需要帮助理解 Ordereddict 的行为

Question

我有一个星期几的有序字典：

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，因此如果我将 n 设置为 4，则第 4 个键是 'Thu' 所以工作日变为：

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，但如果我想更改与第 n 个键之前或之后的键对应的值，则 ordereddict 开始表现得很滑稽。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')])

如何实现我想要的结果？

Answer 1

因为您进行的是词法比较而不是数字排序，'Tue' > 'Thurs':

您可能想要尝试的只是 enumerate() 键并使用数值，例如：

In []:
for i, key in enumerate(weekdays, 1):
    if i < startday_2017:
        weekdays[key] = "applesauce"
    elif i == startday_2017:
        weekdays[key] = 1
    else:
        weekdays[key] = 2
weekdays

Out[]:
OrderedDict([('Mon', 'applesauce'),
             ('Tue', 'applesauce'),
             ('Wed', 'applesauce'),
             ('Thu', 1),
             ('Fri', 2),
             ('Sat', 2),
             ('Sun', 2)])

需要帮助理解 Ordereddict 的行为

Need help understanding behaviour of Ordereddict

ordereddictionary

python-3.x