Python: How to access every kth key in an OrderedDict?

Question

I have an OrderedDict (the keys are ordered):

od1 = {0:10, 1:3, 2:7, 3:11, 4:30, 5:15, 6:19, 7:4, 8:3}

I want a new OrderedDict() object that contains the key:value pairs of every second element of od1 . That is, I want:

od2 = {0:10, 2:7, 4:30, 6:19, 8:3}

I have tried:

od2 = OrderedDict()
od2 = {k:v for k, v in od1.items() if k % 2 == 0}

However, that does not give me an ordered dictionary, and it does not give me all keys. Instead, I get this, which is NOT what I want:

{0: 10, 8: 3, 2: 7, 4: 30, 6: 19}

How can I get a new OrderedDict() object that contains every kth key and its associated value?

Answer 1

Use itertools.islice on the od.items :

>>> od
OrderedDict([(0, 10), (1, 3), (2, 7), (3, 11), (4, 30), (5, 15), (6, 19), (7, 4), (8, 3)])

Then simply:

>>> from itertools import islice
>>> OrderedDict(islice(od.items(), None, None, 2))
OrderedDict([(0, 10), (2, 7), (4, 30), (6, 19), (8, 3)])

Or, for example, every third key:

>>> OrderedDict(islice(od.items(), None, None, 3))
OrderedDict([(0, 10), (3, 11), (6, 19)])

islice works analogously to list slicing, eg. my_list[::2] or my_list[::3] , however, it works on any iterable, doesn't support negative step values, and you must use explicit None values, but the principles are the same.

Answer 2

Since you are starting with ordered dict you can just simply use this snippet (Python2.7):

from collections import OrderedDict

od1 = OrderedDict([(0, 10), (1, 3), (2, 7), (3, 11), (4, 30), (5, 15), (6, 19), (7, 4), (8, 3)])
c = od1.items()[::2]

print OrderedDict(c)  # OrderedDict([(0, 10), (2, 7), (4, 30), (6, 19), (8, 3)])

For Python 3.6 it is a slightly different:

from collections import OrderedDict

od1 = OrderedDict([(0, 10), (1, 3), (2, 7), (3, 11), (4, 30), (5, 15), (6, 19), (7, 4), (8, 3)])
c = list(od1.items())[::2]
print(OrderedDict(c)) # OrderedDict([(0, 10), (2, 7), (4, 30), (6, 19), (8, 3)])

Answer 3

You can use enumerate :

import collections
new_d = collections.OrderedDict([a for i, a in enumerate(od1.items()) if i%2 == 0])

Output:

OrderedDict([(0, 10), (2, 7), (4, 30), (6, 19), (8, 3)])

Answer 4

 od1 = {0:10, 1:3, 2:7, 3:11, 4:30, 5:15, 6:19, 7:4, 8:3}
 od2=sorted([(key,item) for key,item in od1.items() if key%2==0])
 print(od2)

output:

 [(0, 10), (2, 7), (4, 30), (6, 19), (8, 3)]

Answer 5

od2 type is 'dict' that is why you are getting unordered dict.

>>> type(od2)
<class 'dict'>

quick solutions will be like this:

>>> od2 = OrderedDict({k:v for k, v in od.items() if k % 2 == 0})
>>> od2
OrderedDict([(0, 10), (2, 7), (4, 30), (6, 19), (8, 3)])