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# Python program to find N largest and smallest elements from the list

**Python heapq.nlargest() and heapq.nsmallest() functions Examples**: Here, we are going to learn **how to find N largest and smallest elements?**

Submitted by Yash Khandelwal, on April 13, 2019

Here, we learn how to find out the N largest and smallest elements from the list? Where, list and N given by the user, N may be any value but less than the list length.

Description:

There are two ways,

**1. By defining a function:**

Procedure:

- Define the function name largest_ele and smallest_ele.
- Pass two arguments in a function (l,n) : l is the list and n is the number of elements.
- Run the for loop n times
- In the loop find the maximum of the given list and append it to another list
- And after appending to another list remove the maximum element from the list

**By the inbuilt module heapq module**

If you are looking for the N smallest or largest items and N is small compared to the overall size of the collection, these functions provide superior performance.

- Import the heapq module
- Give the list
- Now use the function heapq.nlargest(n,l) and heapq.nsmallest(n,l) from the module to find the largest and the smallest numbers.

**Python code:**

# N largest and smallest element in a list # by function and by the help of heapq module #function to find n largest element def largest_ele(l,n): s=[] for i in range(n): s.append(max(l)) #append max of list in a new list l.remove(max(l)) #remove max of list from the list print('by largest_ele function: ',s) #function to find n largest element def smallest_ele(m,n): t=[] for i in range(n): t.append(min(m))#append min of list in a new list m.remove(min(m))#remove min of list from the list print('by smallest_ele function: ',t) l=[2,4,6,8,10] m=[0,1,2,3,4,5,6] n=2 largest_ele(l,n) smallest_ele(m,n) # using the inbuilt module function # heapq.nlargest and heapq.nsmallest import heapq nums = [1, 8, 2, 23, 7, -4, 18, 23, 42, 37, 2] print('BY heapq.nlargest: ',heapq.nlargest(3, nums)) # Prints [42, 37, 23] print('BY heapq.nsmallest: ',heapq.nsmallest(3, nums)) # Prints [-4, 1, 2]

**Output**

by largest_ele function: [10, 8] by smallest_ele function: [0, 1] BY heapq.nlargest: [42, 37, 23] BY heapq.nsmallest: [-4, 1, 2]

**Note:** The **nlargest() and nsmallest() functions** are most appropriate if you are trying to find a relatively small number of items. If you are simply trying to find the single smallest or largest item (N=1), it is faster to use min() and max().

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