python sort dictionary by key and list by another key

Question

I have a list of dictionaries and need only the items with and unique item names but only the dictionary which has latest endtime.

eg:

query_result = (
{item: 'name1', starttime ='2013-10-29 07:28:14', endtime = '2013-10.29 08:28:14'}, 
{item: 'name1', starttime ='2013-10-29 08:28:14', endtime = '2013-10.29 09:28:14'},
{item: 'name1', starttime ='2013-10-29 07:28:14', endtime = '2013-10.29 08:28:14'},
{item: 'name2', starttime ='2013-10-29 07:28:14', endtime = '2013-10.29 07:29:14'},
{item: 'name2', starttime ='2013-10-29 07:28:14', endtime = '2013-10.29 07:45:14'},
{item: 'name2', starttime ='2013-10-29 07:28:14', endtime = '2013-10.29 08:28:14'} )

so the result should be

only

({item1: 'name1', starttime ='2013-10-29 07:28:14', endtime = '2013-10.29 09:28:14'},
 {item2: 'name2', starttime ='2013-10-29 07:28:14', endtime = '2013-10.29 08:28:14'})

any hints?

UPDATE:

what if the endtime is in datetime format already?

 {item: 'name1', starttime : datetime.datetime(2013, 10, 30, 7, 13, 21), endtime:datetime.datetime(2013, 10, 30, 7, 13, 21}

possible duplicate of [In Python how do I sort a list of dictionaries by values of the dictionary?](http://stackoverflow.com/questions/72899/in-python-how-do-i-sort-a-list-of-dictionaries-by-values-of-the-dictionary) — Joe, Oct 30 '13 at 16:16

score 1 · Answer 1 · answered Oct 30 '13 at 16:17

Sort by descending endtime, then take the first of each name:

import datetime
def endtime_sort_key(item):
    return datetime.datetime.strptime(item['endtime'], "%Y-%m.%d %H:%M:%S")

used_names = set()
final_result = []
for item in sorted(query_result, key=endtime_sort_key, reverse=True):
    if item['item'] not in used_names:
        final_result.append(item)
        used_names.add(item['item'])

oleg · Accepted Answer · 2013-10-30T18:50:21.437

I think itertools.groupby(iterable[, key]) can be helpful here.

You can group data by item with key item And then pick desired item

used_names = set()
final_result = []
for k, g in itertools.groupby(query_result, key=lambda x: x['item']):
    final_result.append(sorted(g, key=lambda item: item['endtime'], reverse=True)[0])

if end_time item is string You should strip first

import datetime
def endtime_sort_key(item):
    return datetime.datetime.strptime(item['endtime'], "%Y-%m.%d %H:%M:%S")

used_names = set()
final_result = []
for k, g in itertools.groupby(query_result, key=lambda x: x['item']):
    final_result.append(sorted(g, key=endtime_sort_key, reverse=True)[0])

thanks oleg, this seems to work, however the endtime key is in datetime format already...any idea? — krisdigitx, Oct 30 '13 at 17:34

score 0 · Answer 3 · answered Oct 30 '13 at 17:02

from time import strptime

result = []
for i in query_result:
    tmp = {}
    for j in (k for k in query_result if k['item'] == i['item']):
        if strptime(j['endtime'], '%Y-%m.%d %I:%M:%S') > strptime(i['endtime'], '%Y-%m.%d %I:%M:%S'):
            tmp['item'] = j['item']
            tmp['starttime'] = j['starttime']
            tmp['endtime'] = j['endtime']
    if(tmp and tmp not in result):
        result.append(tmp)

result = (tuple(result))

python sort dictionary by key and list by another key

3 Answers3