[PY3]——根据某个特定的字段来分组迭代访问一个字段或序列？/ itertools.groupby( )函数

rows = [
    {‘address‘: ‘5412 N CLARK‘, ‘date‘: ‘07/01/2012‘},
    {‘address‘: ‘5148 N CLARK‘, ‘date‘: ‘07/04/2012‘},
    {‘address‘: ‘5800 E 58TH‘, ‘date‘: ‘07/02/2012‘},
    {‘address‘: ‘2122 N CLARK‘, ‘date‘: ‘07/03/2012‘},
    {‘address‘: ‘5645 N RAVENSWOOD‘, ‘date‘: ‘07/02/2012‘},
    {‘address‘: ‘1060 W ADDISON‘, ‘date‘: ‘07/02/2012‘},
    {‘address‘: ‘4801 N BROADWAY‘, ‘date‘: ‘07/01/2012‘},
    {‘address‘: ‘1039 W GRANVILLE‘, ‘date‘: ‘07/04/2012‘},
]
from operator import itemgetter
from itertools import groupby
rows.sort(key=itemgetter(‘date‘))
print(rows)

　　[{‘date‘: ‘07/01/2012‘, ‘address‘: ‘5412 N CLARK‘}, {‘date‘: ‘07/01/2012‘, ‘address‘: ‘4801 N BROADWAY‘}, {‘date‘: ‘07/02/2012‘, ‘address‘: ‘5800 E 58TH‘}, {‘date‘: ‘07/02/2012‘, ‘address‘: ‘5645 N RAVENSWOOD‘}, {‘date‘: ‘07/02/2012‘, ‘address‘: ‘1060 W ADDISON‘}, {‘date‘: ‘07/03/2012‘, ‘address‘: ‘2122 N CLARK‘}, {‘date‘: ‘07/04/2012‘, ‘address‘: ‘5148 N CLARK‘}, {‘date‘: ‘07/04/2012‘, ‘address‘: ‘1039 W GRANVILLE‘}]


for date,items in groupby(rows,key=itemgetter(‘date‘)):
    print(date)
    for i in items:
        print(‘ ‘,i)
07/01/2012
{‘date‘: ‘07/01/2012‘, ‘address‘: ‘5412 N CLARK‘}
{‘date‘: ‘07/01/2012‘, ‘address‘: ‘4801 N BROADWAY‘}
07/02/2012
{‘date‘: ‘07/02/2012‘, ‘address‘: ‘5800 E 58TH‘}
{‘date‘: ‘07/02/2012‘, ‘address‘: ‘5645 N RAVENSWOOD‘}
{‘date‘: ‘07/02/2012‘, ‘address‘: ‘1060 W ADDISON‘}
07/03/2012
{‘date‘: ‘07/03/2012‘, ‘address‘: ‘2122 N CLARK‘}
07/04/2012
{‘date‘: ‘07/04/2012‘, ‘address‘: ‘5148 N CLARK‘}
{‘date‘: ‘07/04/2012‘, ‘address‘: ‘1039 W GRANVILLE‘}