How to use or operator in python

Question

In my project I have used or but in some cases it works and in some cases it fails me.. I cant understand Why?

if (a or b) is not True:
    # This works for me


if (a or b) == "Django" and (c or d) == "Pyramid":
    # This fails me everytime

Answer 1

a or b evaluates to a if a evaluates to True. If a evaluates to False, a or b evaluates to b.

For instance:

[] or 'a' #evaluates to 'a'
'a' or None #evaluates to 'a'
[] or None #evaluates to None
None or [] #evaluates to []

Your (a or b) == 'Django' should be a == 'Django' or b == 'Django'.

('Django' or 'Nomatterwhat') == 'Django' #True
([] or 'Django') == 'Django' #True
('Mono' or 'Django') == 'Django' #False

---

Having said the theoretical part, try running this little program, to see how or and and work (especially the lazy evaluation part):

def x():
    print('X')
    return False

def y():
    print('Y')
    return True

print ('-' * 20)
x() or y()
print ('-' * 20)
y() or x()
print ('-' * 20)
x() and y()
print ('-' * 20)
y() and x()

Answer 2

if a == 'Django' or b == 'Django':

Answer 3

Your parentheses are in the wrong place. (a or b) will always return the first of a or b which evaluates to True, so:

>>> a = "Django"
>>> b = "Pyramid"
>>> (a or b)
"Django"
>>> (a or b) == "Django"
True

but

>>> (b or a)
"Pyramid"
>>> (b or a) == "Django"
False

What you actually want to write is:

>>> a == "Django" or b == "Django"
True

However, an alternative way of writing it which may be better in many cases is:

if "Django" in (a, b) or "Pyramid" in (a, b):
    ...

Or, if there are many options:

if set([a, b, c, ...]) & set(["Django", "Pyramid", ...]):
    ...