从特定键开始迭代有序 dict 项

Question

问题的样式在 python 2.7 中。

我使用OrderedDict来存储一些项目如下：

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

（d 等于 {'a': 0, 'b': 1, 'c': 2, 'd': 3}）

有没有办法从特定键开始迭代字典d？ 例如，我想从键 'b' 开始迭代 d 项目

非常感谢！

Answer 1

这行得通吗？

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

my_key 是您要开始的键

Answer 2

您可以通过使用items() 查找b 的值并在需要的位置进行切片来进行迭代。如果您有自己的方式知道要从哪里开始，请替换 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items() 可以让您像往常一样获取键和值。

Answer 3

在我看来，如果您不想索引元组列表，这是一种选择：

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)

Answer 4

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():: 创建键值列表

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

时间

%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