Creating probability by value from a dict - python-2.7

Question

I'm creating a type of survival game in python, and am creating the different animals you can encounter and "battle" with (users will only encounter one animal at a time). I want certain animals with a lower number associated with them to have a higher probability of showing up then the animals with a higher number associated to them:
POSSIBLE_ANIMALS_ENCOUNTERED = {
"deer": 50, "elk": 75, "mountain lion": 150,
"rat": 5, "raccoon": 15, "squirrel": 5,
"black bear": 120, "grizzly bear": 200, "snake": 5,
}
So, I would like a rat to appear more then a deer and just as much as a squirrel or snake. How can I go about creating probability for the animals inside of the dict based on their value? I'd like it so that the user will see a higher percentage of animals with lower value, and a lower percentage of animals with a higher value.
For example:
The user should see a animal with a value from 1 to 5, 50% of the time (0.5), an animal with a value from 6 to 50 %25 percent of the time (0.25), and any animal with a value higher then that 10 percent of the time (0.10).

Answer 1

You need to codify the percentage likelihood of an encounter based on "health points" (encounter_ranges tuples are from the data in your edit), then do a weighted random choice from your elements. I've put comments inline:
from random import random
from bisect import bisect
POSSIBLE_ANIMALS_ENCOUNTERED = {
"deer": 50, "elk": 75, "mountain lion": 150,
"rat": 5, "raccoon": 15, "squirrel": 5,
"black bear": 120, "grizzly bear": 200, "snake": 5,
}
# codify your ranges. (min, max, encounter %)
encounter_ranges = (
(50, float('inf'), .10),
(6, 50, .25),
(1, 5, .50)
)
# create a list of their probability to be encountered
# replace .items() with .iteritems() if python 2.x
ANIMAL_ENCOUNTER_PERCENTAGE_PER_HP_LIST = []
for k, v in POSSIBLE_ANIMALS_ENCOUNTERED.items():
for encounter_chance in encounter_ranges:
if (v >= encounter_chance[0]) and (v <= encounter_chance[1]):
# multiplied by 100 because we're going to use it in a weighted random below
ANIMAL_ENCOUNTER_PERCENTAGE_PER_HP_LIST.append([k, encounter_chance[2] * 100])
# With our percentages in hand, we need to do a weighted random distribution
# if you're only planning on encountering one animal at a time.
# stolen from http://stackoverflow.com/a/4322940/559633
def weighted_choice(choices):
values, weights = zip(*choices)
total = 0
cum_weights = []
for w in weights:
total += w
cum_weights.append(total)
x = random() * total
i = bisect(cum_weights, x)
return values[i]
# run this function to return the animal you will encounter:
weighted_choice(ANIMAL_ENCOUNTER_PERCENTAGE_PER_HP_LIST)
Note that this approach will always return something -- a 100% chance encounter of some animal. If you want to make the encounters more random, that's a much simpler problem, but I didn't include it as you'd need to specify how you expect this game mechanic to work (random 0-100% chance of any encounter? a percentage chance of the encounter based on the return (50% chance of encounter if rat is returned)? etc).
Note that I wrote this for Python 3, as if you're new to Python, you really should be using 3.x, but I left a comment in what you need to switch if you decide to stick with 2.x.