value - sort dictionary python




How do I sort a dictionary by value? (20)

As of Python 3.6 the built-in dict will be ordered

Good news, so the OP's original use case of mapping pairs retrieved from a database with unique string ids as keys and numeric values as values into a built-in Python v3.6+ dict, should now respect the insert order.

If say the resulting two column table expressions from a database query like:

SELECT a_key, a_value FROM a_table ORDER BY a_value;

would be stored in two Python tuples, k_seq and v_seq (aligned by numerical index and with the same length of course), then:

k_seq = ('foo', 'bar', 'baz')
v_seq = (0, 1, 42)
ordered_map = dict(zip(k_seq, v_seq))

Allow to output later as:

for k, v in ordered_map.items():
    print(k, v)

yielding in this case (for the new Python 3.6+ built-in dict!):

foo 0
bar 1
baz 42

in the same ordering per value of v.

Where in the Python 3.5 install on my machine it currently yields:

bar 1
foo 0
baz 42

Details:

As proposed in 2012 by Raymond Hettinger (cf. mail on python-dev with subject "More compact dictionaries with faster iteration") and now (in 2016) announced in a mail by Victor Stinner to python-dev with subject "Python 3.6 dict becomes compact and gets a private version; and keywords become ordered" due to the fix/implementation of issue 27350 "Compact and ordered dict" in Python 3.6 we will now be able, to use a built-in dict to maintain insert order!!

Hopefully this will lead to a thin layer OrderedDict implementation as a first step. As @JimFasarakis-Hilliard indicated, some see use cases for the OrderedDict type also in the future. I think the Python community at large will carefully inspect, if this will stand the test of time, and what the next steps will be.

Time to rethink our coding habits to not miss the possibilities opened by stable ordering of:

  • Keyword arguments and
  • (intermediate) dict storage

The first because it eases dispatch in the implementation of functions and methods in some cases.

The second as it encourages to more easily use dicts as intermediate storage in processing pipelines.

Raymond Hettinger kindly provided documentation explaining "The Tech Behind Python 3.6 Dictionaries" - from his San Francisco Python Meetup Group presentation 2016-DEC-08.

And maybe quite some Stack Overflow high decorated question and answer pages will receive variants of this information and many high quality answers will require a per version update too.

Caveat Emptor (but also see below update 2017-12-15):

As @ajcr rightfully notes: "The order-preserving aspect of this new implementation is considered an implementation detail and should not be relied upon." (from the whatsnew36) not nit picking, but the citation was cut a bit pessimistic ;-). It continues as " (this may change in the future, but it is desired to have this new dict implementation in the language for a few releases before changing the language spec to mandate order-preserving semantics for all current and future Python implementations; this also helps preserve backwards-compatibility with older versions of the language where random iteration order is still in effect, e.g. Python 3.5)."

So as in some human languages (e.g. German), usage shapes the language, and the will now has been declared ... in whatsnew36.

Update 2017-12-15:

In a mail to the python-dev list, Guido van Rossum declared:

Make it so. "Dict keeps insertion order" is the ruling. Thanks!

So, the version 3.6 CPython side-effect of dict insertion ordering is now becoming part of the language spec (and not anymore only an implementation detail). That mail thread also surfaced some distinguishing design goals for collections.OrderedDict as reminded by Raymond Hettinger during discussion.

I have a dictionary of values read from two fields in a database: a string field and a numeric field. The string field is unique, so that is the key of the dictionary.

I can sort on the keys, but how can I sort based on the values?

Note: I have read Stack Overflow question How do I sort a list of dictionaries by values of the dictionary in Python? and probably could change my code to have a list of dictionaries, but since I do not really need a list of dictionaries I wanted to know if there is a simpler solution.


As simple as: sorted(dict1, key=dict1.get)

Well, it is actually possible to do a "sort by dictionary values". Recently I had to do that in a Code Golf ( question Code golf: Word frequency chart). Abridged, the problem was of the kind: given a text, count how often each word is encountered and display a list of the top words, sorted by decreasing frequency.

If you construct a dictionary with the words as keys and the number of occurrences of each word as value, simplified here as:

from collections import defaultdict
d = defaultdict(int)
for w in text.split():
  d[w] += 1

then you can get a list of the words, ordered by frequency of use with sorted(d, key=d.get) - the sort iterates over the dictionary keys, using the number of word occurrences as a sort key .

for w in sorted(d, key=d.get, reverse=True):
  print w, d[w]

I am writing this detailed explanation to illustrate what people often mean by "I can easily sort a dictionary by key, but how do I sort by value" - and I think the OP was trying to address such an issue. And the solution is to do sort of list of the keys, based on the values, as shown above.


As pointed out by Dilettant, Python 3.6 will now keep the order! I thought I'd share a function I wrote that eases the sorting of an iterable (tuple, list, dict). In the latter case, you can sort either on keys or values, and it can take numeric comparison into account. Only for >= 3.6!

