【发布时间】:2017-08-09 19:40:58
【问题描述】:
我有一本有序的星期几的字典:
weekdays = collections.OrderedDict([ ('Mon', 0), ('Tue', 0), ('Wed', 0), ('Thu', 0), ('Fri', 0), ('Sat', 0), ('Sun', 0) ])
我想将第 n 个键的值更改为 1,所以如果我将 n 设置为 4,则第 4 个键是“星期四”,因此工作日变为:
OrderedDict([('Mon', 0), ('Tue', 0), ('Wed', 0), ('Thu', 1), ('Fri', 0), ('Sat', 0), ('Sun', 0)])
我可以用下面的代码做到这一点:
startday_2017 = 4
weekdays = collections.OrderedDict([ ('Mon', 0), ('Tue', 0), ('Wed', 0), ('Thu', 0), ('Fri', 0), ('Sat', 0), ('Sun', 0) ])
date = list(weekdays.keys())[(startday_2017-1)]
for key in weekdays.keys():
if key == date:
weekdays[key] = 1
这似乎可行,但是如果我想更改与第 n 个键之前或之后的键对应的值,ordereddict 就会开始表现得很有趣。使用此代码:
startday_2017 = 4
weekdays = collections.OrderedDict([ ('Mon', 0), ('Tue', 0), ('Wed', 0), ('Thu', 0), ('Fri', 0), ('Sat', 0), ('Sun', 0) ])
date = list(weekdays.keys())[(startday_2017-1)]
for key in weekdays.keys():
if key < date:
weekdays[key] = "applesauce"
elif key == date:
weekdays[key] = 1
else:
weekdays[key] = 2
print(weekdays)
我得到这个输出:
OrderedDict([('Mon', 'applesauce'), ('Tue', 2), ('Wed', 2), ('Thu', 1), ('Fri', 'applesauce'), ('Sat', 'applesauce'), ('Sun', 'applesauce')])
如何达到我想要的结果?
【问题讨论】:
标签: python-3.x ordereddictionary