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How do I sort a dictionary by its keys?
Example input:
{2:3, 1:89, 4:5, 3:0}
Desired output:
{1:89, 2:3, 3:0, 4:5}
Note: for Python 3.7+, see this answer
Standard Python dictionaries are unordered (until Python 3.7). Even if you sorted the (key,value) pairs, you wouldn't be able to store them in a dict in a way that would preserve the ordering.
The easiest way is to use OrderedDict, which remembers the order in which the elements have been inserted:
In [1]: import collections
In [2]: d = {2:3, 1:89, 4:5, 3:0}
In [3]: od = collections.OrderedDict(sorted(d.items()))
In [4]: od
Out[4]: OrderedDict([(1, 89), (2, 3), (3, 0), (4, 5)])
Never mind the way od is printed out; it'll work as expected:
In [11]: od[1]
Out[11]: 89
In [12]: od[3]
Out[12]: 0
In [13]: for k, v in od.iteritems(): print k, v
....:
1 89
2 3
3 0
4 5
Python 3
For Python 3 users, one needs to use the .items() instead of .iteritems():
In [13]: for k, v in od.items(): print(k, v)
....:
1 89
2 3
3 0
4 5
Dictionaries themselves do not have ordered items as such, should you want to print them etc to some order, here are some examples:
In Python 2.4 and above:
mydict = {'carl':40,
'alan':2,
'bob':1,
'danny':3}
for key in sorted(mydict):
print "%s: %s" % (key, mydict[key])
gives:
alan: 2
bob: 1
carl: 40
danny: 3
(Python below 2.4:)
keylist = mydict.keys()
keylist.sort()
for key in keylist:
print "%s: %s" % (key, mydict[key])
Source: http://www.saltycrane.com/blog/2007/09/how-to-sort-python-dictionary-by-keys/
For CPython/PyPy 3.6, and any Python 3.7 or higher, this is easily done with:
>>> d = {2:3, 1:89, 4:5, 3:0}
>>> dict(sorted(d.items()))
{1: 89, 2: 3, 3: 0, 4: 5}
From Python's collections library documentation:
>>> from collections import OrderedDict
>>> # regular unsorted dictionary
>>> d = {'banana': 3, 'apple':4, 'pear': 1, 'orange': 2}
>>> # dictionary sorted by key -- OrderedDict(sorted(d.items()) also works
>>> 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)])
>>> # dictionary sorted by length of the key string
>>> OrderedDict(sorted(d.items(), key=lambda t: len(t[0])))
OrderedDict([('pear', 1), ('apple', 4), ('orange', 2), ('banana', 3)])
There are a number of Python modules that provide dictionary implementations which automatically maintain the keys in sorted order. Consider the sortedcontainers module which is pure-Python and fast-as-C implementations. There is also a performance comparison with other popular options benchmarked against one another.
Using an ordered dict is an inadequate solution if you need to constantly add and remove key/value pairs while also iterating.
>>> from sortedcontainers import SortedDict
>>> d = {2:3, 1:89, 4:5, 3:0}
>>> s = SortedDict(d)
>>> s.items()
[(1, 89), (2, 3), (3, 0), (4, 5)]
The SortedDict type also supports indexed location lookups and deletion which isn't possible with the built-in dict type.
>>> s.iloc[-1]
4
>>> del s.iloc[2]
>>> s.keys()
SortedSet([1, 2, 4])
Simply:
d = {2:3, 1:89, 4:5, 3:0}
sd = sorted(d.items())
for k,v in sd:
print k, v
Output:
1 89
2 3
3 0
4 5
Python dictionary was unordered before Python 3.6. In CPython implementation of Python 3.6, the dictionary keeps the insertion order.
From Python 3.7, this will become a language feature.
In changelog of Python 3.6 (https://docs.python.org/3.6/whatsnew/3.6.html#whatsnew36-compactdict):
The order-preserving aspect of this new implementation is considered
an implementation detail and should not be relied upon (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
backward-compatibility with older versions of the language where
random iteration order is still in effect, e.g. Python 3.5).
In the document of Python 3.7 (https://docs.python.org/3.7/tutorial/datastructures.html#dictionaries):
Performing list(d) on a dictionary returns a list of all the keys used
in the dictionary, in insertion order (if you want it sorted, just use
sorted(d) instead).
So unlike previous versions, you can sort a dict after Python 3.6/3.7. If you want to sort a nested dict including the sub-dict inside, you can do:
test_dict = {'a': 1, 'c': 3, 'b': {'b2': 2, 'b1': 1}}
def dict_reorder(item):
return {k: dict_reoder(v) if isinstance(v, dict) else v for k, v in sorted(item.items())}
reordered_dict = dict_reorder(test_dict)
https://gist.github.com/ligyxy/f60f0374defc383aa098d44cfbd318eb
Found another way:
import json
print json.dumps(d, sort_keys = True)
upd:
1. this also sorts nested objects (thanks #DanielF).
2. python dictionaries are unordered therefore this is sutable for print or assign to str only.
