Range as dictionary key in Python

Question

So, I had an idea that I could use a range of numbers as a key for a single value in a dictionary.

I wrote the code bellow, but I cannot get it to work. Is it even possible?

    stealth_roll = randint(1, 20)     # select from a dictionary of 4 responses using one of four ranges.     ## not working.     stealth_check = {                     range(1, 6) : 'You are about as stealthy as thunderstorm.',                     range(6, 11) : 'You tip-toe through the crowd of walkers, while loudly calling them names.',                     range(11, 16) : 'You are quiet, and deliberate, but still you smell.',                     range(16, 20) : 'You move like a ninja, but attracting a handful of walkers was inevitable.'                     }      print stealth_check[stealth_roll] 

Answer 1

It is possible on Python 3 — and on Python 2 if you use xrange instead of range:

stealth_check = {                 xrange(1, 6) : 'You are about as stealthy as thunderstorm.', #...                 } 

However, the way you're trying to use it it won't work. You could iterate over the keys, like this:

for key in stealth_check:     if stealth_roll in key:         print stealth_check[key]         break 

Performance of this isn't nice (O(n)) but if it's a small dictionary like you showed it's okay. If you actually want to do that, I'd subclass dict to work like that automatically:

class RangeDict(dict):     def __getitem__(self, item):         if not isinstance(item, range): # or xrange in Python 2             for key in self:                 if item in key:                     return self[key]             raise KeyError(item)         else:             return super().__getitem__(item) # or super(RangeDict, self) for Python 2  stealth_check = RangeDict({range(1,6): 'thunderstorm', range(6,11): 'tip-toe'}) stealth_roll = 8 print(stealth_check[stealth_roll]) # prints 'tip-toe'