# 假如我们有一个记录需要进行分组

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"},
]

# 如果我们想按日期进行分组，那itertools里的groupby会很好用
from operator import itemgetter
from itertools import groupby

# 因为groupby只能检查连续的项，所以我们先对列表进行排序
rows.sort(key=itemgetter("date"))
print(rows)

# 然后我们进行分组操作：
for date, items in groupby(rows, itemgetter("date")):
    print(date)
    for i in items:
        print(' ', i)
# groupby每次返回的是一个值（分组名称）和一个子迭代器（组内数据）


# 当然如果是单纯的分组，一键多值字典是个好东西
from collections import defaultdict
data = defaultdict(list)
for item in rows:
    data[item["date"]].append(item)
    
print(data)

# 在不考虑内存开销的情况下，这东西比先排序再groupby快