As others have mentioned, dictionaries are inherently unordered. However, if the issue is merely displaying dictionaries in an ordered fashion, you can override the __str__ method in a dictionary subclass, and use this dictionary class rather than the builtin dict. Eg.
class SortedDisplayDict(dict):
def __str__(self):
return "{" + ", ".join("%r: %r" % (key, self[key]) for key in sorted(self)) + "}"
>>> d = SortedDisplayDict({2:3, 1:89, 4:5, 3:0})
>>> d
{1: 89, 2: 3, 3: 0, 4: 5}
Note, this changes nothing about how the keys are stored, the order they will come back when you iterate over them etc, just how they're displayed with print or at the python console.
An easy way to do this:
d = {2:3, 1:89, 4:5, 3:0}
s = {k : d[k] for k in sorted(d)}
s
Out[1]: {1: 89, 2: 3, 3: 0, 4: 5}
In Python 3.
>>> D1 = {2:3, 1:89, 4:5, 3:0}
>>> for key in sorted(D1):
print (key, D1[key])
gives
1 89
2 3
3 0
4 5
There are plenty of answers here already showcasing popular ways to sort a Python dictionary. I thought I'd add a few more less-obvious ways for those coming here from Google looking for non-standard ideas.
Sample Dictionary: d = {2: 'c', 1: 'b', 0: 'a', 3: 'd'}
Dictionary Comprehension
# Converts to list, sorts, re-converts to dict
{k: v for k, v in sorted(list(d.items()))}
Using Lambdas
Sorting isn't always intended to order strictly in ascending or descending order. For more conditional sorting, use the above method combined with lamdas:
{k: v for k, v in sorted(d.items(), key=lambda v: ord(v[1]))}
More Examples
This thread is already full enough of good examples. For some more examples, as well as edge-cases and oddities check out this article on sorting dictionaries in Python.
You can create a new dictionary by sorting the current dictionary by key as per your question.
This is your dictionary
d = {2:3, 1:89, 4:5, 3:0}
Create a new dictionary d1 by sorting this d using lambda function
d1 = dict(sorted(d.items(), key = lambda x:x[0]))
d1 should be {1: 89, 2: 3, 3: 0, 4: 5}, sorted based on keys in d.
A simple way I found to sort a dictionary is to create a new one, based on the sorted key:value items of the one you're trying to sort.
If you want to sort dict = {}, retrieve all its items using the associated method, sort them using the sorted() function then create the new dictionary.
Here's the code using dictionary comprehension :
sorted_dict = {k:v for k,v in sorted(dict.items())}
Here I found some simplest solution to sort the python dict by key using pprint.
eg.
>>> x = {'a': 10, 'cd': 20, 'b': 30, 'az': 99}
>>> print x
{'a': 10, 'b': 30, 'az': 99, 'cd': 20}
but while using pprint it will return sorted dict
>>> import pprint
>>> pprint.pprint(x)
{'a': 10, 'az': 99, 'b': 30, 'cd': 20}
There is an easy way to sort a dictionary.
According to your question,
The solution is :
c={2:3, 1:89, 4:5, 3:0}
y=sorted(c.items())
print y
(Where c,is the name of your dictionary.)
This program gives the following output:
[(1, 89), (2, 3), (3, 0), (4, 5)]
like u wanted.
Another example is:
d={"John":36,"Lucy":24,"Albert":32,"Peter":18,"Bill":41}
x=sorted(d.keys())
print x
Gives the output:['Albert', 'Bill', 'John', 'Lucy', 'Peter']
y=sorted(d.values())
print y
Gives the output:[18, 24, 32, 36, 41]
z=sorted(d.items())
print z
Gives the output:
[('Albert', 32), ('Bill', 41), ('John', 36), ('Lucy', 24), ('Peter', 18)]
Hence by changing it into keys, values and items , you can print like what u wanted.Hope this helps!
Will generate exactly what you want:
D1 = {2:3, 1:89, 4:5, 3:0}
sort_dic = {}
for i in sorted(D1):
sort_dic.update({i:D1[i]})
print sort_dic
{1: 89, 2: 3, 3: 0, 4: 5}
But this is not the correct way to do this, because, It could show a distinct behavior with different dictionaries, which I have learned recently. Hence perfect way has been suggested by Tim In the response of my Query which I am sharing here.
from collections import OrderedDict
sorted_dict = OrderedDict(sorted(D1.items(), key=lambda t: t[0]))
Here is the performance of the suggested solutions:
from collections import OrderedDict
from sortedcontainers import SortedDict
import json
keys = np.random.rand(100000)
vals = np.random.rand(100000)
d = dict(zip(keys, vals))
timeit SortedDict(d)
#45.8 ms ± 780 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
timeit sorted(d.items())
#91.9 ms ± 707 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
timeit OrderedDict(sorted(d.items(), key=lambda x: x[0]))
#93.7 ms ± 1.52 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
timeit dict(sorted(dic.items()))
#113 ms ± 824 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
timeit OrderedDict(sorted(dic.items()))
#122 ms ± 2.65 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
timeit json.dumps(d, sort_keys=True)
#259 ms ± 9.42 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
As we see, Grant Jenks's solution is by far the fastest.
