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    in which situation is a python dictionary sorted?

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    How to Sort a Dictionary by Value in Python

    In this article, we'll explore how to sort a dictionary in Python by its value. These solutions use for loops and the sorted() function, as well as lambdas and the operator module.

    How to Sort a Dictionary by Value in Python

    Naazneen Jatu

    Introduction

    A dictionary in Python is a collection of items that stores data as key-value pairs. In Python 3.7 and later versions, dictionaries are sorted by the order of item insertion. In earlier versions, they were unordered.

    Let's have a look at how we can sort a dictionary on basis of the values they contain.

    Sort Dictionary Using a for Loop

    We can sort a dictionary with the help of a for loop. First, we use the sorted() function to order the values of the dictionary. We then loop through the sorted values, finding the keys for each value. We add these keys-value pairs in the sorted order into a new dictionary.

    Note: Sorting does not allow you to re-order the dictionary in-place. We are writing the ordered pairs in a completely new, empty dictionary.

    dict1 = {1: 1, 2: 9, 3: 4}

    sorted_values = sorted(dict1.values()) # Sort the values

    sorted_dict = {}

    for i in sorted_values:

    for k in dict1.keys():

    if dict1[k] == i:

    sorted_dict[k] = dict1[k]

    break print(sorted_dict)

    If you run this with the Python interpreter you would see:

    {1: 1, 3: 4, 2: 9}

    Now that we've seen how to sort with loops, let's look at a more popular alternative that uses the sorted() function.

    Sort Dictionary Using the sorted() Function

    We previously used the sorted() function to sort the values of an array. When sorting a dictionary, we can pass one more argument to the sorted() function like this: sorted(dict1, key=dict1.get).

    Here, key is a function that's called on each element before the values are compared for sorting. The get() method on dictionary objects returns the value of for a dictionary's key.

    The sorted(dict1, key=dict1.get) expression will return the list of keys whose values are sorted in order. From there, we can create a new, sorted dictionary:

    dict1 = {1: 1, 2: 9, 3: 4}

    sorted_dict = {}

    sorted_keys = sorted(dict1, key=dict1.get) # [1, 3, 2]

    for w in sorted_keys:

    sorted_dict[w] = dict1[w]

    print(sorted_dict) # {1: 1, 3: 4, 2: 9}

    Using the sorted() function has reduced the amount of code we had to write when using for loops. However, we can further combine the sorted() function with the itemgetter() function for a more succinct solution to sorting dictionaries by values.

    Sort Dictionary Using the operator Module and itemgetter()

    The operator module includes the itemgetter() function. This function returns a callable object that returns an item from an object.

    For example, let's use to itemgetter() to create a callable object that returns the value of any dictionary with a key that's 2:

    import operator

    dict1 = {1: 1, 2: 9}

    get_item_with_key_2 = operator.itemgetter(2)

    print(get_item_with_key_2(dict1)) # 9

    Every dictionary has access to the items() method. This function returns the key-value pairs of a dictionary as a list of tuples. We can sort the list of tuples by using the itemgetter() function to pull the second value of the tuple i.e. the value of the keys in the dictionary.

    Once it's sorted, we can create a dictionary based on those values:

    import operator

    dict1 = {1: 1, 2: 9, 3: 4}

    sorted_tuples = sorted(dict1.items(), key=operator.itemgetter(1))

    print(sorted_tuples) # [(1, 1), (3, 4), (2, 9)]

    sorted_dict = {k: v for k, v in sorted_tuples}

    print(sorted_dict) # {1: 1, 3: 4, 2: 9}

    With much less effort, we have a dictionary sorted by values!

    As the key argument accepts any function, we can use lambda functions to return dictionary values so they can be sorted. Let's see how.

    Sort Dictionary Using a Lambda Function

    Lambda functions are anonymous, or nameless, functions in Python. We can use lamba functions to get the value of a dictionary item without having to import the operator module for itemgetter(). If you'd like to learn more about lambas, you can read about them in our guide to Lambda Functions in Python.

    Let's sort a dictionary by values using a lambda function in the key argument of sorted():

    dict1 = {1: 1, 2: 9, 3: 4}

    sorted_tuples = sorted(dict1.items(), key=lambda item: item[1])

    print(sorted_tuples) # [(1, 1), (3, 4), (2, 9)]

    sorted_dict = {k: v for k, v in sorted_tuples}

    print(sorted_dict) # {1: 1, 3: 4, 2: 9}

    Note that the methods we've discussed so far only work with Python 3.7 and later. Let's see what we can do for earlier versions of Python.

    Returning a New Dictionary with Sorted Values

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    After sorting a dictionary by values, to keep a sorted dictionary in Python versions before 3.7, you have to use the OrderedDict - available in the collections module. These objects are dictionaries that keep the order of insertion.

