brand new to python - silly/obvious question

Question

I am brand new to coding and trying to learn Python. I cannot work out how to code an elif statement such that if the variable is an integer there is a certain output, but if it is not an integer then something else is printed. I keep getting an error. I am trying to write a program that can analyse the variable "x" and if x is a number it will print saying it is below 2 or 2 or more, but if x is anything else including a string, it will print "something else"

x = 2
if x < 2:
    print("Below 2")
elif x >= 2:
    print("Two or more")
else:
    print("Something else")

x = world

if x < 2:
    print("Below 2")
elif x >= 2:
    print("Two or more")
else:
    print("Something else")

try [`isinstance`](https://docs.python.org/3/library/functions.html#isinstance) — C.Nivs, Feb 26 '21 at 17:17
"_I keep getting an error_" You should tell us what the error is. — takendarkk, Feb 26 '21 at 17:19
This code should throw loads of syntax errors that will point you in the right direction. Additionally, have a look at the `isinstance` function for determining types. — S3DEV, Feb 26 '21 at 17:20
Does this answer your question? [Checking whether a variable is an integer or not](https://stackoverflow.com/questions/3501382/checking-whether-a-variable-is-an-integer-or-not) — wjandrea, Feb 26 '21 at 17:31
Welcome to SO! Please take the [tour] and read [ask]. I suggested an existing thread that should answer your question. If it doesn't help, you'll need to [edit] your question, make the title more descriptive, and provide a [mre] including the full error message. — wjandrea, Feb 26 '21 at 17:32
@S3DEV Syntax errors? No, there's only `NameError: name 'world' is not defined` — wjandrea, Feb 26 '21 at 17:33
@wjandrea - Comparing a string to int using '>' should throw an error as well. (Does on my system) — S3DEV, Feb 26 '21 at 18:01
@S3DEV True, though it doesn't get to that point with OP's code. And that's a `TypeError`. — wjandrea, Feb 26 '21 at 18:19

score 1 · Accepted Answer · answered Feb 26 '21 at 17:21

1

Here are a couple of different ways you might use to determine if x is an int:

x = "a"
# also try with x = 5

# option 1
if type(x) == int:
    print("int")
else:
    print("not int")

# option 2
if isinstance(x, int):
    print("int")
else:
    print("not int")

answered Feb 26 '21 at 17:21

Adam Oellermann

303
1
7

that is very helpful, thank you very much. I think I have now solved it based on your answer. See below, this works. is this most efficient way? `x = "word" if type(x) == int: if x < 2: print("Below 2") elif x >= 2: print("Two or more") else: print ("something else")` – newbie-star Feb 26 '21 at 17:30
1

It's bad practice to use `type()` cause it'll ignore subclasses. Prefer `isinstance()`. – wjandrea Feb 26 '21 at 17:36
@wjandrea - please could you write out how I would change my code to remove `type` and use `isinstance`? – newbie-star Feb 26 '21 at 17:42
1

You're probably OK with type if you're just dealing with "integers" vs "non-integers" - it seems rather unlikely that you will be subclassing int in your code! However, you would just change if type(x) == int: to if isinstance(x, int): - a straight line for line swap. – Adam Oellermann Feb 26 '21 at 17:46

brand new to python - silly/obvious question

1 Answers1