I think the easiest thing is to sort the dict by key and save the sorted key:value pair in a new dict.
dict1 = {'renault': 3, 'ford':4, 'volvo': 1, 'toyota': 2}
dict2 = {} # create an empty dict to store the sorted values
for key in sorted(dict1.keys()):
if not key in dict2: # Depending on the goal, this line may not be neccessary
dict2[key] = dict1[key]
To make it clearer:
dict1 = {'renault': 3, 'ford':4, 'volvo': 1, 'toyota': 2}
dict2 = {} # create an empty dict to store the sorted values
for key in sorted(dict1.keys()):
if not key in dict2: # Depending on the goal, this line may not be neccessary
value = dict1[key]
dict2[key] = value
I come up with single line dict sorting.
>> a = {2:3, 1:89, 4:5, 3:0}
>> c = {i:a[i] for i in sorted(a.keys())}
>> print(c)
{1: 89, 2: 3, 3: 0, 4: 5}
[Finished in 0.4s]
Hope this will be helpful.
Python dicts are un-ordered. Usually, this is not a problem since the most common use case is to do a lookup.
The simplest way to do what you want would be to create a collections.OrderedDict inserting the elements in sorted order.
ordered_dict = collections.OrderedDict([(k, d[k]) for k in sorted(d.keys())])
If you need to iterated, as others above have suggested, the simplest way would be to iterate over sorted keys. Examples-
Print values sorted by keys:
# create the dict
d = {k1:v1, k2:v2,...}
# iterate by keys in sorted order
for k in sorted(d.keys()):
value = d[k]
# do something with k, value like print
print k, value
Get list of values sorted by keys:
values = [d[k] for k in sorted(d.keys())]
This function will sort any dictionary recursively by its key. That is, if any value in the dictionary is also a dictionary, it too will be sorted by its key. If you are running on CPython 3.6 or greater, than a simple change to use a dict rather than an OrderedDict can be made.
from collections import OrderedDict
def sort_dict(d):
items = [[k, v] for k, v in sorted(d.items(), key=lambda x: x[0])]
for item in items:
if isinstance(item[1], dict):
item[1] = sort_dict(item[1])
return OrderedDict(items)
#return dict(items)
Simplest solution is that you should get a list of dict key is sorted order and then iterate over dict. For example
a1 = {'a':1, 'b':13, 'd':4, 'c':2, 'e':30}
a1_sorted_keys = sorted(a1, key=a1.get, reverse=True)
for r in a1_sorted_keys:
print r, a1[r]
Following will be the output (desending order)
e 30
b 13
d 4
c 2
a 1
For the way how question is formulated, the most answers here are answering it correctly.
However, considering how the things should be really done, taking to acount decades and decades of computer science, it comes to my total suprise that there is actually only one answer here (from GrantJ user) suggesting usage of sorted associative containers (sortedcontainers) which sorts elements based on key at their insertions point.
That will avoid massive performance impact per each calling of sort(...) (at minimum O(N*log(N)), where N is in number of elements (logically, this applies for all such solutions here which suggest to use the sort(...)). Take to account that for all such solutions, the sort(...) will need to be called every time when colletion needs to be accessed as sorted AFTER it was modified by adding/removing elements ...
Guys you are making things complicated ... it's really simple
from pprint import pprint
Dict={'B':1,'A':2,'C':3}
pprint(Dict)
The output is:
{'A':2,'B':1,'C':3}
from operator import itemgetter
# if you would like to play with multiple dictionaries then here you go:
# Three dictionaries that are composed of first name and last name.
user = [
{'fname': 'Mo', 'lname': 'Mahjoub'},
{'fname': 'Abdo', 'lname': 'Al-hebashi'},
{'fname': 'Ali', 'lname': 'Muhammad'}
]
# This loop will sort by the first and the last names.
# notice that in a dictionary order doesn't matter. So it could put the first name first or the last name first.
for k in sorted (user, key=itemgetter ('fname', 'lname')):
print (k)
# This one will sort by the first name only.
for x in sorted (user, key=itemgetter ('fname')):
print (x)
dictionary = {1:[2],2:[],5:[4,5],4:[5],3:[1]}
temp=sorted(dictionary)
sorted_dict = dict([(k,dictionary[k]) for i,k in enumerate(temp)])
sorted_dict:
{1: [2], 2: [], 3: [1], 4: [5], 5: [4, 5]}
A timing comparison of the two methods in 2.7 shows them to be virtually identical:
>>> setup_string = "a = sorted(dict({2:3, 1:89, 4:5, 3:0}).items())"
>>> timeit.timeit(stmt="[(k, val) for k, val in a]", setup=setup_string, number=10000)
0.003599141953657181
>>> setup_string = "from collections import OrderedDict\n"
>>> setup_string += "a = OrderedDict({1:89, 2:3, 3:0, 4:5})\n"
>>> setup_string += "b = a.items()"
>>> timeit.timeit(stmt="[(k, val) for k, val in b]", setup=setup_string, number=10000)
0.003581275490432745
Or use pandas,
Demo:
>>> d={'B':1,'A':2,'C':3}
>>> df=pd.DataFrame(d,index=[0]).sort_index(axis=1)
A B C
0 2 1 3
>>> df.to_dict('int')[0]
{'A': 2, 'B': 1, 'C': 3}
>>>
See:
Docs of this
Documentation of whole pandas
l = dict.keys()
l2 = l
l2.append(0)
l3 = []
for repeater in range(0, len(l)):
smallnum = float("inf")
for listitem in l2:
if listitem < smallnum:
smallnum = listitem
l2.remove(smallnum)
l3.append(smallnum)
l3.remove(0)
l = l3
for listitem in l:
print(listitem)
I have some code that iterates over the values of a dictionary. If the value meets certain conditions, it is deleted from the dictionary. Those conditions are contingent on the existence of other values in the dictionary. This is why I don't want to just copy the old dictionary and make deletions, then re-attribute it.
