Home - News ( United States | United Kingdom | Italy | Germany ) - Football scores

Showing content from https://stackoverflow.com/questions/613183/how-do-i-sort-a-dictionary-by-value below:

python - How do I sort a dictionary by value?

Asked 16 years, 5 months ago

Viewed 5.7m times

. Edit existing answers to improve this post. It is not currently accepting new answers or interactions.

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 here How do I sort a list of dictionaries by a value of the dictionary? 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 to sort either in ascending or descending order.

10922 silver badges1212 bronze badges

asked Mar 5, 2009 at 0:49

42.7k1919 gold badges5252 silver badges6464 bronze badges

Dicts preserve insertion order in Python 3.7+. Same in CPython 3.6, but it's an implementation detail.

>>> x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
>>> {k: v for k, v in sorted(x.items(), key=lambda item: item[1])}
{0: 0, 2: 1, 1: 2, 4: 3, 3: 4}

or

>>> dict(sorted(x.items(), key=lambda item: item[1]))
{0: 0, 2: 1, 1: 2, 4: 3, 3: 4}

It is not possible to sort a dictionary, only to get a representation of a dictionary that is sorted. Dictionaries are inherently orderless, but other types, such as lists and tuples, are not. So you need an ordered data type to represent sorted values, which will be a list—probably a list of tuples.

For instance,

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

sorted_x will be a list of tuples sorted by the second element in each tuple. dict(sorted_x) == x.

And for those wishing to sort on keys instead of values:

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

In Python3 since unpacking is not allowed we can use

x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=lambda kv: kv[1])

If you want the output as a dict, you can use collections.OrderedDict:

import collections

sorted_dict = collections.OrderedDict(sorted_x)

33.5k1010 gold badges6969 silver badges103103 bronze badges

answered Mar 5, 2009 at 0:59

95.9k55 gold badges5858 silver badges8080 bronze badges

Well, it is actually possible to do a "sort by dictionary values". Recently I had to do that in a Code Golf (Stack Overflow 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])

or, if we want a dictionary back (since Python 3.6+ preserves insertion order):

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

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 original post 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.

answered Jul 5, 2010 at 8:01

29.1k66 gold badges5050 silver badges6969 bronze badges

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.

To sort it in descending order just add reverse=True:

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

Input:

d = {'one':1,'three':3,'five':5,'two':2,'four':4}
a = sorted(d.items(), key=lambda x: x[1])    
print(a)

Output:

[('one', 1), ('two', 2), ('three', 3), ('four', 4), ('five', 5)]

3,92833 gold badges2020 silver badges2626 bronze badges

answered Feb 13, 2010 at 16:33

11.7k11 gold badge1717 silver badges55 bronze badges

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))

answered Mar 5, 2009 at 1:05

33.6k55 gold badges4444 silver badges5757 bronze badges

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)])

31.6k2222 gold badges110110 silver badges134134 bronze badges

answered Jul 5, 2010 at 2:50

20k1111 gold badges8383 silver badges8484 bronze badges

Whilst I found the accepted answer useful, I was also surprised that it hasn't been updated to reference OrderedDict from the standard library collections module as a viable, modern alternative - designed to solve exactly this type of problem.

from operator import itemgetter
from collections import OrderedDict

x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = OrderedDict(sorted(x.items(), key=itemgetter(1)))
# OrderedDict([(0, 0), (2, 1), (1, 2), (4, 3), (3, 4)])

The official OrderedDict documentation offers a very similar example too, but using a lambda for the sort function:

# regular unsorted dictionary
d = {'banana': 3, 'apple':4, 'pear': 1, 'orange': 2}

# dictionary sorted by value
OrderedDict(sorted(d.items(), key=lambda t: t[1]))
# OrderedDict([('pear', 1), ('orange', 2), ('banana', 3), ('apple', 4)])

31.6k2222 gold badges110110 silver badges134134 bronze badges

answered Dec 5, 2015 at 9:46

37.3k1717 gold badges126126 silver badges127127 bronze badges

Pretty much the same as Hank Gay's answer:

sorted([(value,key) for (key,value) in mydict.items()])

Or optimized slightly as suggested by John Fouhy:

sorted((value,key) for (key,value) in mydict.items())

answered Mar 5, 2009 at 1:06

5,65244 gold badges2424 silver badges1818 bronze badges

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

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.

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.

