Let's create a tutorial on finding the indices of the k smallest elements in a list using Python.
Given a list of numbers and an integer k, the goal is to find the indices of the k smallest numbers in the list.
For instance:
List: [8, 2, 4, 5, 3, 7]
For k=3, the k smallest elements are [2, 3, 4]
and their indices are [1, 4, 2]
.
Using sorted()
with a Custom Key:
The sorted()
function can sort elements based on a custom key. We can use the values as keys to sort a list of indices.
def k_smallest_indices(lst, k): # Create a list of indices from 0 to len(lst) - 1 indices = list(range(len(lst))) # Sort the indices based on the values in lst sorted_indices = sorted(indices, key=lambda x: lst[x]) # Return the first k indices return sorted_indices[:k]
Example Usage:
data = [8, 2, 4, 5, 3, 7] k = 3 print(k_smallest_indices(data, k))
This will output:
[1, 4, 2]
Optimized Approach using Heapq:
When k is much smaller than the size of the list, sorting the entire list might be overkill. Python's heapq
module allows for efficient retrieval of the smallest elements.
import heapq def k_smallest_indices_heapq(lst, k): # Use a heap to get k smallest numbers along with their indices # The heap will contain tuples of (value, index) smallest_items = heapq.nsmallest(k, enumerate(lst), key=lambda x: x[1]) # Extract indices from the tuples indices = [item[0] for item in smallest_items] return indices
Usage:
data = [8, 2, 4, 5, 3, 7] k = 3 print(k_smallest_indices_heapq(data, k))
This will also output:
[1, 4, 2]
heapq
approach is more efficient than sorting the entire list.Finding the indices of the k smallest elements in a list is a common operation in many algorithms, especially in selection and ranking problems. Depending on the relative value of k and the size of the list, different methods, such as sorting or heaps, can be employed to solve the problem efficiently.
psycopg2 timeout numpy jsondecoder dojo-1.8 spring-retry android-appcompat cdata formik large-title