Sort the values of first list using second list in Python

Sort the values of first list using second list in Python

Sorting the values of one list using another list as the reference can be efficiently achieved using Python's built-in sorted() function and the key parameter. Let's dive into the tutorial on how to accomplish this.

Step 1: Understand the Problem

Suppose we have two lists:

list1 = ['a', 'b', 'c', 'd', 'e'] list2 = [3, 2, 4, 1, 5]

We want to sort the elements of list1 based on the corresponding values in list2. The desired output would be:

['d', 'b', 'a', 'c', 'e']

Step 2: Use the zip() Function with sorted()

You can combine the two lists using the zip() function and then sort them based on the values of list2. After sorting, you can unpack the zipped pairs to get the sorted list1.

Here's how you can do it:

sorted_pairs = sorted(zip(list1, list2), key=lambda x: x[1]) sorted_list1 = [item[0] for item in sorted_pairs]

Full Code:

Let's put everything together:

list1 = ['a', 'b', 'c', 'd', 'e'] list2 = [3, 2, 4, 1, 5] # Zip the two lists together and sort by the values of list2 sorted_pairs = sorted(zip(list1, list2), key=lambda x: x[1]) # Extract the sorted values of list1 sorted_list1 = [item[0] for item in sorted_pairs] print(sorted_list1)

Output:

['d', 'b', 'a', 'c', 'e']

Recap:

To sort the values of one list using another list:

  1. Use the zip() function to combine the two lists into pairs.
  2. Use the sorted() function with a custom key function to sort these pairs based on the reference list.
  3. Extract the sorted values of the first list using a list comprehension.

This method allows you to efficiently sort one list based on the values of another.

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