    Here's an example of sorting and using OrderedDict:

    import operator

    from collections import OrderedDict

    dict1 = {1: 1, 2: 9, 3: 4}

    sorted_tuples = sorted(dict1.items(), key=operator.itemgetter(1))

    Source : stackabuse.com

    Python

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    Python | Sort Python Dictionaries by Key or Value

    Difficulty Level : Medium

    Last Updated : 23 Jul, 2021

    Prerequisite:

    Dictionary List

    Merging Two Dictionaries

    Dictionary Methods

    Problem Statement – Here are the major tasks that are needed to be performed.

    Create a dictionary and display its keys alphabetically.

    Display both the keys and values sorted in alphabetical order by the key.

    Same as part (ii), but sorted in alphabetical order by the value.

    Approach –

    Load the Dictionary and perform the following operations:

    First, sort the keys alphabetically using function.

    Second, sort the keys alphabetically using function & print the value corresponding to it.

    Third, sort the values alphabetically using

    Recommended: Please try your approach on first, before moving on to the solution.

    Let’s try performing the above-mentioned tasks:

    Displaying the Keys Alphabetically:

    Examples:

    Input: key_value[2] = '64' key_value[1] = '69' key_value[4] = '23' key_value[5] = '65' key_value[6] = '34' key_value[3] = '76' Output: 1 2 3 4 5 6

    Program:

    Python3

    # Function calling def dictionairy():

    # Declare hash function

    key_value ={}

    # Initializing value

    key_value[2] = 56 key_value[1] = 2 key_value[5] = 12 key_value[4] = 24 key_value[6] = 18 key_value[3] = 323

    print ("Task 1:-\n")

    print ("Keys are")

    # iterkeys() returns an iterator over the

    # dictionary’s keys.

    for i in sorted (key_value.keys()) :

    print(i, end = " ") def main(): # function calling dictionairy()

    # Main function calling

    if __name__=="__main__":

    main()

    Output:

    Task 1:- Keys are 1 2 3 4 5 6

    Sorting the Keys and Values in Alphabetical Order using the Key.

    Examples:

    Input: key_value[2] = '64' key_value[1] = '69' key_value[4] = '23' key_value[5] = '65' key_value[6] = '34' key_value[3] = '76' Output:

    (1, 69) (2, 64) (3, 76) (4, 23) (5, 65) (6, 34)

    Program:

    Python3

    # function calling def dictionairy():

    # Declaring the hash function

    key_value ={} # Initialize value key_value[2] = 56 key_value[1] = 2 key_value[5] = 12 key_value[4] = 24 key_value[6] = 18 key_value[3] = 323

    print ("Task 2:-\nKeys and Values sorted in",

    "alphabetical order by the key  ")

    # sorted(key_value) returns an iterator over the

    # Dictionary’s value sorted in keys.

    for i in sorted (key_value) :

    print ((i, key_value[i]), end =" ")

    def main(): # function calling dictionairy()

    # main function calling

    if __name__=="__main__":

    main()

    Output:

    Task 2:-

    Keys and Values sorted in alphabetical order by the key

    (1, 2) (2, 56) (3, 323) (4, 24) (5, 12) (6, 18)

    Sorting the Keys and Values in alphabetical using the value

    Examples:

    Input: key_value[2] = '64' key_value[1] = '69' key_value[4] = '23' key_value[5] = '65' key_value[6] = '34' key_value[3] = '76' Output:

    (4, 23), (6, 34), (2, 64), (5, 65), (1, 69), (3, 76)

    Program:

    Python3

    # Function calling def dictionairy():

    # Declaring hash function

    key_value ={}

    # Initializing the value

    key_value[2] = 56 key_value[1] = 2 key_value[5] = 12 key_value[4] = 24 key_value[6] = 18 key_value[3] = 323

    print ("Task 3:-\nKeys and Values sorted",

    "in alphabetical order by the value")

    # Note that it will sort in lexicographical order

    # For mathematical way, change it to float

    print(sorted(key_value.items(), key =

    lambda kv:(kv[1], kv[0])))

    def main(): # function calling dictionairy()

    # main function calling

    if __name__=="__main__":

    main()

    Output:

    Task 3:-

    Keys and Values sorted in alphabetical order by the value

    [(1, 2), (5, 12), (6, 18), (4, 24), (2, 56), (3, 323)]

    Sort the dictionary by keyNote:it will sort in lexicographical order

    Taking key’s type as string

    Program:

    Python3

    # Creates a sorted dictionary (sorted by key)

    from collections import OrderedDict

    dict = {'ravi':'10','rajnish':'9','sanjeev':'15','yash':'2','suraj':'32'}

    dict1 = OrderedDict(sorted(dict.items()))

    print(dict1)

    Here, Output is Sorted Dictionary using key in Lexicographical order

    Lexicographical order:https://en.wikipedia.org/wiki/Lexicographical_orderLearning Outcome:

    How to handle a dictionary.

    Dictionary has O(1) search time complexity whereas List has O(n) time complexity. So it is recommended to use the dictionary where ever possible.

    The best application of dictionary can be seen in where analysis Twitter sentiments are analysed, using Lexicon approach.

    Source : www.geeksforgeeks.org

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