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Python | Sort a List according to the Length of the Elements
  • Last Updated : 19 Apr, 2020

In this program, we need to accept a list and sort it based on the length of the elements present within.
Examples:

Input : list = ["rohan", "amy", "sapna", "muhammad",
                "aakash", "raunak", "chinmoy"]
Output : ['amy', 'rohan', 'sapna', 'aakash', 'raunak', 
         'chinmoy', 'muhammad']

Input : list = [["ram", "mohan", "aman"], ["gaurav"], 
                 ["amy", "sima", "ankita", "rinku"]]
Output : [['gaurav'], ['ram', 'mohan', 'aman'], 
          ['amy', 'sima', 'ankita', 'rinku']]

Note: The first example comprises of Strings whose 
length can be calculated. The second example comprises 
of sublists, which is also arranged according to there 
length. 

There are many ways of performing this. Anyone can use there own algorithmic technique, but Python provides us with various built-in functions to perform these. The built-in functions include sort() and sorted() along with the key parameter. We can perform these in two ways. One way is to sort the list by creating a new list and another way is to sort within the given list, saving space.

The syntax for sorting by creating a new list is:

sorted_list = sorted(unsorted_list, key=len)
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# Python code to sort a list by creating 
# another list Use of sorted()
def Sorting(lst):
    lst2 = sorted(lst, key=len)
    return lst2
      
# Driver code
lst = ["rohan", "amy", "sapna", "muhammad"
       "aakash", "raunak", "chinmoy"]
print(Sorting(lst))

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The syntax for sorting without creating a new list is:

unsorted_list.sort(key=len)
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# Python code to sort a list without 
# creating another list Use of sort()
def Sorting(lst):
    lst.sort(key=len)
    return lst
      
# Driver code
lst = ["rohan", "amy", "sapna", "muhammad"
       "aakash", "raunak", "chinmoy"]
print(Sorting(lst))

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Output:



['amy', 'rohan', 'sapna', 'aakash', 'raunak', 'chinmoy', 'muhammad']

Working:
These key function of Python’s while sorting implemented is known as the decorate-sort-undecorate design pattern. It follows the following steps:

  1. Each element of the list is temporarily replaced with a “decorated” version that includes the result of the key function applied to the element.
  2. The list is sorted based upon the natural order of the keys.
  3. The decorated elements are replaced by the original elements.

The code for sorting by creating a new dummy list is:

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import numpy
  
def Sorting(lst):
  
    # list for storing the length of each string in list 
    lenlist=[]   
    for x in lst:
         lenlist.append(len(x))     
  
    # return a list with the index of the sorted
    # items in the list
    sortedindex = numpy.argsort(lenlist)  
  
    # creating a dummy list where we will place the 
    # word according to the sortedindex list 
    lst2 = ['dummy']*len(lst)   
  
    # print(sortedindex,lenlist)
    for i in range(len(lst)):    
  
        # placing element in the lst2 list by taking the
        # value from original list lst where it should belong 
        # in the sorted list by taking its index from sortedindex
        lst2[i] = lst[sortedindex[i]]     
                                          
    return lst2
      
# Driver code
lst = ["rohan", "amy", "sapna", "muhammad"
       "aakash", "raunak", "chinmoy"]
print(Sorting(lst))

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Output:

['amy', 'rohan', 'sapna', 'aakash', 'raunak', 'chinmoy', 'muhammad']


Reference:
stackoverflow

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