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This is the dictionary:

lettersandnumbers = {'Z': 1, 'Y': 0, 'X': 1, 'W': 17, 'V': 4, 'U': 0,\
 'T': 22, 'S': 21, 'R': 31, 'Q': 0, 'P': 12, 'O': 8,\
 'N': 10, 'M': 29, 'L': 27, 'K': 14, 'J': 51, 'I': 7,\
 'H': 14, 'G': 21, 'F': 12, 'E': 27, 'D': 40, 'C': 43,\
 'B': 28, 'A': 34}

I want it to be sorted alphabetically. I've tried making it into a list so I could use the sorted function, but I'm completely lost. Any ideas?

ajkey94
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  • Shockingly, this seems to not be a duplicate. All other questions I can find are about sorting by *value*. :-) – Lennart Regebro Apr 11 '13 at 04:24
  • @ajkey94 No need for backslashes!!! It just makes it look ugly, you can still split it into lines without them just using the commas – jamylak Apr 11 '13 at 04:26
  • possible duplicate of [In Python, how to I iterate over a dictionary in sorted order?](http://stackoverflow.com/questions/364519/in-python-how-to-i-iterate-over-a-dictionary-in-sorted-order) – dawg Apr 11 '13 at 15:54

5 Answers5

3

There are a couple of ways to do the iteration, the links above described methods of sorting the dictionary for one time use. If you need persistent ordered dictionary, with Python 2.7 and 3.x onward, you could used an OrderedDict collections, see example below:

>>> from collections import OrderedDict
# Random dictionary
>>> foo = {'a': 1, 'b':2, 'e':4, 'd':6, 'f': 7}
>>> foo
{'a': 1, 'b':2, 'e':4, 'd':6, 'f': 7}
# OrderedDict remember order of insert, hence to maintain a sorted
# dictionary entry, we must supplies a sorted tuple the example below
# sorted by entry key
>>> bar = OrderedDict(sorted(foo.items(), key=lambda t: t[0]))
>>> bar
OrderedDict([('a', 1), ('b', 2), ('d', 6), ('e', 4), ('f', 7)])

Despite the difference in notation the bar object functions identical to a dictionary.

nqngo
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    There is no need for a key function in the call to sorted - the default tuple ordering will give the same result. – lvc Apr 11 '13 at 02:58
  • Indeed, it is. I usually use explicit key and forgot all about default behaviour. Edit: I figure it would be useful to leave the key example in there. – nqngo Apr 11 '13 at 03:00
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    @Nerolynx simple is better than complex! And I will upvote you if you take it out ;) – jamylak Apr 11 '13 at 03:36
  • @jamylak Not that I refused to update the answer. It just that it is a nice little example of how to sort keys using sorted as well. Might as well leave it. – nqngo Apr 11 '13 at 04:58
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Dictionary is just a container. If you want to iterate keys in sorted order, see In Python, how do I iterate over a dictionary in sorted order?

Community
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monoid
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1

Since @Nerolynx refuses to use this simpler and faster version of his method

>>> from collections import OrderedDict
>>> d = {'Z': 1, 'Y': 0, 'X': 1, 'W': 17, 'V': 4, 'U': 0, 'T': 22, 'S': 21, 'R': 31, 'Q': 0, 'P': 12, 'O': 8,'N': 10, 'M': 29, 'L': 27, 'K': 14, 'J': 51, 'I': 7,      'H': 14, 'G': 21, 'F': 12, 'E': 27, 'D': 40, 'C': 43, 'B': 28, 'A': 34}
>>> OrderedDict(sorted(d.items()))
OrderedDict([('A', 34), ('B', 28), ('C', 43), ('D', 40), ('E', 27), ('F', 12), ('G', 21), ('H', 14), ('I', 7), ('J', 51), ('K', 14), ('L', 27), ('M', 29), ('N', 10), ('O', 8), ('P', 12), ('Q', 0), ('R', 31), ('S', 21), ('T', 22), ('U', 0), ('V', 4), ('W', 17), ('X', 1), ('Y', 0), ('Z', 1)])
jamylak
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0
>>> from collections import OrderedDict
>>> from string import ascii_uppercase as alphabet
>>> d = {'Z': 1, 'Y': 0, 'X': 1, 'W': 17, 'V': 4, 'U': 0, 'T': 22, 'S': 21, 'R': 31, 'Q': 0, 'P': 12, 'O': 8,'N': 10, 'M': 29, 'L': 27, 'K': 14, 'J': 51, 'I': 7,      'H': 14, 'G': 21, 'F': 12, 'E': 27, 'D': 40, 'C': 43, 'B': 28, 'A': 34}
>>> OrderedDict((k, d[k]) for k in alphabet if k in d)
OrderedDict([('A', 34), ('B', 28), ('C', 43), ('D', 40), ('E', 27), ('F', 12), ('G', 21), ('H', 14), ('I', 7), ('J', 51), ('K', 14), ('L', 27), ('M', 29), ('N', 10), ('O', 8), ('P', 12), ('Q', 0), ('R', 31), ('S', 21), ('T', 22), ('U', 0), ('V', 4), ('W', 17), ('X', 1), ('Y', 0), ('Z', 1)])
jamylak
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0

If you use [some dict].items(), it produces a list of tuples:

>>> d={3:'three',35:'thirty five',100:'one hindred'}
>>> d
{35: 'thirty five', 3: 'three', 100: 'one hindred'}
>>> d.items()
[(35, 'thirty five'), (3, 'three'), (100, 'one hindred')]

Tuples are sortable in Python:

>>> sorted(d.items())
[(3, 'three'), (35, 'thirty five'), (100, 'one hindred')]

Therefore, your dictionary sorts just as you desire with Python's default sort order:

>>> lettersandnumbers = {'Z': 1, 'Y': 0, 'X': 1, 'W': 17, 'V': 4, 'U': 0,'T': 22, 'S': 21, 'R': 31, 'Q': 0, 'P': 12, 'O': 8,'N': 10, 'M': 29, 'L': 27, 'K': 14, 'J': 51, 'I': 7,'H': 14, 'G': 21, 'F': 12, 'E': 27, 'D': 40, 'C': 43,'B': 28, 'A': 34}
>>> sorted(lettersandnumbers.items())
[('A', 34), ('B', 28), ('C', 43), ('D', 40), ('E', 27), ('F', 12), ('G', 21), ('H', 14), ('I', 7), ('J', 51), ('K', 14), ('L', 27), ('M', 29), ('N', 10), ('O', 8), ('P', 12), ('Q', 0), ('R', 31), ('S', 21), ('T', 22), ('U', 0), ('V', 4), ('W', 17), ('X', 1), ('Y', 0), ('Z', 1)]

(You may need to use sorted's key parameter if the desired order is different that the default. In the case, the default works.)

Then if you want a dict that keeps this order:

>>> from collections import OrderedDict
>>> od=OrderedDict(sorted(lettersandnumbers.items()))
>>> od.items()[0]
('A', 34)
>>> od.items()[-1]
('Z', 1)

But ordered dict is based on insertion order, so it will be unsorted if you add anything that is not in sort order.

Better still, avoid the making an unnecessary copy of the dict, and just iterate on the dict as needed in the then current sort order:

for k,v in sorted(d.items()): 
    # deal with the key and value in sorted order at that moment...

With Python 2, you need to use d.iteritems() or you will produce 2 temp lists: 1 for the items and 1 to sort. With Py 3 (or with Py 2 and d.iteritems() method) only 1 temp list is produced for the sort.

dawg
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