从特定键开始迭代有序的dict项

Question

The question is styled in python 2.7 .

I'm using OrderedDict to store some items as follows:

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

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

Is there a way iterating dictionary d , starting from specific key? For instance, I'd like to iterate d items starting from key 'b'

Many thanks in advance!

Answer 1

You can iterate by looking up the value of b by using items() and slicing off where you need to be. Replace d.keys().index('b') if you have your own way of knowing where you want to start.

from collections import OrderedDict

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

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

Using items() allow you to get the key and value off like normally.

Answer 2

A solution that works for Python 2 and 3, using 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)

Output:

b 1
c 2
d 3

Python 2, avoiding the creation of the key-value list with .items() ::

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

Timing

%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

Answer 3

Would this work?

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

where my_key is the key you want to start from

Answer 4

It seems to me this is one option if you don't want to index a list of tuples:

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)