python – 从特定键开始迭代有序的dict项

2023年7月27日 267次阅读

这个问题在
python 2.7中设计.

我正在使用OrderedDict存储一些项目,如下所示：

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

提前谢谢了！

最佳答案 使用itertools.dropwhile()的适用于Python 2和3的解决方案：

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)

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
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