When I try to run it, I get an error that the size of the dictionary changed while iterating it.
Is there a way to iterate over a dictionary that allows it to change size, and the existence of keys and values, while it is iterating?
Build a new dictionary which contains the keys you want to keep. This can be done with a dictionary comprehension, or a manual for loop.
Here's a comprehension:
return {k: v for k, v in my_dict.items() if some-condition}
Here's a manual loop:
result = {}
for k, v in my_dict.items():
if some-condition:
result[k] = v
return result
Well, yes you can iterate on by the keys (Python3)! Take a look:
>>> dc
{1: 'aze', 3: 'poi', 4: 'mlk'}
>>> dc = {1:"aze", 2:"qsd", 3:"poi", 4:"mlk"}
>>> dc
{1: 'aze', 2: 'qsd', 3: 'poi', 4: 'mlk'}
>>> keys = list(dc.keys())
>>> keys
[1, 2, 3, 4]
>>> for k in keys:
if "q" in dc[k]:
del dc[k]
>>> dc
{1: 'aze', 3: 'poi', 4: 'mlk'}
>>>
You can iterate over the keys instead of over the dict itself. In the following example, all values that are odd-numbered are removed from the dict:
>>> a = {'a': 12, 'b': 3, 'c': 14}
>>> for key in list(a.keys()):
if a[key] % 2 == 0:
del a[key]
>>> a
{'b': 3}
def invert_dict(d):
inv = dict()
for key in d:
val = d[key]
if val not in inv:
inv[val] = [key]
else:
inv[val].append(key)
return inv
This is an example from Think Python book, a function for inverting(swapping) keys and values in a dictionary. New values (former keys) are stored as lists, so if there was multiple dictionary values (bound to a different keys) that were equal before inverting, then this function simply appends them to the list of former keys.
Example:
somedict = {'one': 1, 'two': 2, 'doubletwo': 2, 'three': 3}
invert_dict(somedict) ---> {1: ['one'], 2: ['doubletwo', 'two'], 3: ['three']}
My question is, can the same be done with dictionary comprehensions? This function creates an empty dict inv = dict(), which is then checked later in the function with if/else for the presence of values. Dict comprehension, in this case, should check itself. Is that possible, and how the syntax should look like?
General dict comprehension syntax for swapping values is:
{value:key for key, value in somedict.items()}
but if I want to add an 'if' clausule, what it should look like? if value not in (what)?
Thanks.
I don't think it's possible with simple dict comprehension without using other functions.
Following code uses itertools.groupby to group keys that have same values.
>>> import itertools
>>> {k: [x[1] for x in grp]
for k, grp in itertools.groupby(
sorted((v,k) for k, v in somedict.iteritems()),
key=lambda x: x[0])
}
{1: ['one'], 2: ['doubletwo', 'two'], 3: ['three']}
You can use a set comprehension side effect:
somedict = {'one': 1, 'two': 2, 'doubletwo': 2, 'three': 3}
invert_dict={}
{invert_dict.setdefault(v, []).append(k) for k, v in somedict.items()}
print invert_dict
# {1: ['one'], 2: ['doubletwo', 'two'], 3: ['three']}
Here is a good answer:
fts = {1:1,2:1,3:2,4:1}
new_dict = {dest: [k for k, v in fts.items() if v == dest] for dest in set(fts.values())}
Reference: Head First Python ,2nd Edition, Page(502)
How do I sort a dictionary by its keys?
Example input:
{2:3, 1:89, 4:5, 3:0}
Desired output:
{1:89, 2:3, 3:0, 4:5}
Note: for Python 3.7+, see this answer
Standard Python dictionaries are unordered (until Python 3.7). Even if you sorted the (key,value) pairs, you wouldn't be able to store them in a dict in a way that would preserve the ordering.
