Sorting a Dictionary in Place with Respect to Keys

Sorting a Dictionary in place with respect to keys

You can't sort a Dictionary<TKey, TValue> - it's inherently unordered. (Or rather, the order in which entries are retrieved is implementation-specific. You shouldn't rely on it working the same way between versions, as ordering isn't part of its designed functionality.)

You can use SortedList<TKey, TValue> or SortedDictionary<TKey, TValue>, both of which sort by the key (in a configurable way, if you pass an IEqualityComparer<T> into the constructor) - might those be of use to you?

Pay little attention to the word "list" in the name SortedList - it's still a dictionary in that it maps keys to values. It's implemented using a list internally, effectively - so instead of looking up by hash code, it does a binary search. SortedDictionary is similarly based on binary searches, but via a tree instead of a list.

How do I sort a dictionary by key?

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

How do I sort a dictionary by value?

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)

How to sort dictionaries by keys in Python

Dicts don't have an order.

You can call sorted but this just gives you a sorted list of the keys:

>>> sorted(d)
['a', 'b', 'c', 'd']

You can treat it as an iterable and sort the key-value tuples, but then you've just got a list of tuples. That's not the same as a dict.

>>> sorted(d.items())
[
 ('a', [1, 2, 3]),
 ('b', ['blah', 'bhasdf', 'asdf']),
 ('c', ['one', 'two']),
 ('d', ['asdf', 'wer', 'asdf', 'zxcv'])
]

If you are using Python 2.7 or newer you could also consider using an OrderedDict.

dict subclass that remembers the order entries were added

For example:

>>> d = collections.OrderedDict(sorted(d.items()))
>>> for k, v in d.items():
>>>     print k, v

a [1, 2, 3]
b ['blah', 'bhasdf', 'asdf']
c ['one', 'two']
d ['asdf', 'wer', 'asdf', 'zxcv']

How do you sort a dictionary by value?

Use:

using System.Linq.Enumerable;
...
List<KeyValuePair<string, string>> myList = aDictionary.ToList();

myList.Sort(
    delegate(KeyValuePair<string, string> pair1,
    KeyValuePair<string, string> pair2)
    {
        return pair1.Value.CompareTo(pair2.Value);
    }
);

Since you're targeting .NET 2.0 or above, you can simplify this into lambda syntax -- it's equivalent, but shorter. If you're targeting .NET 2.0 you can only use this syntax if you're using the compiler from Visual Studio 2008 (or above).

var myList = aDictionary.ToList();

myList.Sort((pair1,pair2) => pair1.Value.CompareTo(pair2.Value));

Sorting a dictionary by value then by key

In [62]: y={100:1, 90:4, 99:3, 92:1, 101:1}
In [63]: sorted(y.items(), key=lambda x: (x[1],x[0]), reverse=True)
Out[63]: [(90, 4), (99, 3), (101, 1), (100, 1), (92, 1)]

The key=lambda x: (x[1],x[0]) tells sorted that for each item x in y.items(), use (x[1],x[0]) as the proxy value to be sorted. Since x is of the form (key,value), (x[1],x[0]) yields (value,key). This causes sorted to sort by value first, then by key for tie-breakers.

reverse=True tells sorted to present the result in descending, rather than ascending order.

See this wiki page for a great tutorial on sorting in Python.

PS. I tried using key=reversed instead, but reversed(x) returns an iterator, which does not compare as needed here.

Sort dictionary alphabetically when the key is a string (name)

simple algorithm to sort dictonary keys in alphabetical order, First sort the keys using sorted

sortednames=sorted(dictUsers.keys(), key=lambda x:x.lower())

for each key name retreive the values from the dict

for i in sortednames:
   values=dictUsers[i]
   print("Name= " + i)
   print ("   Age= " + values.age)
   print ("   Address= " + values.address)
   print ("   Phone Number= " + values.phone)

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