! So you need nothing fancy:

import operator
yourlistofdicts.sort(key=operator.itemgetter('date'))
Alex Martelli
`operator` to the rescue!!
jathanism
BTW, I totally want this to work but it's not working for me w/ Py2.6.2 and the input list from the OP. It returns `TypeError: itemgetter expected 1 arguments, got 2`. If I run it using `sorted` it works: `sorted(yourlistofdicts, key=operator.itemgetter('date'))`. Thoughts?
jathanism
Oops, typo: I missed the `key=`, editing to fix, thanks.
Alex Martelli
+2  A: 

In python 2.6 you can use soerted w/operator.itemgetter. Since date is YYYY-MM-DD it is sorted even though its a string cause its largest to smallest - i use that format all the time for this reason

>>> import operator
>>> l = [{'date': '2010-04-01','people': 1047, 'hits': 4522}, 
         {'date': '2010-04-03', 'people': 617, 'hits': 2582}, 
         {'date': '2010-04-02', 'people': 736, 'hits': 3277}]
>>> sorted( l, key = operator.itemgetter('date') )
[{'date': '2010-04-01', 'hits': 4522, 'people': 1047}, {'date': '2010-04-02', 'hits': 3277, 'people': 736}, {'date': '2010-04-03', 'hits': 2582, 'people': 617}]
Rescommunes
+3  A: 

Satoru.Logic's solution is clean and simple. But, per Alex's post, you don't need to manipulate the date string to get the sort order right...so lose the .split('-')

This code will suffice:

records.sort(key=lambda x:x['date'])
damzam
...And once again the simplest, most Pythonic, and correct answer is hastily brushed aside when the asker is too hasty *sighs*.
Beau Martínez
Upvoted for simplicity, but definitely requires dates to be in YYYY-MM-DD format (which therefore sort similarly as strings).
dkamins