Iterate over OrderedDict in Python

Question

I have the following OrderedDict:

OrderedDict([('r', 1), ('s', 1), ('a', 1), ('n', 1), ('y', 1)])

This actually presents a frequency of a letter in a word.

In the first step - I would take the last two elements to create a union tuple like this;

 pair1 = list.popitem()
    pair2 = list.popitem()
    merge_list = (pair1[0],pair2[0])
    new_pair = {}
    new_pair[merge_list] = str(pair1[1] + pair2[1])
    list.update(new_pair);

This created for me the following OrderedList:

OrderedDict([('r', 1), ('s', 1), ('a', 1), (('y', 'n'), '2')])

I would like now to iterate over the elements, each time taking the last three and deciding based on the lower sum of the values what is the union object.

For instance the above list will turn to;

OrderedDict([('r', 1), (('s', 'a'), '2'), (('y', 'n'), '2')])

but the above was:

OrderedDict([ ('r', 1), ('s', 2), ('a', 1), (('y', 'n'), '2')])

The result would be:

OrderedDict([('r', 1), ('s', 2), (('a','y', 'n'), '3')])

as I want the left ones to have the smaller value

I tried to do it myself but doesn't understand how to iterate from end to beginning over an OrderedDict.

How can I do it?

EDITED Answering the comment:

I get a dictionary of frequency of a letter in a sentence:

{ 's':1, 'a':1, 'n':1, 'y': 1}

and need to create a huffman tree from it.

for instance:

((s,a),(n,y))

I am using python 3.3

Mussolini

Solve your XY Problem and this will be easier to get an answer for! In layman's terms -- tell us what your broad goal is, not how to solve your particular way of doing it. – Bushed

You mean reversed(OrderedDict.items())? – Poly

@adsmith you are right - I edited my question with actually what I need to do – Mussolini

Answer 1

Simple example

from collections import OrderedDict

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

for key, value in d.items():
    print key, value

Output:

a 1
b 2
c 3

Passus

Is this for python 3 or 2? – Latishalatitude

@CharlieParker 2 – Passus

for me to work I had to use d = OrderedDict() instead of d = collections.OrderedDict() – Barvick

@Csaba depends on your import form, if you do import collections you'd have to, otherwise as the example shoes it, OrderedDict only is imported from collections – Fratricide

@Fratricide yep, the code was edited recently and contains my suggestions but it wasn't correct before that – Barvick

Answer 2

how to iterate from end to beginning over an OrderedDict ?

Either:

z = OrderedDict( ... )
for item in z.items()[::-1]:
   # operate on item

Or:

z = OrderedDict( ... )
for item in reversed(z.items()):
   # operate on item

Clayborn

would reversed(z.items()) change the order? as I want to keep it for more iterations – Mussolini

reversed() creates a new list, which is the reverse of the passed-in list. It will not change any aspect of the original OrderedDict. – Headset

This results in 'dict_items is not subscriptable' or 'dict_items is not reversible', respectively, for me (python 3). I had to do: for key in reversed(z.keys()): # get value using key then do stuff – Sewerage

Yes, the first example won't work in Python3. The second example, however, works well for me in Python 3.5.2. @NicholasMorley – Headset

Answer 3

You can iterate using enumerate and iteritems:

dict = OrderedDict()
# ...

for i, (key, value) in enumerate(dict.iteritems()):
    # Do what you want here

Expeditionary

It works for python 2, but not for python 3.7: AttributeError: 'collections.OrderedDict' object has no attribute 'iteritems' – Kattiekatuscha

Try in Python 3.x with dict.items() instead of dict.iteritems(). – Genie

Answer 4

For Python 3.x

d = OrderedDict( ... )

for key, value in d.items():
    print(key, value)

---

For Python 2.x

d = OrderedDict( ... )

for key, value in d.iteritems():
    print key, value

Pipeline

Answer 5

Note that, as noted in the comments by adsmith, this is probably an instance of an XY Problem and you should reconsider your data structures.

Having said that, if you need to operate only on last three elements, then you don't need to iterate. For example:

MergeInfo = namedtuple('MergeInfo', ['sum', 'toMerge1', 'toMerge2', 'toCopy'])

def mergeLastThree(letters):
    if len(letters) < 3:
        return False

    last = letters.popitem()
    last_1 = letters.popitem()
    last_2 = letters.popitem()

    sum01 = MergeInfo(int(last[1]) + int(last_1[1]), last, last_1, last_2)
    sum12 = MergeInfo(int(last_1[1]) + int(last_2[1]), last_1, last_2, last)
    sum02 = MergeInfo(int(last[1]) + int(last_2[1]), last, last_2, last_1)

    mergeInfo = min((sum01, sum12, sum02), key = lambda s: s.sum)

    merged = ((mergeInfo.toMerge1[0], mergeInfo.toMerge2[0]), str(mergeInfo.sum))

    letters[merged[0]] = merged[1]
    letters[mergeInfo.toCopy[0]] = mergeInfo.toCopy[1]

    return True

Then having:

letters = OrderedDict([('r', 1), ('s', 1), ('a', 1), ('n', 1), ('y', 1)])

print letters
mergeLastThree(letters)
print letters
mergeLastThree(letters)
print letters

Produces:

>>> OrderedDict([('r', 1), ('s', 1), ('a', 1), ('n', 1), ('y', 1)])
OrderedDict([('r', 1), ('s', 1), (('y', 'n'), '2'), ('a', 1)])
OrderedDict([('r', 1), (('a', 's'), '2'), (('y', 'n'), '2')])

And to merge the whole structure completely you need to just:

print letters
while mergeLastThree(letters):
    pass
print letters

Which gives:

>>> OrderedDict([('r', 1), ('s', 1), ('a', 1), ('n', 1), ('y', 1)])
OrderedDict([((('a', 's'), 'r'), '3'), (('y', 'n'), '2')])
>>> 

Blemish