When you try using sorted on an iterable that holds e.g. strings as well as ints, sorted() will fail. Of course you can force string comparison with str(). However, in some cases you want to do actual numeric comparison where 12 is smaller than 20 (which is not the case in string comparison). So I came up with the following. When you want explicit numeric comparison you can use the flag num_as_num which will try to do explicit numeric sorting by trying to convert all values to floats. If that succeeds, it will do numeric sorting, otherwise it'll resort to string comparison.

Comments for improvement or push requests welcome.

def sort_iterable(iterable, sort_on=None, reverse=False, num_as_num=False):
    def _sort(i):
      # sort by 0 = keys, 1 values, None for lists and tuples
      try:
        if num_as_num:
          if i is None:
            _sorted = sorted(iterable, key=lambda v: float(v), reverse=reverse)
          else:
            _sorted = dict(sorted(iterable.items(), key=lambda v: float(v[i]), reverse=reverse))
        else:
          raise TypeError
      except (TypeError, ValueError):
        if i is None:
          _sorted = sorted(iterable, key=lambda v: str(v), reverse=reverse)
        else:
          _sorted = dict(sorted(iterable.items(), key=lambda v: str(v[i]), reverse=reverse))

      return _sorted

    if isinstance(iterable, list):
      sorted_list = _sort(None)
      return sorted_list
    elif isinstance(iterable, tuple):
      sorted_list = tuple(_sort(None))
      return sorted_list
    elif isinstance(iterable, dict):
      if sort_on == 'keys':
        sorted_dict = _sort(0)
        return sorted_dict
      elif sort_on == 'values':
        sorted_dict = _sort(1)
        return sorted_dict
      elif sort_on is not None:
        raise ValueError(f"Unexpected value {sort_on} for sort_on. When sorting a dict, use key or values")
    else:
      raise TypeError(f"Unexpected type {type(iterable)} for iterable. Expected a list, tuple, or dict")

Dicts can't be sorted, but you can build a sorted list from them.

A sorted list of dict values:

sorted(d.values())

A list of (key, value) pairs, sorted by value:

from operator import itemgetter
sorted(d.items(), key=itemgetter(1))

Here is a solution using zip on d.values() and d.keys(). A few lines down this link (on Dictionary view objects) is:

This allows the creation of (value, key) pairs using zip(): pairs = zip(d.values(), d.keys()).

So we can do the following:

d = {'key1': 874.7, 'key2': 5, 'key3': 8.1}

d_sorted = sorted(zip(d.values(), d.keys()))

print d_sorted 
# prints: [(5, 'key2'), (8.1, 'key3'), (874.7, 'key1')]

I came up with this one,

import operator    
x = {1: 2, 3: 4, 4:3, 2:1, 0:0}
sorted_x = {k[0]:k[1] for k in sorted(x.items(), key=operator.itemgetter(1))}

For Python 3.x: x.items() replacing iteritems().

>>> sorted_x
{0: 0, 1: 2, 2: 1, 3: 4, 4: 3}

Or try with collections.OrderedDict!

x = {1: 2, 3: 4, 4:3, 2:1, 0:0}
from collections import OrderedDict

od1 = OrderedDict(sorted(x.items(), key=lambda t: t[1]))

If values are numeric you may also use Counter from collections

from collections import Counter

x={'hello':1,'python':5, 'world':3}
c=Counter(x)
print c.most_common()


>> [('python', 5), ('world', 3), ('hello', 1)]    

If your values are integers, and you use Python 2.7 or newer, you can use collections.Counter instead of dict. The most_common method will give you all items, sorted by the value.


In recent Python 2.7, we have the new OrderedDict type, which remembers the order in which the items were added.

>>> d = {"third": 3, "first": 1, "fourth": 4, "second": 2}

>>> for k, v in d.items():
...     print "%s: %s" % (k, v)
...
second: 2
fourth: 4
third: 3
first: 1

>>> d
{'second': 2, 'fourth': 4, 'third': 3, 'first': 1}

To make a new ordered dictionary from the original, sorting by the values:

>>> from collections import OrderedDict
>>> d_sorted_by_value = OrderedDict(sorted(d.items(), key=lambda x: x[1]))

The OrderedDict behaves like a normal dict:

>>> for k, v in d_sorted_by_value.items():
...     print "%s: %s" % (k, v)
...
first: 1
second: 2
third: 3
fourth: 4

>>> d_sorted_by_value
OrderedDict([('first': 1), ('second': 2), ('third': 3), ('fourth': 4)])

It can often be very handy to use namedtuple. For example, you have a dictionary of 'name' as keys and 'score' as values and you want to sort on 'score':

import collections
Player = collections.namedtuple('Player', 'score name')
d = {'John':5, 'Alex':10, 'Richard': 7}

sorting with lowest score first:

worst = sorted(Player(v,k) for (k,v) in d.items())

sorting with highest score first:

best = sorted([Player(v,k) for (k,v) in d.items()], reverse=True)

Now you can get the name and score of, let's say the second-best player (index=1) very Pythonically like this:

player = best[1]
player.name
    'Richard'
player.score
    7

Iterate through a dict and sort it by its values in descending order:

