**π‘ Problem Formulation:** When computing the average salary from a set of values, it can be more representative to exclude the highest (maximum) and lowest (minimum) salaries to prevent outliers from skewing the data. Consider a list of salary amounts: `[45000, 52000, 60000, 75000, 80000]`

. The task is to calculate the average salary after excluding the minimum (`45000`

) and the maximum (`80000`

) salaries. The desired output for this set would be `62333.33`

(i.e., the average of `[52000, 60000, 75000]`

).

## Method 1: Basic Iterative Approach

This first method uses a straightforward iterative approach. After sorting the salary list, it calculates the average by summing the salaries, excluding the first and last elements which are the minimum and maximum values respectively, and then dividing by the count of remaining elements.

Here’s an example:

salaries = [45000, 52000, 60000, 75000, 80000] sorted_salaries = sorted(salaries) average_salary = sum(sorted_salaries[1:-1]) / (len(sorted_salaries) - 2) print(average_salary)

Output: `62333.333333333336`

In this snippet, the `sorted()`

function sorts the list of salaries, and the sum is taken from the second to the second-to-last item in this sorted list. The average is then calculated by dividing by the length of the reduced list. It’s a practical approach, but not very efficient for very large datasets where sorting could be costly.

## Method 2: Using Min and Max Functions

The second method emphasizes efficiency by avoiding the full sort of the salary list. It calculates the sum of all salaries and then subtracts the minimum and maximum salary values directly, dividing by the length of the list minus 2 for average calculation.

Here’s an example:

salaries = [45000, 52000, 60000, 75000, 80000] total_sum = sum(salaries) average_salary = (total_sum - min(salaries) - max(salaries)) / (len(salaries) - 2) print(average_salary)

Output: `62333.333333333336`

Here, `min()`

and `max()`

functions find the smallest and largest values in the list of salaries. This method avoids sorting and is therefore more efficient for larger datasets than the previous method, especially if salaries are not required to be sorted for any other purpose.

## Method 3: Using Heapq to Find Min and Max Efficiently

The third method is similar to the second but uses the `heapq`

module for obtaining the smallest and largest values more efficiently, which is more performant on very large datasets.

Here’s an example:

import heapq salaries = [45000, 52000, 60000, 75000, 80000] total_sum = sum(salaries) min_salary = heapq.nsmallest(1, salaries)[0] max_salary = heapq.nlargest(1, salaries)[0] average_salary = (total_sum - min_salary - max_salary) / (len(salaries) - 2) print(average_salary)

Output: `62333.333333333336`

By utilizing `heapq.nsmallest()`

and `heapq.nlargest()`

, this code efficiently retrieves the minimum and maximum salaries. It reduces the time complexity compared to sorting or scanning the entire list, particularly when the dataset is large.

## Method 4: Pandas DataFrame Approach

For those working with larger datasets, especially those stored as a DataFrame in pandas, this method provides a robust option. It uses the capabilities of pandas to filter out the min and max before calculating the average.

Here’s an example:

import pandas as pd data = {'salaries': [45000, 52000, 60000, 75000, 80000]} df = pd.DataFrame(data) filtered_df = df[(df['salaries'] > df['salaries'].min()) & (df['salaries'] < df['salaries'].max())] average_salary = filtered_df['salaries'].mean() print(average_salary)

Output: `62333.333333333336`

The pandas library’s `DataFrame`

allows for quick manipulation of the salaries, filtering out the min and max values. The mean is then easily computed with the `mean()`

function. This method is powerful when dealing with very structured data and provides additional tools for analysis.

## Bonus One-Liner Method 5: Using List Comprehensions

A one-liner solution in Python which succinctly combines list comprehensions with the min and max functions to filter out unwanted salaries and compute the average.

Here’s an example:

salaries = [45000, 52000, 60000, 75000, 80000] average_salary = sum(s for s in salaries if s not in (min(salaries), max(salaries))) / (len(salaries) - 2) print(average_salary)

Output: `62333.333333333336`

This compact solution elegantly filters out the minimum and maximum salaries using a condition within the sum’s list comprehension. While concise, it will not perform optimally on large datasets due to evaluating `min`

and `max`

for each element.

## Summary/Discussion

**Method 1:**Basic Iterative Approach. Simple and easy to understand. Not the most efficient due to sorting requirement.**Method 2:**Using Min and Max Functions. More efficient than Method 1. It avoids sorting but still has to scan the entire list twice to find min and max salaries.**Method 3:**Using Heapq. Optimized for larger datasets. Uses a heap to efficiently find the smallest and largest values without scanning the entire list.**Method 4:**Pandas DataFrame Approach. Best for structured data in tables. It integrates well with other data analysis processes and tools, although it may be overkill for simple tasks or small datasets.**Method 5:**Bonus One-Liner. It provides a Pythonic and concise approach. However, it’s not suitable for very large datasets due to the inefficiency of checking min and max for each element.