Logical or operator not behaving as expected [duplicate]

Question

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• How to test multiple variables against a value? 23 answers

def Forest(Health,Hunger):
    print'You wake up in the middle of the forest'
    Inventory = 'Inventory: '
    Squirrel =  'Squirrel'
    while True:
        Choice1 = raw_input('You...\n')
        if Choice1 == 'Life' or 'life':
            print('Health: '+str(Health))
            print('Hunger: '+str(Hunger))
        elif Choice1 == 'Look' or 'look':
            print 'You see many trees, and what looks like an edible dead Squirrel, \na waterfall to the north and a village to the south.'
        elif Choice1 == 'Pickup' or 'pickup':
            p1 = raw_input('Pickup what?\n')
            if p1 == Squirrel:
                if Inventory == 'Inventory: ':
                    print'You picked up a Squirrel!'
                    Inventory = Inventory + Squirrel + ', '
                elif Inventory == 'Inventory: Squirrel, ':
                        print'You already picked that up!'
            else:
                print"You can't find a "+str(p1)+"."
        elif Choice1 == 'Inventory' or 'inventory':
            print Inventory

I am trying to make it so when it says You... you can type either Life, Pickup, Look, or Inventory. I have way more code on this program I am just showing you a portion. But every time I run it, it always shows the "Life" portion even if you type "Pickup" or "Look" or "Inventory". Please help! Thanks, John

EDIT: I think it's just a spacing issue but I am not sure it was running fine earlier...

Answer 1

You are misunderstanding the or expression. Use this instead:

if Choice1.lower() == 'life':

or, if you must test against multiple options, use in:

if Choice1 in ('Life', 'life'):

or, if you must use or then use it like this:

if Choice1 == 'Life' or Choice1 == 'life':

and expand this to your other Choice1 tests.

Choice1 == 'Life' or 'life' is interpreted as (Choice1 == 'Life') or ('life'), with the latter part always being True. Even if it was interpreted as Choice1 == ('Life' or 'life') then the latter part would evaluate to 'Life' only (it being True as far as boolean tests go), so you'd be testing if Choice1 == 'Life' instead, and setting Choice to 'life' would never make the test pass.

Answer 2

You have:

    if Choice1 == 'Life' or 'life':

Which is actually the equivalent of:

    if (Choice1 == 'Life') or 'life':

A non-empty/non-zero string ('life') will always be treated as true, hence why you end up there.

You either want:

    if Choice1 == 'Life' or Choice1 == 'life':

or:

    if Choice1.lower() == 'life':

Answer 3

Use in:

elif Choice1 in ('Pickup', 'pickup'):

Alternatively, you can use regular expressions:

import re

elif re.match("[Pp]ickup", Choice1):

Separately, I'd use a set for your inventory:

Inventory = set()
Squirrel =  'Squirrel'
while True:
...
        if p1 == Squirrel:
            if not Inventory:
                print'You picked up a Squirrel!'
                Inventory.add(Squirrel)
            elif Squirrel in Inventory:
                print'You already picked that up!'