In this article, we will learn how to sort any list, according to the second element of the sublist present within the main list. We will see two methods of doing this. We will learn three methods of performing this sort. One by the use of Bubble Sort, the second by using the sort() method, and last but not the least by the use of the sorted() method. In this program, we have sorted the list in ascending order.
Illustration:
Input : [['rishav', 10], ['akash', 5], ['ram', 20], ['gaurav', 15]] Output : [['akash', 5], ['rishav', 10], ['gaurav', 15], ['ram', 20]]
Input : [['452', 10], ['256', 5], ['100', 20], ['135', 15]] Output : [['256', 5], ['452', 10], ['135', 15], ['100', 20]]
Method 1: Using the Bubble Sort Technique
Here we have used the technique of Bubble Sort to perform the sorting. We have tried to access the second element of the sublists using the nested loops. This performs the in-place method of sorting. The time complexity is similar to the Bubble Sort i.e., O(n^2)
Python3
# Python code to sort the lists using the second element# of sublists inplace way to sort, use of third variabledef Sort(sub_li): l = len(sub_li) for i in range(0, l): for j in range(0, l-i-1): if (sub_li[j][1] > sub_li[j + 1][1]): tempo = sub_li[j] sub_li[j] = sub_li[j + 1] sub_li[j + 1] = tempo return sub_li# Input listsub_li = [['rishav', 10], ['akash', 5], ['ram', 20], ['gaurav', 15]]# Printing the listprint(Sort(sub_li)) |
Output:
[['akash', 5], ['rishav', 10], ['gaurav', 15], ['ram', 20]]
Time Complexity: O(n*n), where n is the length of the given sub_li.
Auxiliary Space: O(1)
Method 2: Sorting by the use of sort() method
While sorting via this method the actual content of the tuple is changed, and just like the previous method, an in-place method of sort is performed.
Python3
# Python code to sort the tuples using second element# of sublist Inplace way to sort using sort()def Sort(sub_li): # reverse = None (Sorts in Ascending order) # key is set to sort using second element of # sublist lambda has been used sub_li.sort(key = lambda x: x[1]) return sub_li# Input listsub_li =[['rishav', 10], ['akash', 5], ['ram', 20], ['gaurav', 15]]# Printing the sub listprint(Sort(sub_li)) |
[['akash', 5], ['rishav', 10], ['gaurav', 15], ['ram', 20]]
Time complexity: O(n*logn)
Auxiliary space: O(1) as the sorting is performed in place using the sort() function. The sort() function has a time complexity of O(nlogn) for the average and worst-case scenarios.
Method 3: Sorting by the use of sorted() method
Sorted() sorts a list and always returns a list with the elements in a sorted manner, without modifying the original sequence. It takes three parameters of which two are optional, here we tried to use all three:
- Iterable : sequence (list, tuple, string) or collection (dictionary, set, frozenset) or any other iterator that needs to be sorted.
- Key(optional): A function that would serve as a key or a basis of sort comparison.
- Reverse(optional): To sort this in ascending order we could have just ignored the third parameter, which we did in this program. If set true, then the iterable would be sorted in reverse (descending) order, by default it is set as false.
Python3
# Python code to sort the tuples using second element# of sublist Function to sort using sorted()def Sort(sub_li): # reverse = None (Sorts in Ascending order) # key is set to sort using second element of # sublist lambda has been used return(sorted(sub_li, key=lambda x: x[1]))# Input listsub_li = [['rishav', 10], ['akash', 5], ['ram', 20], ['gaurav', 15]]# Printing resultant listprint(Sort(sub_li)) |
[['akash', 5], ['rishav', 10], ['gaurav', 15], ['ram', 20]]
Time complexity: O(n log n) where ‘n’ is the number of elements in the list. This is because the sorting is done using the built-in Python function sorted() which has a time complexity of O(n log n) in the average and worst case.
Auxiliary space: O(n) because a new list is created and returned by the sorted() function which has the same number of elements as the input list.
Method 4: Use orderedDIct
Python3
from collections import OrderedDictdef Sort(sub_li): # create an ordered dictionary sub_li_dict = OrderedDict() for i in sub_li: sub_li_dict[i[1]] = i # sorting the dictionary by key sorted_dict = sorted(sub_li_dict.items()) # extracting the values from the sorted dictionary sort_sub_li = [value for key, value in sorted_dict] return sort_sub_li # Driver Codesub_li =[['rishav', 10], ['akash', 5], ['ram', 20], ['gaurav', 15]]print(Sort(sub_li)) |
[['akash', 5], ['rishav', 10], ['gaurav', 15], ['ram', 20]]
The above approach implements an ordered dictionary that stores the sublist element as key-value pairs. The dictionary is then sorted by the keys and the list is returned with the elements sorted according to the second element in the sublist.
Time complexity: O(n log n), where n is the number of sublists in the input list.
