Don't understand why 1 not print

Question

#Not print element from list

lst = ["#",1,2,"4","a",3]
num = []
for n in lst:
if type(n) == int:
try:
num = num.append(n)
except:
print(n)

#number 1 not print. Output only 2 and 3 from list "lst". I do not understand why

Answer 1

When n was 1, num was still a list, that's why 1 wasn't printed.

statement "for n in lst" broken in iterations:
n = "#"
if type(n) == int:
~skip~

n = 1
if type(n) == int:
#num = []
num = num.append(n)
#num = None

n = 2
if type(n) == int:
#num = None
num = num.append(n)
#None.append(2)?, error raised.

Answer 2

append is a method from class 'list' and return None.
In case of n = 1 (first integer of lst), then n is append to num but num is now None (because append returns None), so list num doesn't exist anymore. No error is raised so 1 isn't printed.

For the two other integer, because num is None, it produces an error and print n.

So output is 2 and 3.

Answer 3

#Sorry. For more readable must be:

#Not print element from list

lst = ["#",1,2,"4","a",3]
num = []
for n in lst:
if type(n) == int:
print(n)

#but there output is 1 2 and 3

Answer 4

Why this:......num = num.append(n)??
shouldn't it be just ......num.append(n)
also:-
Not sure why your using a try except block.
Look up the builtin function isinstance().