从特定键开始迭代有序的字典项

这行得通吗？

for x in list(a.keys())[a.index(my_key):]:
    print(a[x])

要从何处开始的密钥在哪里my_key

您可以通过查找 by use 的值并切掉您需要的位置来进行迭代。如果您有自己的方式知道要从哪里开始，请替换。bitems()d.keys().index('b')

from collections import OrderedDict

d = OrderedDict(zip(['a', 'b', 'c', 'd'], range(4)))

for each in d.items()[d.keys().index('b'):]:
    print(each)

使用允许您像往常一样关闭键和值。items()

在我看来，如果您不想索引元组列表，这是一个选项：

from collections import OrderedDict
from itertools import islice

d = OrderedDict(zip(['a', 'b', 'c', 'd'], range(4)))

for each in islice(d.iteritems(), d.keys().index('b'), None):
    print(each)

适用于 Python 2 和 3 的解决方案，使用：itertools.dropwhile()

from __future__ import print_function

from collections import OrderedDict
from itertools import dropwhile

d = OrderedDict(zip(['a', 'b', 'c', 'd'], range(4)))

for k, v in dropwhile(lambda x: x[0] != 'b', d.items()):
    print(k, v)

输出：

b 1
c 2
d 3

Python 2，避免使用 ：.items()

for k, v in dropwhile(lambda x: x[0] != 'b', d.iteritems()):
    print(k, v)

定时

%timeit
for each in d.items()[d.keys().index('b'):]:
    pass
The slowest run took 5.18 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 3.27 µs per loop

%%timeit
for each in islice(d.iteritems(), d.keys().index('b'), None):
    pass
The slowest run took 5.23 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 3.05 µs per loop

%%timeit
for k, v in dropwhile(lambda x: x[0] != 'b', d.iteritems()):
    pass
The slowest run took 4.92 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 2.23 µs per loop