The easiest way is to use OrderedDict, which remembers the order in which the elements have been inserted:
In [1]: import collections
In [2]: d = {2:3, 1:89, 4:5, 3:0}
In [3]: od = collections.OrderedDict(sorted(d.items()))
In [4]: od
Out[4]: OrderedDict([(1, 89), (2, 3), (3, 0), (4, 5)])
Never mind the way od is printed out; it'll work as expected:
In [11]: od[1]
Out[11]: 89
In [12]: od[3]
Out[12]: 0
In [13]: for k, v in od.iteritems(): print k, v
....:
1 89
2 3
3 0
4 5
Python 3
For Python 3 users, one needs to use the .items() instead of .iteritems():
In [13]: for k, v in od.items(): print(k, v)
....:
1 89
2 3
3 0
4 5
Dictionaries themselves do not have ordered items as such, should you want to print them etc to some order, here are some examples:
In Python 2.4 and above:
mydict = {'carl':40,
'alan':2,
'bob':1,
'danny':3}
for key in sorted(mydict):
print "%s: %s" % (key, mydict[key])
gives:
alan: 2
bob: 1
carl: 40
danny: 3
(Python below 2.4:)
keylist = mydict.keys()
keylist.sort()
for key in keylist:
print "%s: %s" % (key, mydict[key])
Source: http://www.saltycrane.com/blog/2007/09/how-to-sort-python-dictionary-by-keys/
For CPython/PyPy 3.6, and any Python 3.7 or higher, this is easily done with:
>>> d = {2:3, 1:89, 4:5, 3:0}
>>> dict(sorted(d.items()))
{1: 89, 2: 3, 3: 0, 4: 5}
From Python's collections library documentation:
>>> from collections import OrderedDict
>>> # regular unsorted dictionary
>>> d = {'banana': 3, 'apple':4, 'pear': 1, 'orange': 2}
>>> # dictionary sorted by key -- OrderedDict(sorted(d.items()) also works
>>> 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)])
>>> # dictionary sorted by length of the key string
>>> OrderedDict(sorted(d.items(), key=lambda t: len(t[0])))
OrderedDict([('pear', 1), ('apple', 4), ('orange', 2), ('banana', 3)])
There are a number of Python modules that provide dictionary implementations which automatically maintain the keys in sorted order. Consider the sortedcontainers module which is pure-Python and fast-as-C implementations. There is also a performance comparison with other popular options benchmarked against one another.
Using an ordered dict is an inadequate solution if you need to constantly add and remove key/value pairs while also iterating.
>>> from sortedcontainers import SortedDict
>>> d = {2:3, 1:89, 4:5, 3:0}
>>> s = SortedDict(d)
>>> s.items()
[(1, 89), (2, 3), (3, 0), (4, 5)]
The SortedDict type also supports indexed location lookups and deletion which isn't possible with the built-in dict type.
>>> s.iloc[-1]
4
>>> del s.iloc[2]
>>> s.keys()
SortedSet([1, 2, 4])
Simply:
d = {2:3, 1:89, 4:5, 3:0}
sd = sorted(d.items())
for k,v in sd:
print k, v
Output:
1 89
2 3
3 0
4 5
Python dictionary was unordered before Python 3.6. In CPython implementation of Python 3.6, the dictionary keeps the insertion order.
From Python 3.7, this will become a language feature.
In changelog of Python 3.6 (https://docs.python.org/3.6/whatsnew/3.6.html#whatsnew36-compactdict):
The order-preserving aspect of this new implementation is considered
an implementation detail and should not be relied upon (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
backward-compatibility with older versions of the language where
random iteration order is still in effect, e.g. Python 3.5).
In the document of Python 3.7 (https://docs.python.org/3.7/tutorial/datastructures.html#dictionaries):
Performing list(d) on a dictionary returns a list of all the keys used
in the dictionary, in insertion order (if you want it sorted, just use
sorted(d) instead).
So unlike previous versions, you can sort a dict after Python 3.6/3.7. If you want to sort a nested dict including the sub-dict inside, you can do:
test_dict = {'a': 1, 'c': 3, 'b': {'b2': 2, 'b1': 1}}
def dict_reorder(item):
return {k: dict_reoder(v) if isinstance(v, dict) else v for k, v in sorted(item.items())}
reordered_dict = dict_reorder(test_dict)
https://gist.github.com/ligyxy/f60f0374defc383aa098d44cfbd318eb
Found another way:
import json
print json.dumps(d, sort_keys = True)
upd:
1. this also sorts nested objects (thanks #DanielF).
2. python dictionaries are unordered therefore this is sutable for print or assign to str only.
As others have mentioned, dictionaries are inherently unordered. However, if the issue is merely displaying dictionaries in an ordered fashion, you can override the __str__ method in a dictionary subclass, and use this dictionary class rather than the builtin dict. Eg.
class SortedDisplayDict(dict):
def __str__(self):
return "{" + ", ".join("%r: %r" % (key, self[key]) for key in sorted(self)) + "}"
>>> d = SortedDisplayDict({2:3, 1:89, 4:5, 3:0})
>>> d
{1: 89, 2: 3, 3: 0, 4: 5}
Note, this changes nothing about how the keys are stored, the order they will come back when you iterate over them etc, just how they're displayed with print or at the python console.
An easy way to do this:
d = {2:3, 1:89, 4:5, 3:0}
s = {k : d[k] for k in sorted(d)}
s
Out[1]: {1: 89, 2: 3, 3: 0, 4: 5}
In Python 3.
>>> D1 = {2:3, 1:89, 4:5, 3:0}
>>> for key in sorted(D1):
print (key, D1[key])
gives
1 89
2 3
3 0
4 5
There are plenty of answers here already showcasing popular ways to sort a Python dictionary. I thought I'd add a few more less-obvious ways for those coming here from Google looking for non-standard ideas.
