Python return elif statement

Question

Im trying to check is the number passed is a valid perfect square by

return True if(math.sqrt(num)%2 == 0 or \
 (math.sqrt(num)+1) % 2 == 0 ) else False

I want to know if I could write this better by breaking the if statement into two parts like

return True if(math.sqrt(num)%2 == 0) 
 elif ((math.sqrt(num)+1) % 2 == 0 ) 
 else False

Could someone help me with how to use elif here or if there is a better approach.

Thank you.

• Andrew Li: Don't use inline if statements if they are going to be very complicated. Use full if statements with blocks...
• Carcigenicate: Can't you just write return (math.sqrt(num)%2 == 0 or (math.sqrt(num)+1) % 2 == 0)?
• Tarun Kolla: Yes i can just write return (math.sqrt(num)%2 == 0 or (math.sqrt(num)+1) % 2 == 0) but i want to break this as I don't have to compute the or case if the first one is true.
• John Coleman: That isn't a reliable test. It would call 1 + math.factorial(25) a perfect square even though it is manifestly not.
• Carcigenicate: @Tarun Run def f(x): print(x); return x; f(True) or f(False);. False won't be printed. or short circuits.

Answer 1

No need to use if statements at all

return math.sqrt(num) % 2 == 0 or (math.sqrt(num)+1) % 2 == 0 

• user10356004: These type of codes are hard to read and are prone to unintended consequences. Although python interpreter doesn't need if, humans do.
• michael.schuett: Sure for some stuff but this is a pretty straightforward and easy to read statement. All the extra syntax in other answers just makes it harder to read.
• Sean Peters: The ternary if statement adds an extra layer of indirection which in my opinion makes it harder to read. If the complexity of the boolean equation becomes too large for a single line (I don't think that's the case here), my first step would be to seperate that equation into variables e.g. a = math.sqrt(num) % 2 == 0 b = (math.sqrt(num)+1) % 2 == 0 return a or b It'd have to be a rather large boolean equation for me to resort to an if, and I would never use a ternary if.
• Stack Overflown: even simpler: return math.sqrt(num) % 1 == 0

Answer 2

Simple one-liner with readability in mind.

def check(num): 
 return True if math.sqrt(num) % 2 == 0 or (math.sqrt(num)+1) % 2 == 0 else False

Result

>>> check(25)
True
>>> check(23)
False

• Prune: You and I differ on "readability". I find this redundant, and thus distracting from readability. I'm not saying you're wrong, just that we differ. When I see something like this, I have to take time to see what the coder intended that could not be handled with the expression itself.
• user10356004: @Prune I think we are not that different. I too look for exceptions in the expression. The reason that I think this is more readable is it clearly states the return values for the values that match the expression, and value that does not match the expression.

Answer 3

First of all, there is no elif in the ternary operator. Second, your elif doesn't specify a return value (one reason it was left out of the ternary expression). Third, your check for the root being an integer is susceptible to round-off error. Instead, try rounding teh root to an integer, square that, and compare against the original value.

Answer 4

Just answering your question on how to use elif and do it in more than a single line by breaking the if statement. I assume you put these in a function. There could be better ways though as pointed out in other answers

def check_sqrt(num):
 if math.sqrt(num)%2 == 0:
 return True
 elif (math.sqrt(num)+1) % 2 == 0:
 return True
 else:
 return False
print (check_sqrt(23)) 
> False 
print (check_sqrt(25))
> True 

Answer 5

You should always write clearer code that everyone could understand. And you should make it a habit.

if math.sqrt(num)%2 == 0:
 return True
if (math.sqrt(num)+1) % 2 == 0:
 return True
return False

Answer 6

A cleaner version with Python 3.8+ would be:

import math
def is_square(number):
 return math.isqrt(number) ** 2 == number if number >= 0 else False

As an example:

is_square(0)
True
is_square(4)
True
is_square(1)
True
is_square(3)
False
is_square(-4)
False
is_square(3.5)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 2, in is_square
TypeError: 'float' object cannot be interpreted as an integer