Python select ith element in OrderedDict

Question

I have a snippet of code which orders a dictionary alphabetically. Is there a way to select the ith key in the ordered dictionary and return its corresponding value? i.e.

import collections
initial = dict(a=1, b=2, c=2, d=1, e=3)
ordered_dict = collections.OrderedDict(sorted(initial.items(), key=lambda t: t[0]))
print(ordered_dict)

OrderedDict([('a', 1), ('b', 2), ('c', 2), ('d', 1), ('e', 3)])

I want to have some function along the vein of...

select = int(input("Input dictionary index"))
#User inputs 2
#Program looks up the 2nd entry in ordered_dict (c in this case)
#And then returns the value of c (2 in this case)

How can this be achieved? Thanks.

(Similar to Accessing Items In a ordereddict, but I only want to output the value of the key-value pair.)

Answer 1

In Python 2:

If you want to access the key:

>>> ordered_dict = OrderedDict([('a', 1), ('b', 2), ('c', 2), ('d', 1), ('e', 3)])
>>> ordered_dict.keys()[2]
'c'

If want to access the value:

>>> ordered_dict.values()[2]
2

If you're using Python 3, you can convert the KeysView object returned by the keys method by wrapping it as a list:

>>> list(ordered_dict.keys())[2]
'c'
>>> list(ordered_dict.values())[2]
2

Not the prettiest solution, but it works.

Answer 2

Using itertools.islice is efficient here, because we don't have to create any intermediate lists, for the sake of subscripting.

from itertools import islice
print(next(islice(ordered_dict.items(), 2, None)))

If you want just the value, you can do

print ordered_dict[next(islice(ordered_dict, 2, None))]

Answer 3

Do you have to use an OrderedDict or do you just want a dict-like type that supports indexing? If the latter, then consider a sorted dict object. Some implementations of SortedDict (which orders pairs based on the key sort order) support fast n-th indexing. For example, the sortedcontainers project has a SortedDict type with random-access indexing.

In your case it would look something like:

>>> from sortedcontainers import SortedDict
>>> sorted_dict = SortedDict(a=1, b=2, c=2, d=1, e=3)
>>> print sorted_dict.iloc[2]
'c'

If you do a lot of lookups, this will be a lot faster than repeatedly iterating to the desired index.

Answer 4

You could do something along these lines (od is the ordered dict):

def get_idx(od, idx):
   from itertools import islice
   idx = (idx + len(od)) % len(od)
   t = islice(od.items(), idx, idx + 1)
   return next(t)

>>>x

OrderedDict([('a', 2), ('b', 3), ('k', 23), ('t', 41), ('q', 23)])

>>>get_idx(x, 1)
('b', 3)
>>>get_idx(x, 2)
('k', 23)
>>>get_idx(x, 4)
('q', 23)
>>>get_idx(x, -1)
('q', 23)
>>>get_idx(x, -2)
('t', 41)

Answer 5

Don't underestimate just a plain 'ole for loop:

from collections import OrderedDict

od=OrderedDict([('a', 1), ('b', 2), ('c', 2), ('d', 1), ('e', 3)])

def ith(od, tgt):
    for i, t in enumerate(od.items()):
        if i==tgt:
            print('element {}\'s key is "{}"'.format(i,t[0]))
            break
    else:
        print('element {} not found'.format(tgt)) 

ith(od, 2)
# element 2's key is "c"
ith(od, 20) 
# element 20 not found

The advantage here is that the loop will break as soon as the desired element is found and returns a sensible result if not found...

The disadvantage is that relative slices are not supported.