How does Python insertion sort work?

Question

Here\'s a Python implementation of insertion sort, I tried to follow the values on paper but once the counting variable i gets bigger than len(s) I don\'t know what to do, h

Answer 1

a recursive implementation

def insert(x, L):
    if [] == L:      return [x]
    elif x <= L[0]:  return [x] + L
    else:            return [L[0]] + insert(x,L[1:])

def insertion_sort(L):
    if [] == L:  return []
    else:        return insert(L[0], insertion_sort(L[1:]))

# test
import random

L = [random.randint(1,50) for _ in range(10)]

print L
print insertion_sort(L)

Answer 2

Consider [3, 2, 1]

The loop starts with 3. Since it is the first item in the list there is nothing else to do.

[3, 2, 1]

The next item is 2. It compares 2 to 3 and since 2 is less than 3 it swaps them, first putting 3 in the second position and then placing 2 in the first position.

[2, 3, 1]

The last item is 1. Since 1 is less than 3 it moves 3 over.

[2, 3, 3]

Since 1 is less than 2 it swaps moves 2 over.

[2, 2, 3]

Then it inserts 1 at the beginning.

[1, 2, 3]

Answer 3

def sort_numbers(list):
    for i in range(1, len(list)):
        val = list[i]
        j = i - 1
        while (j >= 0) and (list[j] > val):
            list[j+1] = list[j]
            j = j - 1
        list[j+1] = val

n = int(input("Enter the no. of elements"))
list = []
for i in range(0,n):
  t = int(input())
  list.append(t)

sort_numbers(list)

print list