>>> d = {'a':1, 'b':2, 'c':3}
>>> d.get('a')
1
>>> d.get('d', 'not found')
'not found'

Sometimes, I find something like this would be useful for lists, using the list’s index in place of the dict’s key:

>>> l = [0,1,2,3]
>>> l.get(2)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: 'list' object has no attribute 'get'

Here’s a quick lambda that accomplishes that:

list_get = lambda l,i,a=False: (len(l) <= i and a) or (len(l) > i and l[i])

Usage:

>>> list_get = lambda l,i,a=False: (len(l) <= i and a) or (len(l) > i and l[i])
>>> my_list = [0,False,None,3]
>>> list_get(my_list,0)
0
>>> list_get(my_list,1)
False
>>> list_get(my_list,2)  # returns None
>>> list_get(my_list,3)
3
>>> list_get(my_list,4)
False
>>> list_get(my_list,4,0)  # Note: returns False instead of 0
False
>>> list_get(my_list,4,1)
1

It’s not perfect. Be careful with False-equivalent values in the alt parameters as noted above. But otherwise it should work for most practical cases were such usage is desired.

What I wanted was something like the choice function, but one to which I could pass a dictionary with the keys representing the choice list and the values representing their relative weights so that I could get a representative distribution. Since the random module doesn’t offer one itself, I was left to my own devices, which are admittedly clumsy and slow. The situation begged for a stupid lambda trick, but I was pressed for time and so I just threw this together:

def weighted_choice(ChoiceDict):
    wsum = sum([w for w in ChoiceDict.values()])
    n = random.uniform(0, wsum)
    for k in ChoiceDict:
        if n < ChoiceDict[k]: break
        n = n - ChoiceDict[k]
    return k

A little later, I passed the lambda challenge off to a couple of my colleagues. Later in the day, I had a chance to rattle off my own:

weighted_choice = lambda d: random.choice(reduce(list.__add__, [[a for a in k for n in range(d[k])] for k in d.keys()]))

But the Letterman spot goes to my colleague Leonard, for the conciseness and nice symmetry of his solution:

weighted_choice = lambda d: random.choice([k_ for s in [[k]*w for k,w in d.iteritems()] for k_ in s])

Find a script testing these and a couple other variations on the function here:

http://klenwell.googlecode.com/svn/trunk/pastebin/weighted_choice.py