Sample Dictionary: d = {2: 'c', 1: 'b', 0: 'a', 3: 'd'}
Dictionary Comprehension
# Converts to list, sorts, re-converts to dict
{k: v for k, v in sorted(list(d.items()))}
Using Lambdas
Sorting isn't always intended to order strictly in ascending or descending order. For more conditional sorting, use the above method combined with lamdas:
{k: v for k, v in sorted(d.items(), key=lambda v: ord(v[1]))}
More Examples
This thread is already full enough of good examples. For some more examples, as well as edge-cases and oddities check out this article on sorting dictionaries in Python.
You can create a new dictionary by sorting the current dictionary by key as per your question.
This is your dictionary
d = {2:3, 1:89, 4:5, 3:0}
Create a new dictionary d1 by sorting this d using lambda function
d1 = dict(sorted(d.items(), key = lambda x:x[0]))
d1 should be {1: 89, 2: 3, 3: 0, 4: 5}, sorted based on keys in d.
A simple way I found to sort a dictionary is to create a new one, based on the sorted key:value items of the one you're trying to sort.
If you want to sort dict = {}, retrieve all its items using the associated method, sort them using the sorted() function then create the new dictionary.
Here's the code using dictionary comprehension :
sorted_dict = {k:v for k,v in sorted(dict.items())}
Here I found some simplest solution to sort the python dict by key using pprint.
eg.
>>> x = {'a': 10, 'cd': 20, 'b': 30, 'az': 99}
>>> print x
{'a': 10, 'b': 30, 'az': 99, 'cd': 20}
but while using pprint it will return sorted dict
>>> import pprint
>>> pprint.pprint(x)
{'a': 10, 'az': 99, 'b': 30, 'cd': 20}
There is an easy way to sort a dictionary.
According to your question,
The solution is :
c={2:3, 1:89, 4:5, 3:0}
y=sorted(c.items())
print y
(Where c,is the name of your dictionary.)
This program gives the following output:
[(1, 89), (2, 3), (3, 0), (4, 5)]
like u wanted.
Another example is:
d={"John":36,"Lucy":24,"Albert":32,"Peter":18,"Bill":41}
x=sorted(d.keys())
print x
Gives the output:['Albert', 'Bill', 'John', 'Lucy', 'Peter']
y=sorted(d.values())
print y
Gives the output:[18, 24, 32, 36, 41]
z=sorted(d.items())
print z
Gives the output:
[('Albert', 32), ('Bill', 41), ('John', 36), ('Lucy', 24), ('Peter', 18)]
Hence by changing it into keys, values and items , you can print like what u wanted.Hope this helps!
Will generate exactly what you want:
D1 = {2:3, 1:89, 4:5, 3:0}
sort_dic = {}
for i in sorted(D1):
sort_dic.update({i:D1[i]})
print sort_dic
{1: 89, 2: 3, 3: 0, 4: 5}
But this is not the correct way to do this, because, It could show a distinct behavior with different dictionaries, which I have learned recently. Hence perfect way has been suggested by Tim In the response of my Query which I am sharing here.
from collections import OrderedDict
sorted_dict = OrderedDict(sorted(D1.items(), key=lambda t: t[0]))
Here is the performance of the suggested solutions:
from collections import OrderedDict
from sortedcontainers import SortedDict
import json
keys = np.random.rand(100000)
vals = np.random.rand(100000)
d = dict(zip(keys, vals))
timeit SortedDict(d)
#45.8 ms ± 780 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
timeit sorted(d.items())
#91.9 ms ± 707 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
timeit OrderedDict(sorted(d.items(), key=lambda x: x[0]))
#93.7 ms ± 1.52 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
timeit dict(sorted(dic.items()))
#113 ms ± 824 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
timeit OrderedDict(sorted(dic.items()))
#122 ms ± 2.65 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
timeit json.dumps(d, sort_keys=True)
#259 ms ± 9.42 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
As we see, Grant Jenks's solution is by far the fastest.
I think the easiest thing is to sort the dict by key and save the sorted key:value pair in a new dict.
dict1 = {'renault': 3, 'ford':4, 'volvo': 1, 'toyota': 2}
dict2 = {} # create an empty dict to store the sorted values
for key in sorted(dict1.keys()):
if not key in dict2: # Depending on the goal, this line may not be neccessary
dict2[key] = dict1[key]
To make it clearer:
dict1 = {'renault': 3, 'ford':4, 'volvo': 1, 'toyota': 2}
dict2 = {} # create an empty dict to store the sorted values
for key in sorted(dict1.keys()):
if not key in dict2: # Depending on the goal, this line may not be neccessary
value = dict1[key]
dict2[key] = value
I come up with single line dict sorting.
>> a = {2:3, 1:89, 4:5, 3:0}
>> c = {i:a[i] for i in sorted(a.keys())}
>> print(c)
{1: 89, 2: 3, 3: 0, 4: 5}
[Finished in 0.4s]
Hope this will be helpful.
Python dicts are un-ordered. Usually, this is not a problem since the most common use case is to do a lookup.
