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From version 3.6, the package random has the function https://docs.python.org/3/library/random.html#random.choices which allows you to define the weights of each value within a population.from random import choices
population = [1, 7, 15]
weights = [40, 30, 30]
samples = choices(population, weights, k=100)
The above code will generate 100 random values following the defined weights.Based on https://pt.stackoverflow.com/q/147884/5878 The easiest way to implement is to generate a structure that already has these odds. Since the number 1 must have a probability equal to 40%, while the numbers 7 and 15 must have a probability equal to 30%, you can generate a list that has 10 elements in all, repeating the number 1 four times and the numbers 7 and 15 three times each.v = [1, 1, 1, 1, 7, 7, 7, 15, 15, 15]
So, in doing random.choice(v), the desired odds will be respected.Implementing a genetic form, and quite simple, you can define a function that generates this list as your need. For example:from typing import Tuple, Any, List
from itertools import starmap, chain, repeat
from random import choice
def sample_space(*values: Tuple[Any, float]) -> List:
if sum(value[1] for value in values) != 1.0:
raise ValueError('Soma das probabilidades não é 1.0')
while True:
if all(probability.is_integer() for _, probability in values):
break
values = [(value, probability * 10) for value, probability in values]
values = [(value, int(probability)) for value, probability in values]
return list(chain(*starmap(lambda a, b: list(repeat(a, b)), values)))
The function sample_space will generate for you a list that defines exactly the sample space you want, just pass as parameter a set of tuples with desired value and probability. For the data presented in the question, it would be:>>> space = sample_space((1, 0.4), (7, 0.3), (15, 0.3))
>>> print(space)
[1, 1, 1, 1, 7, 7, 7, 15, 15, 15]
If you draw 100 numbers from this sample space and check how many times you repeat (you can use https://docs.python.org/2/library/collections.html#collections.Counter for this), you will see that the odds tend to be followed:>>> samples = [choice(space) for _ in range(100)]
>>> print(Counter(samples))
Counter({1: 40, 7: 32, 15: 28})
See working on https://repl.it/@acwoss/StarchyLoathsomeAttribute | https://ideone.com/MrsKS6 It will work for any odds as long as the sum is always 1.0. For example, for a sample space that is 99% True and 1% False I would:>>> space = sample_space((True, 0.99), (False, 0.01))
However, this would generate a list of 100 values, 99 being True and 1 False; therefore, for this solution, given its simplicity, be careful not to generate too large lists and affect application memory. The more decimal houses possess the odds, the more elements in memory will be needed.