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.

answered Sep 10, 2016 at 10:05

3,36533 gold badges3333 silver badges3030 bronze badges

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

26.4k66 gold badges6969 silver badges9292 bronze badges

answered Aug 30, 2011 at 0:30

21.2k99 gold badges5959 silver badges4141 bronze badges

Starting from Python 3.6, dict objects are now ordered by insertion order. It's officially in the specifications of Python 3.7.

>>> words = {"python": 2, "blah": 4, "alice": 3}
>>> dict(sorted(words.items(), key=lambda x: x[1]))
{'python': 2, 'alice': 3, 'blah': 4}

Before that, you had to use OrderedDict.

Python 3.7 documentation says:

Changed in version 3.7: Dictionary order is guaranteed to be insertion order. This behavior was implementation detail of CPython from 3.6.

31.6k2222 gold badges110110 silver badges134134 bronze badges

answered Sep 15, 2018 at 13:37

19k1010 gold badges5757 silver badges7777 bronze badges

I had the same problem, and I solved it like this:

WantedOutput = sorted(MyDict, key=lambda x : MyDict[x]) 

(People who answer "It is not possible to sort a dict" did not read the question! In fact, "I can sort on the keys, but how can I sort based on the values?" clearly means that he wants a list of the keys sorted according to the value of their values.)

Please notice that the order is not well defined (keys with the same value will be in an arbitrary order in the output list).

31.6k2222 gold badges110110 silver badges134134 bronze badges

answered Nov 18, 2010 at 14:19

5,56322 gold badges4141 silver badges6060 bronze badges

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)]    

13.9k77 gold badges6969 silver badges7777 bronze badges

answered Jun 27, 2012 at 15:43

1,03322 gold badges1010 silver badges1414 bronze badges

In Python 2.7, simply do:

from collections import OrderedDict
# regular unsorted dictionary
d = {'banana': 3, 'apple':4, 'pear': 1, 'orange': 2}

# dictionary sorted by key
OrderedDict(sorted(d.items(), key=lambda t: t[0]))
OrderedDict([('apple', 4), ('banana', 3), ('orange', 2), ('pear', 1)])

# dictionary sorted by value
OrderedDict(sorted(d.items(), key=lambda t: t[1]))
OrderedDict([('pear', 1), ('orange', 2), ('banana', 3), ('apple', 4)])

copy-paste from : http://docs.python.org/dev/library/collections.html#ordereddict-examples-and-recipes

Enjoy ;-)

answered Aug 22, 2013 at 8:38

1,41922 gold badges1616 silver badges1717 bronze badges

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}

4,89633 gold badges3939 silver badges5454 bronze badges

answered Mar 8, 2011 at 2:06

9361212 silver badges1515 bronze badges

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

31.6k2222 gold badges110110 silver badges134134 bronze badges

answered Apr 7, 2014 at 4:46

2,83722 gold badges3131 silver badges3434 bronze badges

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]

answered Mar 5, 2009 at 1:52

393k8383 gold badges518518 silver badges791791 bronze badges

You can use the collections.Counter. Note, this will work for both numeric and non-numeric values.

>>> x = {1: 2, 3: 4, 4:3, 2:1, 0:0}
>>> from collections import Counter
>>> #To sort in reverse order
>>> Counter(x).most_common()
[(3, 4), (4, 3), (1, 2), (2, 1), (0, 0)]
>>> #To sort in ascending order
>>> Counter(x).most_common()[::-1]
[(0, 0), (2, 1), (1, 2), (4, 3), (3, 4)]
>>> #To get a dictionary sorted by values
>>> from collections import OrderedDict
>>> OrderedDict(Counter(x).most_common()[::-1])
OrderedDict([(0, 0), (2, 1), (1, 2), (4, 3), (3, 4)])

31.6k2222 gold badges110110 silver badges134134 bronze badges

answered Mar 9, 2013 at 12:30

63.9k1919 gold badges142142 silver badges209209 bronze badges

The collections solution mentioned in another answer is absolutely superb, because you retain a connection between the key and value which in the case of dictionaries is extremely important.

I don't agree with the number one choice presented in another answer, because it throws away the keys.

