The ternary operator returns `True` instead of the given value

Question

I use the ternary operator to define short conditional variables. I was wondering when the expression returns True instead of the value in the expression.

>>> digits = '123456' >>> conv_d = digits != None if int(digits) else None >>> conv_d >>> True >>> int(digits) >>> 123456 

Please explain to me how this happens? What is the logical difference between a ternary operator and a regular conditional expression in Python?

Answer 1

int(digits) == 123456 , which is a true ish value. So conv_d = digits != None . Because digits not None , conv_d set to true.

You probably wanted to:

 conv_d = int(digits) if digits is not None else None 

Remember that a string containing something other than a number throws an exception! If you prefer 0 or None for these values, write a small function:

 def toint(s): try: return int(s) except (ValueError, TypeError): return None # or 0 

Answer 2

The Python conditional statement is not in the same order as in other languages. And you should never compare equality with None unless you are sure what you need.

 conv_d = int(digits) if digits is not None else None 

Answer 3

Alternatives to triple operators:

conv_d = digits! = None and int (digits) or None # short-circuiting; just wrong in this context as it does not work for numbers = "0" - see comment below

or

conv_d = int (digits) if the digits are not None else None

The second expression is clearer and therefore preferable.

Answer 4

Pay attention to how keywords are placed. "X, if Y is still Z." "Y" is the part that follows the "if", so this is a condition. "X if Y" is a perfectly valid (albeit less common) English construction, meaning the same as "if Y, X", so X is an expression evaluated when Y is executed. Z is an expression evaluated when Y is not satisfied, similarly.