Lambda Sorted won't sort - class

Question

I'm trying create a lambda Sorted by age, but it's not sorting, would appreciate if someone spots the error. Thank you
I made a class and a menu to input employees with name and age, objective is to print the list sorted by age
Here I show the code I have so far
class employee:
def __init__(self, name, age):
self.name = name
self.age = age
i = 0
list = []
def show():
k = 0
while k < len(list):
list2 =[
{'Name': list[k].name,
'Age' : list[k].age}]
_sorted = sorted(list2, key=lambda x: x['Age'], reverse=True)
print(_sorted)
k += 1
while i == 0:
print("Menu")
print("1. Register")
print("2. Show")
print("3. Exit")
option = int(input())
if option == 1:
print("Register")
n = str(input("Ingrese el nombre del empleado: "))
e = int(input("Ingrese la edad del empleado: "))
emp = employee(n, e)
list.append(emp)
print("Empleado guardado con éxito!")
elif option == 2:
print("Mostrar")
mostar()
elif option == 3:
exit()
else:
print("Option inválid")

Answer 1

I've made some minor changes to make the program run:
class employee:
def __init__(self, name, age):
self.name = name
self.age = age
def show(lst): # <--- make function with parameter, don't use global values
for emp in sorted(lst, key=lambda x: x.age, reverse=True): # <--- iterate over sorted list
print(emp.name, emp.age) # <-- print name, age
lst = []
while True: # <-- you don't need `i` variable here, just break when you want to exit
print("Menu")
print("1. Register")
print("2. Show")
print("3. Exit")
option = int(input())
if option == 1:
print("Register")
n = str(input("Ingrese el nombre del empleado: "))
e = int(input("Ingrese la edad del empleado: "))
emp = employee(n, e)
lst.append(emp)
print("Empleado guardado con éxito!")
elif option == 2:
print("Mostrar")
show(lst) # <--- use show(), not mostrar()
elif option == 3:
break # <--- to exit, break
else:
print("Option inválid")