What would be a nice way to go from {2:3, 1:89, 4:5, 3:0} to {1:89, 2:3, 3:0, 4:5}?
I checked some posts but they all use the "sorted" operator that returns tuples.
This question is related to
python
sorting
dictionary
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}
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}
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.
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:
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
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])
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())]
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 ...
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!
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
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)
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}
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
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.
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.
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)
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
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
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}
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/
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.
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]))
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]}
Python dictionary was unordered before Python 3.6. In CPython implementation of Python 3.6, 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 backwards-compatibility with older versions of the language where random iteration order is still in effect, e.g. Python 3.5).
In 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: sort_dict(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
My suggestion is this as it allows you to sort a dict or keep a dict sorted as you are adding items and might need to add items in the future:
Build a dict from scratch as you go along. Have a second data structure, a list, with your list of keys. The bisect package has an insort function which allows inserting into a sorted list, or sort your list after completely populating your dict. Now, when you iterate over your dict, you instead iterate over the list to access each key in an in-order fashion without worrying about the representation of the dict structure (which was not made for sorting).
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)
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)])
Source: Stackoverflow.com