increment int within an inline if in python

Question

I have a value that starts as None but should become an int that gets incremented in a loop. I tried using an inline if to check if it is None and increment otherwise. However the inline if statement throws:

x = None
for _ in range(5):
    if x is None:
        x = 1
    else:
        x +=1
print(x)

x= None
for _ in range(5):
    x = 1 if x is None else +=1 #SyntaxError: invalid syntax
    x +=1 if x is not None else 1 #TypeError unsupported operand type(s) for +=: 'NoneType' and 'int'
print(x)

It works if you don't use the += and operator and x explicitly: x = 1 if x is None else x+1. But I was wandering if or how it is possible to use += in an inline if.

What you call "inline if" is a *conditional expression*. A `+=` statement is not an expression. — khelwood, Oct 22 '20 at 12:26
@UnholySheep the `x` is an attribute of an object from a third party module. — Finn, Oct 22 '20 at 12:43
@ChristopherPeisert is does! But i also needed the information that it is not an actual "inline if". — Finn, Oct 22 '20 at 12:51

score 0 · Answer 1 · answered Oct 22 '20 at 12:27

You are using the inline code properly with x = 1 if x is None else x+1, you can't use += in inline. To understand, let's add some parenthesis without affecting code:

x = None
for _ in range(5):
    x = (1 if x is None else x+1)
    # same as
    if x is None:
        x = 1
    else:
        x = x + 1

increment int within an inline if in python

1 Answers1