Python - Why is 'b' inserted in array and 1 isn't?

Question

try:
s = ['a','c']
n = [0, 2]
s[1:1] = 'b'
n[1:1] = 1
except:
pass
print (s[1], n[1])

Answer 1

b is a string and so an iterable but 1 is an integer and no iterable.
Sushi operator needs iterable.

Answer 2

I understood))
you can change the code like this:
n[1:1] = [1]

use this code to check the object:

from collections.abc import Iterable

var_1 = "b"
var_2 = 1

if isinstance(var_1, Iterable):
print(f"var_1 (value: {var_1}) is iterable")
else:
print(f"var_1 (value: {var_1}) is no iterable")

if isinstance(var_2, Iterable):
print(f"var_2 (value: {var_2}) is iterable")
else:
print(f"var_2 (value: {var_2}) is no iterable")

Answer 3

Interest question!
type(s) and type(n) returned list...
I don't know answer.