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Python Python Collections (Retired) Dungeon Game Building the Game: Part 1

Sam Dale
Sam Dale
6,136 Points

Using Multiple "== statements" vs "or statements"

Hi there. This is a simple question, but Python has the weird and unique ability to do crazy stuff like 3 < x < 5 checks. So Kenneth has a line:

if monster == door or monster == start or start == door:

The line that follows could do the same thing in a more concise way:

if monster == door == start:

I know functionality-wise it doesn't work and this is an oddity that applies only to Python, but is the second way more Pythonic? Or is the first one more clear? Thanks!

2 Answers

Chris Freeman
MOD
Chris Freeman
Treehouse Moderator 68,064 Points

Your first condition will stop evaluating at the first matching pair since the other matches can not make the result any more True:

monster == door or monster == start or start == door

The second condition

monster == door == start

This is equivalent to

# monster == door and monster == start and start == door  # not correct
monster == door and door == start  # correction by pointed out by @[Jason Anello](https://teamtreehouse.com/jasonanello)

That is, it's True only if they all match. But if you reverse the comparison, it will be True only if they all differ:

>>> start = 2
>>> door = 4
>>> monster = 6
>>> start, door, monster
(2, 4, 6)
>>> start == door == monster
False
>>> start != door != monster
True

EDIT: as pointed out by Jason Anello this is equivalent to

start != door != monster
# is equivalent to
start != door and door != monster

As mentioned further down, adding an additional round-robin case fixes it:

start != door != monster != start
# is equivalent to
start != door and door != monster and monster != start

Note: This is a only a syntax curiosity. The Pythonic way is to spell it out with the long form.

start != door and door != monster and monster != start
# OR
start == door and door == monster and monster == start

depending on your goals.

Sam Dale
Sam Dale
6,136 Points

Right. I totally overlooked that. Thanks!

Hey Chris,

I think

monster == door == start

would be equivalent to

monster == door and door == start

monster and start wouldn't be compared against each other.

Also, your final example where you reverse the comparisons will also be True if start and monster are equal but different from door.

It's equivalent to

start != door and door != monster

This doesn't restrict start and monster from being equal to each other.

>>> start = 2
>>> door = 3
>>> monster = 2
>>> start != door != monster
True
>>>
Chris Freeman
Chris Freeman
Treehouse Moderator 68,064 Points

Ack! You're right! Thanks Jason. I was too quick with insufficient tests. Adding one more round-robin test fixes it:

>>> start, door, monster = (2, 2, 2)
>>> start, door, monster
(2, 2, 2)
>>> start != door != monster != start
False

>>> start, door, monster = (2, 2, 3)
>>> start, door, monster
(2, 2, 3)
>>> start != door != monster != start
False

>>> start, door, monster = (2, 3, 2)
>>> start, door, monster
(2, 3, 2)
>>> start != door != monster != start
False

>>> start, door, monster = (3, 2, 2)
>>> start, door, monster
(3, 2, 2)
>>> start != door != monster != start
False

>>> start, door, monster = (3, 2, 1)
>>> start, door, monster
(3, 2, 1)
>>> start != door != monster != start
True

I'll amend my answer above

Nathan Tallack
Nathan Tallack
22,159 Points

Just remember, the moment it gets match it will stop evaluating.

So for your first one it will never check your third conditional if the first or second match.

For your second one it will only work if all three match.