I used the solution mentioned above (code shown below) and retained access to both keys and values and in my case the ordering was on the values, but the importance was the ordering of the keys after ordering the values.

from collections import Counter

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


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

answered Mar 3, 2014 at 14:58

2,0922121 silver badges2121 bronze badges

You can also use a custom function that can be passed to parameter key.

def dict_val(x):
    return x[1]

x = {1: 2, 3: 4, 4: 3, 2: 1, 0: 0}
sorted_x = sorted(x.items(), key=dict_val)

31.6k2222 gold badges110110 silver badges134134 bronze badges

answered May 25, 2017 at 18:13

49711 gold badge66 silver badges1212 bronze badges

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.

answered Sep 25, 2014 at 22:56

1,4591818 silver badges3434 bronze badges

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

13.9k77 gold badges6969 silver badges7777 bronze badges

answered Jul 31, 2015 at 8:08

3,56011 gold badge3333 silver badges4444 bronze badges

I just learned a relevant skill from Python for Everybody.

You may use a temporary list to help you to sort the dictionary:

# Assume dictionary to be:
d = {'apple': 500.1, 'banana': 1500.2, 'orange': 1.0, 'pineapple': 789.0}

# Create a temporary list
tmp = []

# Iterate through the dictionary and append each tuple into the temporary list
for key, value in d.items():
    tmptuple = (value, key)
    tmp.append(tmptuple)

# Sort the list in ascending order
tmp = sorted(tmp)

print (tmp)

If you want to sort the list in descending order, simply change the original sorting line to:

tmp = sorted(tmp, reverse=True)

Using list comprehension, the one-liner would be:

# Assuming the dictionary looks like
d = {'apple': 500.1, 'banana': 1500.2, 'orange': 1.0, 'pineapple': 789.0}
# One-liner for sorting in ascending order
print (sorted([(v, k) for k, v in d.items()]))
# One-liner for sorting in descending order
print (sorted([(v, k) for k, v in d.items()], reverse=True))

Sample Output:

# Ascending order
[(1.0, 'orange'), (500.1, 'apple'), (789.0, 'pineapple'), (1500.2, 'banana')]
# Descending order
[(1500.2, 'banana'), (789.0, 'pineapple'), (500.1, 'apple'), (1.0, 'orange')]

31.6k2222 gold badges110110 silver badges134134 bronze badges

answered May 27, 2018 at 17:45

35522 silver badges1414 bronze badges

from django.utils.datastructures import SortedDict

def sortedDictByKey(self,data):
    """Sorted dictionary order by key"""
    sortedDict = SortedDict()
    if data:
        if isinstance(data, dict):
            sortedKey = sorted(data.keys())
            for k in sortedKey:
                sortedDict[k] = data[k]
    return sortedDict

answered Nov 1, 2010 at 12:16

41555 silver badges2020 bronze badges

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')]

answered Jun 20, 2015 at 1:44

6,40955 gold badges3737 silver badges5252 bronze badges

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 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")

answered Mar 2, 2018 at 16:48

28.7k2727 gold badges149149 silver badges265265 bronze badges

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)]

answered Oct 19, 2011 at 6:25

60488 silver badges88 bronze badges

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

answered Oct 30, 2011 at 19:42

10911 silver badge22 bronze badges

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.

answered Jan 24, 2012 at 19:28

67.2k99 gold badges8585 silver badges8383 bronze badges

This works in 3.1.x:

import operator
slovar_sorted=sorted(slovar.items(), key=operator.itemgetter(1), reverse=True)
print(slovar_sorted)

21.3k2020 gold badges9797 silver badges112112 bronze badges

answered Nov 16, 2011 at 7:32

8911 silver badge11 bronze badge

For the sake of completeness, I am posting a solution using heapq. Note, this method will work for both numeric and non-numeric values

>>> x = {1: 2, 3: 4, 4:3, 2:1, 0:0}
>>> x_items = x.items()
>>> heapq.heapify(x_items)
>>> #To sort in reverse order
>>> heapq.nlargest(len(x_items),x_items, operator.itemgetter(1))
[(3, 4), (4, 3), (1, 2), (2, 1), (0, 0)]
>>> #To sort in ascending order
>>> heapq.nsmallest(len(x_items),x_items, operator.itemgetter(1))
[(0, 0), (2, 1), (1, 2), (4, 3), (3, 4)]

answered Mar 23, 2013 at 14:19

63.9k1919 gold badges142142 silver badges209209 bronze badges

Start asking to get answers

Find the answer to your question by asking.

Explore related questions

See similar questions with these tags.

RetroSearch is an open source project built by @garambo | Open a GitHub Issue

Search and Browse the WWW like it's 1997 | Search results from DuckDuckGo

HTML: 3.2 | Encoding: UTF-8 | Version: 0.7.4