How to sort the Python list with an index

How to sort the Python list with an index using Ready-to-use function?

Ready-to-use Python function to sort the list with an index:

def sort_list_with_index(li1, idx):
    # Sort a List with given index
    tmp = li1[:] #clone list

    # verify list and index size
    if len(li1) != len(idx):
        messagebox.showinfo('Message', "The sorting list and index size doesn't match!")
    # sort list based on index number
    for i in range(0, len(idx)):
        tmp[i] = li1[idx[i]]
    return tmp

Write your main code as a sample below,

print(sort_list_with_index(['Mon', 'Fri', 'Wed'], [0, 2, 1]))

The output of the code is,

['Mon', 'Wed', 'Fri']
How does the Python function work?

This function sorts a given list (li1) based on a specified order (idx) by creating a copy of the list and rearranging the elements in the copy according to the order specified by the index list.

This Python function, sort_list_with_index, takes two parameters: li1, which is a list to be sorted, and idx, which is a list of indices that specify the order in which li1 should be sorted. Here’s how it works:

  1. The function starts by creating a copy of the input list using the slice operator (tmp = li1[:]). This is done to avoid modifying the original list while sorting it.
  2. The function then checks if the length of the input list li1 is equal to the length of the index list idx. If they are not the same length, the function displays an error message and returns None.
  3. If the length of li1 is equal to the length of idx, the function proceeds to sort li1 based on the order specified by the indices in idx.
  4. To sort the list, the function uses a for loop that iterates over the length of idx. In each iteration, the function sets the value of the i-th element in the copy of li1 (tmp[i]) to the value of the element in li1 at the index specified by idx[i].
  5. After the loop has finished iterating, the function returns the sorted list tmp.

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