CODEWITHSUNDEEP
pythonpython-3.x
ajkey94
ajkey94

Reputation: 431

How can I sort this dictionary with alphabetically?

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?

Upvotes: 2

Views: 4545

Answers (5)

dawg
dawg

Reputation: 104032

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.

Upvotes: 0

jamylak
jamylak

Reputation: 133634

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)])

Upvotes: 1

jamylak
jamylak

Reputation: 133634

>>> 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)])

Upvotes: 0

nqngo
nqngo

Reputation: 528

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.

Upvotes: 3

monoid
monoid

Reputation: 1671

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?

Upvotes: 2

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