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How to sort a Python dict (dictionary) by keys or values

Date:  |  Modified:  |  Tags: datastructures, python  |  28 Comments

Updated to work with both Python 2 and 3

How to sort a dict by key

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

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

Results:

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

To sort the keys in reverse, add reverse=True as a keyword argument to the sorted function.

How to sort a dict by value

for key, value in sorted(mydict.items(), key=lambda item: item[1]):
    print("%s: %s" % (key, value))

Results:

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

Originally taken from Nick Galbreath's Digital Sanitation Engineering blog article

See also

Comments


#1 Marius commented on :

Thank you.


#2 Sven Siegmund commented on :

Thanks a lot. Exactly what I was looking for.


#3 Mike commented on :

Thanks, "How to sort a dict by value" was helpful. :)


#4 Gregg Lind commented on :

Thanks for the link to my article. If you need faster dictionary sorting than the ones you describe, it has some tips, as Nick's updated blog also shows.


#5 Eliot commented on :

Gregg, Good work-- your article shows that you have obviously explored this topic much more thoroughly than I have. I will definitely try your method the next chance I get.


#6 Rama krishna commented on :

tnx


#7 gcd0318 commented on :

are you sure this works? please see my result (I did in python shell):

>>> d={80: '0101', 75: '1011', 85: '0001', 70: '1111'}
>>> d
{80: '0101', 75: '1011', 85: '0001', 70: '1111'}
>>> kl=list(d.keys())
>>> kl
[80, 75, 85, 70]
>>> new_kl = kl + []
>>> new_kl
[80, 75, 85, 70]
>>> new_kl.sort()
>>> new_kl
[70, 75, 80, 85]
>>> new_d = {}
>>> for k in new_kl:
    new_d[k] = d[k]
    print(new_d)


{70: '1111'}
{75: '1011', 70: '1111'}
{80: '0101', 75: '1011', 70: '1111'}
{80: '0101', 75: '1011', 85: '0001', 70: '1111'}
>>> new_d
{80: '0101', 75: '1011', 85: '0001', 70: '1111'}
>>>

#8 Eliot commented on :

gcd0318:
The items in a dict are stored in an undefined order. Printing the dict only confirms this fact.

If you need to operate on items in a dict in a sorted order, you can operate on each item within the for loop.

Or you can use a data structure that preserves order such as a list or the new OrderedDict data structure in the collections module added in Python 2.7. See http://docs.python.org/library/collections.html#collections.OrderedDict


#9 ovi commented on :

Can someone explain the following behavior:

DICT = {'C':'C','G':'G','S':'S','E':'E','GE':'GE','fRrc':'GE','fAct':'FO'}

print DICT {'C': 'C', 'E': 'E', 'G': 'G', 'S': 'S', 'fRrc': 'GE', 'GE': 'GE', 'fAct': 'FO'}

Why is the order of the elements in the dictionary changed?

In fact, from what I've understood, there is no defined order about how the elements are stored.

So, the question is ... there is such a structure in python? (I don't want the dictionary to be sorted, I just want to preserve the order how they were initially added).

Thanks


#10 Joe commented on :

Sweet, thanks for the simple solution!


#11 Abi commented on :

Thank you. Perfect. Just what I was looking for!


#12 Terry commented on :

Thanks for that! I was going wild on python.org trying to figure out how to do this.


#13 adithya commented on :

thanks..:)


#14 Zoynels commented on :
mydict = {'carl':40, 'alan':2, 'bob':1, 'danny':3}

print ('Sort by keys:')
for key in sorted(mydict.keys()):
    print ("%s: %s" % (key, mydict[key]))

print ('Sort by items:')
for key, value in sorted(mydict.items(), key=lambda item: (item[1], item[0])):
    print ("%s: %s" % (key, value))

#15 Zoynels commented on :

my code works in Python 3.3


#16 Pere Vilas commented on :

It works perfect on 2.7. Thanks!


#17 Roberto commented on :

this works


#18 asdasd commented on :

Thanks


#19 emma lee commented on :

it work well, thx a lot


#20 Grant Jenks commented on :

Its worth noting that Python has a number of dictionary implementations that maintain the order based on the key. Consider the sortedcontainers module which is pure-Python and fast-as-C implementations. Iteration is really fast and automatically yields the keys/items in sorted order. There's also a performance comparison that benchmarks popular options against one another.


#21 Nick commented on :

what does the "%s: %s" % mean?

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#22 erm3nda commented on :

Refer to the docs when you have syntax questions. Does transliteration of symbol for a string in the given order. This is why %s is always same, because the iteration. Also "%s: %s" %" means nothing because it's a half of the function. The key is %, and %s is the ocurrence of a previous rule, in this case, a list and a list["item"], in that order because are positional.

disqus:2230871515


#23 Kevin Sweeney commented on :

Great stuff!

The sorting dictionary by key snippet does produce an array of tuples, which is interesting. Also, you only need to return the value that you are comparing in the lambda, as opposed to flipping the key and value in a tuple.

`

>>> mydict = {'carl':40, 'alan':2, 'bob':1, 'danny':3}  
>>>   
>>> sorted(mydict.iteritems(), key=lambda (k,v): (v,k))  
[('bob', 1), ('alan', 2), ('danny', 3), ('carl', 40)]  
>>>   
>>> sorted(mydict.iteritems(), key=lambda (k,v): v)  
[('bob', 1), ('alan', 2), ('danny', 3), ('carl', 40)]  
>>>

`

disqus:3288786392


#26 Ali Hasan commented on :

Thanks

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#28 sunny arora commented on :

You are printing "key" instead of "k" within the for loop. It should throw an error, you must have defined "key" somewhere above.
it should be print "%s: %s" % (k, value)

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