$ python --version
Python 3.2.2

$ cat sort_dict_by_val_desc.py 
dictionary = dict(siis = 1, sana = 2, joka = 3, tuli = 4, aina = 5)
for word in sorted(dictionary, key=dictionary.get, reverse=True):
  print(word, dictionary[word])

$ python sort_dict_by_val_desc.py 
aina 5
tuli 4
joka 3
sana 2
siis 1

Of course, remember, you need to use OrderedDict because regular Python dictionaries don't keep the original order.

from collections import OrderedDict
a = OrderedDict(sorted(originalDict.items(), key = lambda x: x[1]))

If you do not have Python 2.7 or higher, the best you can do is iterate over the values in a generator function. (There is an OrderedDict for 2.4 and 2.6 here, but

a) I don't know about how well it works 

and

b) You have to download and install it of course. If you do not have administrative access, then I'm afraid the option's out.)

def gen(originalDict):
    for x,y in sorted(zip(originalDict.keys(), originalDict.values()), key = lambda z: z[1]):
        yield (x, y)
    #Yields as a tuple with (key, value). You can iterate with conditional clauses to get what you want. 

for bleh, meh in gen(myDict):
    if bleh == "foo":
        print(myDict[bleh])

You can also print out every value

for bleh, meh in gen(myDict):
    print(bleh,meh)

Please remember to remove the parentheses after print if not using Python 3.0 or above


Technically, dictionaries aren't sequences, and therefore can't be sorted. You can do something like

sorted(a_dictionary.values())

assuming performance isn't a huge deal.


This is the code:

import operator
origin_list = [
    {"name": "foo", "rank": 0, "rofl": 20000},
    {"name": "Silly", "rank": 15, "rofl": 1000},
    {"name": "Baa", "rank": 300, "rofl": 20},
    {"name": "Zoo", "rank": 10, "rofl": 200},
    {"name": "Penguin", "rank": -1, "rofl": 10000}
]
print ">> Original >>"
for foo in origin_list:
    print foo

print "\n>> Rofl sort >>"
for foo in sorted(origin_list, key=operator.itemgetter("rofl")):
    print foo

print "\n>> Rank sort >>"
for foo in sorted(origin_list, key=operator.itemgetter("rank")):
    print foo

Here are the results:

Original

{'name': 'foo', 'rank': 0, 'rofl': 20000}
{'name': 'Silly', 'rank': 15, 'rofl': 1000}
{'name': 'Baa', 'rank': 300, 'rofl': 20}
{'name': 'Zoo', 'rank': 10, 'rofl': 200}
{'name': 'Penguin', 'rank': -1, 'rofl': 10000}

Rofl

{'name': 'Baa', 'rank': 300, 'rofl': 20}
{'name': 'Zoo', 'rank': 10, 'rofl': 200}
{'name': 'Silly', 'rank': 15, 'rofl': 1000}
{'name': 'Penguin', 'rank': -1, 'rofl': 10000}
{'name': 'foo', 'rank': 0, 'rofl': 20000}

Rank

{'name': 'Penguin', 'rank': -1, 'rofl': 10000}
{'name': 'foo', 'rank': 0, 'rofl': 20000}
{'name': 'Zoo', 'rank': 10, 'rofl': 200}
{'name': 'Silly', 'rank': 15, 'rofl': 1000}
{'name': 'Baa', 'rank': 300, 'rofl': 20}

Try the following approach. Let us define a dictionary called mydict with the following data:

mydict = {'carl':40,
          'alan':2,
          'bob':1,
          'danny':3}

If one wanted to sort the dictionary by keys, one could do something like:

for key in sorted(mydict.iterkeys()):
    print "%s: %s" % (key, mydict[key])

This should return the following output:

alan: 2
bob: 1
carl: 40
danny: 3

On the other hand, if one wanted to sort a dictionary by value (as is asked in the question), one could do the following:

for key, value in sorted(mydict.iteritems(), key=lambda (k,v): (v,k)):
    print "%s: %s" % (key, value)

The result of this command (sorting the dictionary by value) should return the following:

bob: 1
alan: 2
danny: 3
carl: 40

Use ValueSortedDict from dicts:

from dicts.sorteddict import ValueSortedDict
d = {1: 2, 3: 4, 4:3, 2:1, 0:0}
sorted_dict = ValueSortedDict(d)
print sorted_dict.items() 

[(0, 0), (2, 1), (1, 2), (4, 3), (3, 4)]

You can create an "inverted index", also

from collections import defaultdict
inverse= defaultdict( list )
for k, v in originalDict.items():
    inverse[v].append( k )

Now your inverse has the values; each value has a list of applicable keys.

for k in sorted(inverse):
    print k, inverse[k]

You can use a skip dict which is a dictionary that's permanently sorted by value.

>>> data = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
>>> SkipDict(data)
{0: 0.0, 2: 1.0, 1: 2.0, 4: 3.0, 3: 4.0}

If you use keys(), values() or items() then you'll iterate in sorted order by value.

It's implemented using the skip list datastructure.



You could use:

sorted(d.items(), key=lambda x: x[1])

This will sort the dictionary by the values of each entry within the dictionary from smallest to largest.





dictionary