The simplest way to do what you want would be to create a collections.OrderedDict inserting the elements in sorted order.
ordered_dict = collections.OrderedDict([(k, d[k]) for k in sorted(d.keys())])
If you need to iterated, as others above have suggested, the simplest way would be to iterate over sorted keys. Examples-
Print values sorted by keys:
# create the dict
d = {k1:v1, k2:v2,...}
# iterate by keys in sorted order
for k in sorted(d.keys()):
value = d[k]
# do something with k, value like print
print k, value
Get list of values sorted by keys:
values = [d[k] for k in sorted(d.keys())]
This function will sort any dictionary recursively by its key. That is, if any value in the dictionary is also a dictionary, it too will be sorted by its key. If you are running on CPython 3.6 or greater, than a simple change to use a dict rather than an OrderedDict can be made.
from collections import OrderedDict
def sort_dict(d):
items = [[k, v] for k, v in sorted(d.items(), key=lambda x: x[0])]
for item in items:
if isinstance(item[1], dict):
item[1] = sort_dict(item[1])
return OrderedDict(items)
#return dict(items)
Simplest solution is that you should get a list of dict key is sorted order and then iterate over dict. For example
a1 = {'a':1, 'b':13, 'd':4, 'c':2, 'e':30}
a1_sorted_keys = sorted(a1, key=a1.get, reverse=True)
for r in a1_sorted_keys:
print r, a1[r]
Following will be the output (desending order)
e 30
b 13
d 4
c 2
a 1
For the way how question is formulated, the most answers here are answering it correctly.
However, considering how the things should be really done, taking to acount decades and decades of computer science, it comes to my total suprise that there is actually only one answer here (from GrantJ user) suggesting usage of sorted associative containers (sortedcontainers) which sorts elements based on key at their insertions point.
That will avoid massive performance impact per each calling of sort(...) (at minimum O(N*log(N)), where N is in number of elements (logically, this applies for all such solutions here which suggest to use the sort(...)). Take to account that for all such solutions, the sort(...) will need to be called every time when colletion needs to be accessed as sorted AFTER it was modified by adding/removing elements ...
Guys you are making things complicated ... it's really simple
from pprint import pprint
Dict={'B':1,'A':2,'C':3}
pprint(Dict)
The output is:
{'A':2,'B':1,'C':3}
from operator import itemgetter
# if you would like to play with multiple dictionaries then here you go:
# Three dictionaries that are composed of first name and last name.
user = [
{'fname': 'Mo', 'lname': 'Mahjoub'},
{'fname': 'Abdo', 'lname': 'Al-hebashi'},
{'fname': 'Ali', 'lname': 'Muhammad'}
]
# This loop will sort by the first and the last names.
# notice that in a dictionary order doesn't matter. So it could put the first name first or the last name first.
for k in sorted (user, key=itemgetter ('fname', 'lname')):
print (k)
# This one will sort by the first name only.
for x in sorted (user, key=itemgetter ('fname')):
print (x)
dictionary = {1:[2],2:[],5:[4,5],4:[5],3:[1]}
temp=sorted(dictionary)
sorted_dict = dict([(k,dictionary[k]) for i,k in enumerate(temp)])
sorted_dict:
{1: [2], 2: [], 3: [1], 4: [5], 5: [4, 5]}
A timing comparison of the two methods in 2.7 shows them to be virtually identical:
>>> setup_string = "a = sorted(dict({2:3, 1:89, 4:5, 3:0}).items())"
>>> timeit.timeit(stmt="[(k, val) for k, val in a]", setup=setup_string, number=10000)
0.003599141953657181
>>> setup_string = "from collections import OrderedDict\n"
>>> setup_string += "a = OrderedDict({1:89, 2:3, 3:0, 4:5})\n"
>>> setup_string += "b = a.items()"
>>> timeit.timeit(stmt="[(k, val) for k, val in b]", setup=setup_string, number=10000)
0.003581275490432745
Or use pandas,
Demo:
>>> d={'B':1,'A':2,'C':3}
>>> df=pd.DataFrame(d,index=[0]).sort_index(axis=1)
A B C
0 2 1 3
>>> df.to_dict('int')[0]
{'A': 2, 'B': 1, 'C': 3}
>>>
See:
Docs of this
Documentation of whole pandas
l = dict.keys()
l2 = l
l2.append(0)
l3 = []
for repeater in range(0, len(l)):
smallnum = float("inf")
for listitem in l2:
if listitem < smallnum:
smallnum = listitem
l2.remove(smallnum)
l3.append(smallnum)
l3.remove(0)
l = l3
for listitem in l:
print(listitem)
This question already has answers here:
Reverse / invert a dictionary mapping
(32 answers)
Closed 10 months ago.
I receive a dictionary as input, and would like to to return a dictionary whose keys will be the input's values and whose value will be the corresponding input keys. Values are unique.
For example, say my input is:
a = dict()
a['one']=1
a['two']=2
I would like my output to be:
{1: 'one', 2: 'two'}
To clarify I would like my result to be the equivalent of the following:
res = dict()
res[1] = 'one'
res[2] = 'two'
Any neat Pythonic way to achieve this?
