問題

你有一個字典或者實例的序列，然後你想根據某個特定的字段(比如‘date’)來分組叠代訪問。

解決方案

itertools.groupby( )函數

itertools.groupby(rows,key=itemgetter(‘字段‘))

• groupby( )函數掃描整個序列並且查找連續相同值（或者根據指定 key函數返回值相同）的元素序列
• 在調用groupby( )函數前，我們首先需要按照這個字段來排序（這和SQL語句中的group by的使用是一個道理）
• 因為groupby( )僅僅檢查連續的元素，如果事先並沒有排序完成的話，分組函數將得不到想要的結果

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‘}

[PY3]——根據某個特定的字段來分組叠代訪問一個字段或序列？/ itertools.groupby( )函數