Auxiliary space: O(n), where n is the number of sublists in the input list
Method 4: Using itemgetter() function from the operator module:
- We first import the itemgetter() function from the operator module. This function is used to retrieve the elements of a tuple based on their index.
- We define the sort_tuples() function that takes a list of sublists as input.
- Inside the function, we use the key parameter of the sorted() function to specify the sorting criteria. We pass itemgetter(1) as the key, which returns a function that retrieves the second element of a tuple.
- Finally, we call the sorted() function with the input list sub_li and return the sorted list.
Python3
from operator import itemgetterdef sort_tuples(sub_li): # itemgetter(1) returns a function that can be used to retrieve the # second element of a tuple (i.e., the element at index 1) # this function is used as the key for sorting the sublists return sorted(sub_li, key=itemgetter(1))# Input listsub_li = [['rishav', 10], ['akash', 5], ['ram', 20], ['gaurav', 15]]# Printing resultant listprint(sort_tuples(sub_li)) |
[['akash', 5], ['rishav', 10], ['gaurav', 15], ['ram', 20]]
Time complexity: O(n log n), where n is the number of sublists in the input list. This is because the sorted() function uses a variant of the Merge Sort algorithm, which has a worst-case time complexity of O(n log n).
Auxiliary space: O(n), where n is the number of sublists in the input list.
Method 5: Using numpy’s argsort():
Note: install numpy module using command “pip install numpy”
Algorithm:
- Create an input list of sublists.
- Convert the list to a NumPy array.
- Extract the second column of the array and convert it to integers.
- Sort the array based on the second column using the argsort() method.
- Convert the sorted array back to a list.
- Print the sorted list.
Python3
import numpy as np# Define the input listsub_li = [['rishav', 10], ['akash', 5], ['ram', 20], ['gaurav', 15]]# Converting the list to a NumPy arraysub_arr = np.array(sub_li)# Extracting the second column and convert it to integersvalues = sub_arr[:, 1].astype(int)# Sort the array by the second column (index 1)sorted_arr = sub_arr[values.argsort()]# Converting the sorted array back to a listsorted_li = sorted_arr.tolist()# Printing sorted listprint(sorted_li) |
Output:
[['akash', '5'], ['rishav', '10'], ['gaurav', '15'], ['ram', '20']]
Time complexity: O(N log N), where n is the number of sublists in the input list. The sorting operation is the most time-consuming operation and has a time complexity of O(n log n).
Auxiliary Space: O(N), where n is the number of sublists in the input list. The space complexity is dominated by the NumPy array created from the input list.
Method#6: Using the Recursive method.
- Define a `merge()` function that takes in two lists (`left` and `right`) and merges them while sorting based on the second element of the sublists.
- In the `merge()` function, initialize an empty list `result` to store the merged and sorted sublists.
- Initialize two pointers `i` and `j` to 0, representing the current index in `left` and `right` respectively.
- Compare the second element of the sublists at `left[i]` and `right[j]`.
- If the second element of `left[i]` is less than or equal to the second element of `right[j]`, append `left[i]` to `result` and increment `i`.
- Otherwise, append `right[j]` to `result` and increment `j`.
- Repeat steps 4-6 until either `left` or `right` is exhausted.
- Append the remaining elements in `left` and `right` to `result` if any.
- Return `result` as the merged and sorted list.
Python3
def merge(left, right): # Empty list to store merge and sorted list result = [] i = 0 j = 0 while i < len(left) and j < len(right): if left[i][1] <= right[j][1]: result.append(left[i]) i += 1 else: result.append(right[j]) j += 1 result.extend(left[i:]) result.extend(right[j:]) return result# Recursive function to sort sub listdef sort_recursive(sub_li): if len(sub_li) <= 1: return sub_li mid = len(sub_li) // 2 left = sub_li[:mid] right = sub_li[mid:] left = sort_recursive(left) right = sort_recursive(right) return merge(left, right)# Input sub listsub_li = [['rishav', 10], ['akash', 5], ['ram', 20], ['gaurav', 15]]# Calling function and printing sub listprint(sort_recursive(sub_li)) |
[['akash', 5], ['rishav', 10], ['gaurav', 15], ['ram', 20]]
Time Complexity: O(N logN)
The time complexity of the merge sort algorithm is O(n log n), where n is the length of the input list `sub_li`. This is because the list is recursively divided into halves until the base case of a single element or an empty list is reached, and then merged back up the recursion tree while sorting, taking log n levels. At each level, the merging operation takes O(n) time.
Auxiliary Space: O(1)
The space complexity of the merge sort algorithm is O(n), where n is the length of the input list `sub_li`. This is because additional space is required to store the merged and sorted sublists during the merging operation. However, this implementation uses an in-place merge operation that does not require additional space, so the space complexity is effectively O(1).
