Random Python dictionary key, weighted by values

Question

I have a dictionary where each key has a list of variable length, eg:

d = {
 \'a\': [1, 3, 2],
 \'b\': [6],
 \'c\': [0, 0]
}

Is there a cle

Answer 1

I modified some of the other answers to come up with this. It's a bit more configurable. It takes 2 arguments, a list and a lambda function to tell it how to generate a key.

def select_weighted(lst, weight):
   """ Usage: select_weighted([0,1,10], weight=lambda x: x) """
   thesum = sum([weight(x) for x in lst])
   if thesum == 0:
      return random.choice(lst)
   offset = random.randint(0, thesum - 1)

   for k in lst:
      v = weight(k)
      if offset < v:
         return k
      offset -= v

Thanks to sth for the base code for this.

Answer 2

Make a list in which each key is repeated a number of times equal to the length of its value. In your example: ['a', 'a', 'a', 'b', 'c', 'c']. Then use random.choice().

Edit: or, less elegantly but more efficiently, try this: take the sum of the lengths of all values in the dictionary, S (you can cache and invalidate this value, or keep it up to date as you edit the dictionary, depending on the exact usage pattern you anticipate). Generate a random number from 0 to S, and do a linear search through the dictionary keys to find the range into which your random number falls.

I think that's the best you can do without changing or adding to your data representation.

Answer 3

I'd say this:

random.choice("".join([k * len(d[k]) for k in d]))

This makes it clear that each k in d gets as many chances as the length of its value. Of course, it is relying on dictionary keys of length 1 that are characters....

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Much later:

table = "".join([key * len(value) for key, value in d.iteritems()])
random.choice(table)