【python cookbook】【数据结构与算法】15.根据字段将记录分组

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‘)) #首先以date字段进行排序
for date, items in groupby(rows, key=itemgetter(‘date‘)):  #再以date进行分组
    print(date)
    for i in items:
        print(‘    ‘, i)

# 如果只是简单地根据日期将数据分组到一起，放进一个大的数据结构中以允许进行随机访问，那么可以利用defaultdict构建一个一键多值的字典会更好
#Example of building a multidict

from collections import defaultdict

rows_by_date = defaultdict(list) #创建一个一键多值的字典，

for row in rows:
    rows_by_date[row[‘date‘]].append(row)

for r in rows_by_date[‘07/01/2012‘]:
    print(r)

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