Callable as the default argument to dict.get without it being called if the key exists

Question

I am trying to provide a function as the default argument for the dictionary\'s get function, like this

def run():
   print \"RUNNING\"

test = {\'store\':1         


        

Answer 1

See the discussion in the answers and comments of dict.get() method returns a pointer. You have to break it into two steps.

Your options are:

1. Use a defaultdict with the callable if you always want that value as the default, and want to store it in the dict.

2. Use a conditional expression:

   item = test['store'] if 'store' in test else run()
   

3. Use try / except:

   try:
       item = test['store']
   except KeyError:
       item = run()
   

4. Use get:

   item = test.get('store')
   if item is None:
       item = run()
   

And variations on those themes.

glglgl shows a way to subclass defaultdict, you can also just subclass dict for some situations:

def run():
    print "RUNNING"
    return 1

class dict_nokeyerror(dict):
    def __missing__(self, key):
        return run()

test = dict_nokeyerror()

print test['a']
# RUNNING
# 1

Subclassing only really makes sense if you always want the dict to have some nonstandard behavior; if you generally want it to behave like a normal dict and just want a lazy get in one place, use one of my methods 2-4.