Given two lists, sort the values of one list using the second list.
Examples:
Input : list1 = [“a”, “b”, “c”, “d”, “e”, “f”, “g”, “h”, “i”]
list2 = [ 0, 1, 1, 0, 1, 2, 2, 0, 1]
Output : [‘a’, ‘d’, ‘h’, ‘b’, ‘c’, ‘e’, ‘i’, ‘f’, ‘g’]Input : list1 = [“g”, “e”, “e”, “k”, “s”, “f”, “o”, “r”, “g”, “e”, “e”, “k”, “s”]
list2 = [ 0, 1, 1, 0, 1, 2, 2, 0, 1]
Output : [‘g’, ‘k’, ‘r’, ‘e’, ‘e’, ‘g’, ‘s’, ‘f’, ‘o’]
Approach :
- Zip the two lists.
- Create a new, sorted list based on the zip using sorted().
- Using a list comprehension extract the first elements of each pair from the sorted, zipped list.
Concept :
The purpose of zip() is to map a similar index of multiple containers so that they can be used just using as a single entity.
Below is the implementation of the above approach:
Python
# Python program to sort# one list using# the other listdef sort_list(list1, list2): zipped_pairs = zip(list2, list1) z = [x for _, x in sorted(zipped_pairs)] return z# driver codex = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]y = [0, 1, 1, 0, 1, 2, 2, 0, 1]print(sort_list(x, y))x = ["g", "e", "e", "k", "s", "f", "o", "r", "g", "e", "e", "k", "s"]y = [0, 1, 1, 0, 1, 2, 2, 0, 1]print(sort_list(x, y)) |
['a', 'd', 'h', 'b', 'c', 'e', 'i', 'f', 'g'] ['g', 'k', 'r', 'e', 'e', 'g', 's', 'f', 'o']
Time Complexity: O(nlogn)
Auxiliary Space: O(n)
In the above code, we have two lists, the first list is being sorted with respect to the values of the second list.
y = [ 0, 1, 1, 0, 1, 2, 2, 0, 1]
Here first the lowest value is checked. Like in this list, 0 is the lowest, so starting from the first index, 0 is the lowest and it is at index 0. So the value of index 0 is stored at index 0 in the first list. Similarly, 0 is again found at index 3 and so the value of index 3 in the first list is index 1. The same goes until the list is not completed.
Approach 2: By using Dictionary, list comprehension, lambda function
Python3
def sorting_of_element(list1, list2): # initializing blank dictionary f_1 = {} # initializing blank list final_list = [] # Addition of two list in one dictionary f_1 = {list1[i]: list2[i] for i in range(len(list2))} # sorting of dictionary based on value f_lst = {k: v for k, v in sorted(f_1.items(), key=lambda item: item[1])} # Element addition in the list for i in f_lst.keys(): final_list.append(i) return final_listlist1 = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]list2 = [0, 1, 1, 0, 1, 2, 2, 0, 1]list3 = sorting_of_element(list1, list2)print(list3) |
['a', 'd', 'h', 'b', 'c', 'e', 'i', 'f', 'g']
Time Complexity: O(nlogn)
Auxiliary Space: O(n)
Approach 3: Using sort(),list() and set() methods
Python3
# Python program to sort values of first list based on second listlist1 = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]list2 = [0, 1, 1, 0, 1, 2, 2, 0, 1]a = list(set(list2))a.sort()res = []for i in a: for j in range(0, len(list2)): if(list2[j] == i): res.append(list1[j])print(res) |
['a', 'd', 'h', 'b', 'c', 'e', 'i', 'f', 'g']
Time Complexity: O(nlogn)
Auxiliary Space: O(n)
New Approach : Using collections.OrderedDict()
Using OrderedDict() to store the mapping of list1 and list2, sorted() is used to sort the OrderedDict() on the basis of values from the second list and finally list() is used to get the result.
Python3
# Python program to sort values of first list based on second listfrom collections import OrderedDictlist1 = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]list2 = [0, 1, 1, 0, 1, 2, 2, 0, 1]d = OrderedDict(zip(list1, list2))res = list(OrderedDict(sorted(d.items(), key=lambda x: x[1])))print(res) |
['a', 'd', 'h', 'b', 'c', 'e', 'i', 'f', 'g']
Time Complexity: O(nlogn)
Auxiliary Space: O(n)
Approach : Using the numpy.argsort() function with fancy indexing
note: install numpy module using command “pip install numpy”
The given methods are quite efficient, but we can use the numpy.argsort() function with fancy indexing to solve the problem more efficiently. Here is how we can implement it:
Algorithm:
Import numpy library.
Define a function sort_list that takes two lists list1 and list2 as input.
Get the indices that would sort the list2 using np.argsort(list2) and store it in idx.
Convert list1 to a numpy array, then use idx to sort it and return it.
Python3
import numpy as np# Define a function that takes two lists as input, sorts the first list based on# the order of the second list, and returns the sorted list as a numpy arraydef sort_list(list1, list2): # Use numpy's argsort function to get the indices that would sort the second list idx = np.argsort(list2) # Index the first list using the sorted indices and return the result as a numpy array return np.array(list1)[idx]# Define two lists of strings and integers to be used as inputs for the sort_list functionx = ["a", "b", "c", "d", "e", "f", "g", "h", "i"]y = [0, 1, 1, 0, 1, 2, 2, 0, 1]# Call the sort_list function with the two lists and print the resultprint(sort_list(x, y))# Define another two lists of strings and integers to be used as inputs for the sort_list functionx = ["g", "e", "e", "k", "s", "f", "o", "r", "g", "e", "e", "k", "s"]y = [0, 1, 1, 0, 1, 2, 2, 0, 1]# Call the sort_list function with the two lists and print the resultprint(sort_list(x, y)) |
Output:
[‘a’ ‘d’ ‘h’ ‘b’ ‘c’ ‘e’ ‘i’ ‘f’ ‘g’]
[‘g’ ‘k’ ‘r’ ‘e’ ‘e’ ‘s’ ‘g’ ‘f’ ‘o’]
Time Complexity: O(nlogn) due to sorting operation using np.argsort().
Auxiliary Space: O(n) as we create a numpy array with the same size as list1.
This implementation is efficient as it leverages the optimized numpy library for sorting the array using argsort() function.