Python 2:
res = dict((v,k) for k,v in a.iteritems())
Python 3 (thanks to #erik):
res = dict((v,k) for k,v in a.items())
new_dict = dict(zip(my_dict.values(), my_dict.keys()))
From Python 2.7 on, including 3.0+, there's an arguably shorter, more readable version:
>>> my_dict = {'x':1, 'y':2, 'z':3}
>>> {v: k for k, v in my_dict.items()}
{1: 'x', 2: 'y', 3: 'z'}
You can make use of dict comprehensions:
Python 3
res = {v: k for k, v in a.items()}
Python 2
res = {v: k for k, v in a.iteritems()}
Edited: For Python 3, use a.items() instead of a.iteritems(). Discussions about the differences between them can be found in iteritems in Python on SO.
In [1]: my_dict = {'x':1, 'y':2, 'z':3}
Python 3
In [2]: dict((value, key) for key, value in my_dict.items())
Out[2]: {1: 'x', 2: 'y', 3: 'z'}
Python 2
In [2]: dict((value, key) for key, value in my_dict.iteritems())
Out[2]: {1: 'x', 2: 'y', 3: 'z'}
The current leading answer assumes values are unique which is not always the case. What if values are not unique? You will loose information!
For example:
d = {'a':3, 'b': 2, 'c': 2}
{v:k for k,v in d.iteritems()}
returns {2: 'b', 3: 'a'}.
The information about 'c' was completely ignored.
Ideally it should had be something like {2: ['b','c'], 3: ['a']}. This is what the bottom implementation does.
Python 2.x
def reverse_non_unique_mapping(d):
dinv = {}
for k, v in d.iteritems():
if v in dinv:
dinv[v].append(k)
else:
dinv[v] = [k]
return dinv
Python 3.x
def reverse_non_unique_mapping(d):
dinv = {}
for k, v in d.items():
if v in dinv:
dinv[v].append(k)
else:
dinv[v] = [k]
return dinv
You could try:
Python 3
d={'one':1,'two':2}
d2=dict((value,key) for key,value in d.items())
d2
{'two': 2, 'one': 1}
Python 2
d={'one':1,'two':2}
d2=dict((value,key) for key,value in d.iteritems())
d2
{'two': 2, 'one': 1}
Beware that you cannot 'reverse' a dictionary if
More than one key shares the same value. For example {'one':1,'two':1}. The new dictionary can only have one item with key 1.
One or more of the values is unhashable. For example {'one':[1]}. [1] is a valid value but not a valid key.
See this thread on the python mailing list for a discussion on the subject.
res = dict(zip(a.values(), a.keys()))
new_dict = dict( (my_dict[k], k) for k in my_dict)
or even better, but only works in Python 3:
new_dict = { my_dict[k]: k for k in my_dict}
Another way to expand on Ilya Prokin's response is to actually use the reversed function.
dict(map(reversed, my_dict.items()))
In essence, your dictionary is iterated through (using .items()) where each item is a key/value pair, and those items are swapped with the reversed function. When this is passed to the dict constructor, it turns them into value/key pairs which is what you want.
Suggestion for an improvement for Javier answer :
dict(zip(d.values(),d))
Instead of d.keys() you can write just d, because if you go through dictionary with an iterator, it will return the keys of the relevant dictionary.
Ex. for this behavior :
d = {'a':1,'b':2}
for k in d:
k
'a'
'b'
Can be done easily with dictionary comprehension:
{d[i]:i for i in d}
dict(map(lambda x: x[::-1], YourDict.items()))
.items() returns a list of tuples of (key, value). map() goes through elements of the list and applies lambda x:[::-1] to each its element (tuple) to reverse it, so each tuple becomes (value, key) in the new list spitted out of map. Finally, dict() makes a dict from the new list.
Hanan's answer is the correct one as it covers more general case (the other answers are kind of misleading for someone unaware of the duplicate situation). An improvement to Hanan's answer is using setdefault:
mydict = {1:a, 2:a, 3:b}
result = {}
for i in mydict:
result.setdefault(mydict[i],[]).append(i)
print(result)
>>> result = {a:[1,2], b:[3]}
Using loop:-
newdict = {} #Will contain reversed key:value pairs.
for key, value in zip(my_dict.keys(), my_dict.values()):
# Operations on key/value can also be performed.
newdict[value] = key
If you're using Python3, it's slightly different:
res = dict((v,k) for k,v in a.items())
Adding an in-place solution:
>>> d = {1: 'one', 2: 'two', 3: 'three', 4: 'four'}
>>> for k in list(d.keys()):
... d[d.pop(k)] = k
...
>>> d
{'two': 2, 'one': 1, 'four': 4, 'three': 3}
In Python3, it is critical that you use list(d.keys()) because dict.keys returns a view of the keys. If you are using Python2, d.keys() is enough.
I find this version the most comprehensive one:
a = {1: 'one', 2: 'two'}
swapped_a = {value : key for key, value in a.items()}
print(swapped_a)
output :
{'one': 1, 'two': 2}
An alternative that is not quite as readable (in my opinion) as some of the other answers:
new_dict = dict(zip(*list(zip(*old_dict.items()))[::-1]))
where list(zip(*old_dict.items()))[::-1] gives a list of 2 tuples, old_dict's values and keys, respectively.