#!/usr/bin/python
# 1.15. Grouping Records Together Based on a Field
# Problem: You have a sequence of dictionaries or instances and you want to iterate over the data
# in groups based on the value of a particular field, such as date.
from operator import itemgetter
from itertools import groupby
# To iterate over the data in chunks grouped by date.
# First, sort by the desired field (in this case, date) and
# then use itertools.groupby():
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'},
]
# Sort by the desired field first
rows.sort(key=itemgetter('date'))
print (rows)
for date, items in groupby(rows, key=itemgetter('date')):
print(date)
for i in items:
print(' ', i)
上述代码的输出如下:
[{'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'}]
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'}
“日期”位于“地址”前面。
但是,如果我通过在第24行添加print (rows)
来更改代码,如下所示:
#!/usr/bin/python
# 1.15. Grouping Records Together Based on a Field
# Problem: You have a sequence of dictionaries or instances and you want to iterate over the data
# in groups based on the value of a particular field, such as date.
from operator import itemgetter
from itertools import groupby
# To iterate over the data in chunks grouped by date.
# First, sort by the desired field (in this case, date) and
# then use itertools.groupby():
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'},
]
print (rows)
# Sort by the desired field first
rows.sort(key=itemgetter('date'))
print (rows)
for date, items in groupby(rows, key=itemgetter('date')):
print(date)
for i in items:
print(' ', i)
上述代码的输出如下:
[{'address': '5412 N CLARK', 'date': '07/01/2012'}, {'address': '4801 N BROADWAY', 'date': '07/01/2012'}, {'address': '5800 E 58TH', 'date': '07/02/2012'}, {'address': '5645 N RAVENSWOOD', 'date': '07/02/2012'}, {'address': '1060 W ADDISON', 'date': '07/02/2012'}, {'address': '2122 N CLARK', 'date': '07/03/2012'}, {'address': '5148 N CLARK', 'date': '07/04/2012'}, {'address': '1039 W GRANVILLE', 'date': '07/04/2012'}]
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'}
“地址”位于“日期”前面。
为什么键的顺序会改变?
答案 0 :(得分:6)
订单的变化不是因为您添加了一行代码,而是因为hash randomization。实现散列随机化使用数万个值的破碎序列来减轻DoS攻击,这些值散列到例如相同的值中。 HTTP POST请求。
答案 1 :(得分:2)
如果您希望订单保持不变,则需要使用collections中的OrderedDict
。
from collections import OrderedDict
row = OrderedDict([('address', '5412 N CLARK'), ('date', '07/01/2012')])
>>> row
OrderedDict([('address', '5412 N CLARK'), ('date', '07/01/2012')])
>>> rows.keys()
['address', 'date']