How do I sort a dictionary by value?

Question

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.

Answer 1

Python 3.7+ or CPython 3.6

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}

Older Python

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)

Answer 2

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 (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])

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.

Answer 3

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)]

Answer 4

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))

Answer 5

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)])

Answer 6

UPDATE: 5 DECEMBER 2015 using Python 3.5

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)])

Answer 7

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())

Answer 8

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.

Answer 9

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

Answer 10

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).

Answer 11

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)]    

Answer 12

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 ;-)

Answer 13

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

Answer 14

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}

Answer 15

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

Answer 16

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]

Answer 17

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)])

Answer 18

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.

Answer 19

You can also use custom function that can be passed to 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)

Answer 20

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

Answer 21

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

Answer 22

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")

Answer 23

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')]

Answer 24

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)]

Answer 25

Just learned 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:

#Asending 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')]

Answer 26

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

Answer 27

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)]

Answer 28

This works in 3.1.x:

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

Answer 29

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.

Answer 30

months = {"January": 31, "February": 28, "March": 31, "April": 30, "May": 31,
          "June": 30, "July": 31, "August": 31, "September": 30, "October": 31,
          "November": 30, "December": 31}

def mykey(t):
    """ Customize your sorting logic using this function.  The parameter to
    this function is a tuple.  Comment/uncomment the return statements to test
    different logics.
    """
    return t[1]              # sort by number of days in the month
    #return t[1], t[0]       # sort by number of days, then by month name
    #return len(t[0])        # sort by length of month name
    #return t[0][-1]         # sort by last character of month name


# Since a dictionary can't be sorted by value, what you can do is to convert
# it into a list of tuples with tuple length 2.
# You can then do custom sorts by passing your own function to sorted().
months_as_list = sorted(months.items(), key=mykey, reverse=False)

for month in months_as_list:
    print month

Answer 31

Because of requirements to retain backward compatability with older versions of Python I think the OrderedDict solution is very unwise. You want something that works with Python 2.7 and older versions.

But the collections solution mentioned in another answer is absolutely superb, because you retrain 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)]

Answer 32

Beside using builtin module etc, i try solve it manually..... first i made a function whose job is to returning minimal value each items of the dict :

def returnminDict(_dct):
    dict_items = _dct.items()
    list_items = list(dict_items)
    init_items = list_items[0]
    for i in range(len(list_items)):
        if list_items[i][1] > init_items[1]:
           continue
        else:
           init_items = list_items[i]
    return init_items

Second, now we have a function which return item that has minimal value, then i make a new dict then loop over the dict :

def SelectDictSort(_dct):
    new_dict = {}
    while _dct:
        mindict = returnminDict(_dct)
        new_dict.update(dict((mindict,)))
        _dct.pop(mindict[0])
    return new_dict

i try this SelectDictSort({2: 5, 5: 1, 4: 3, 1: 1, 0: 1, 9: 2, 8: 2}) will return :

{0: 1, 1: 1, 5: 1, 8: 2, 9: 2, 4: 3, 2: 5}

Hmmm... i dont know which is correct but this what i have tried....

(updated code from retrun new_dct to return new_dict)

Answer 33

Using Python 3.2:

x = {"b":4, "a":3, "c":1}
for i in sorted(x.values()):
    print(list(x.keys())[list(x.values()).index(i)])

Answer 34

This method will not use lambda and works well on Python 3.6:

 # sort dictionary by value
d = {'a1': 'fsdfds', 'g5': 'aa3432ff', 'ca':'zz23432'}
def getkeybyvalue(d,i):
    for k, v in d.items():
        if v == i:
            return (k)

sortvaluelist = sorted(d.values())

# In >> Python 3.6+ << the INSERTION-ORDER of a dict is preserved. That is,
# when creating a NEW dictionary and filling it 'in sorted order',
# that order will be maintained.
sortresult ={}
for i1 in sortvaluelist:   
    key = getkeybyvalue(d,i1)
    sortresult[key] = i1
print ('=====sort by value=====')
print (sortresult)
print ('=======================')