3
\$\begingroup\$

I haven't done any research into how you actually write a genetic algorithm, so this is my best guess. What I'm really curious to know is two-fold:

  1. Have I created a genetic algorithm?
  2. If I have, how can I continue to explore the subject more?

I hope this code and docstrings are enough explanation for how the code works.

from random import randint

import time


class Organism(object):

    def __init__(self, r=0, g=0, b=0):
        """Initialize the class with the RGB color values."""
        self.r = r
        self.g = g
        self.b = b

    @property
    def fitness(self):
        """The lower the fitness the better."""
        total = self.r + self.g + self.b
        return abs(765 - total)

    @classmethod
    def spawn(cls, parent):
        """Return a mutated generation of ten members."""
        generation = []
        for number in range(0, 10):
            r = cls.mutate(parent.r)
            g = cls.mutate(parent.g)
            b = cls.mutate(parent.b)
            generation.append(cls(r, g, b))
        return generation

    @staticmethod
    def mutate(value):
        """Mutate the value by 10 points in either direction."""
        min_, max_ = value - 10, value + 10
        return randint(min_, max_)


def breed_and_select_fittest(individual):
    """Return the fittest member of a generation."""
    generation = Organism.spawn(individual)
    fittest = generation[0]
    for member in generation:
        if member.fitness < fittest.fitness:
            fittest = member
    return fittest


if __name__ == '__main__':
    individual = Organism()  # abiogenesis!

    while True:
        individual = breed_and_select_fittest(individual)
        print individual.fitness

        time.sleep(0.2)
\$\endgroup\$
1
  • \$\begingroup\$ The only "issue" is that the correct solutions are really easy to find. :) But working on toy problems is a good way to get started! \$\endgroup\$ Sep 8, 2016 at 4:29

2 Answers 2

6
\$\begingroup\$

That is, more or less, a genetic algorithm, though it only implements mutation and not recombination:

@classmethod
def recombine(cls, parentA, parentB):
    traits = {}
    for t in ("r", "g", "b"):
        traits[t] = (getattr(parentA, t) + getattr(parentB, t))/2
    return cls(**traits)

There are, of course, many ways to implement recombination. For example, instad of the simple averaging I do above, you could instead choose a random parent to take each trait from. The best solution depends on your problem space.

To learn more, I'd suggest googling and finding tutorials or examples, and then trying to solve some more realistic problems with them. Can you adapt them to solve any ProjectEuler problems? Or find another application where they're useful?

\$\endgroup\$
1
\$\begingroup\$

Your breed_and_select function can be more succinctly written as:

import operator

def breed_and_select_fittest(individual):
    """Return the fittest member of a generation."""
    generation = Organism.spawn(individual)
    return min(generation, key=operator.attrgetter("fitness"))

It is better to use operator.attrgetter than lambda here, because it is faster (even though performance is not your limiting factor, your execution time is dominated by the time.sleep(0.2) right now).

I would also make this slightly more customizable by adding the size of each generation and the variation during the spawning as parameters.

I'm also not convinced spawn should really be a @classmethod. After all it is the individual spawning the next generation. Consider this:

import operator

class Organism:
    ...
    def spawn(self, n=10, variance=10):
        """Return a mutated generation of `n` members."""
        return [self.__class__(*self.mutate(variance)) for _ in range(n)]

    def mutate(self, variance):
        for value in (self.r, self.g, self.b):
            yield randint(value - variance, value + variance) 
    ...

def breed_and_select_fittest(individual):
    """Return the fittest member of a generation."""
    generation = individual.spawn()
    return min(generation, key=operator.attrgetter("fitness"))

Note that the default start value for range is already 0, so no need for that.

\$\endgroup\$
4
  • \$\begingroup\$ I would use operator.attrgetter('fitness') instead of a lambda for the key. \$\endgroup\$ Sep 8, 2016 at 10:13
  • \$\begingroup\$ @MathiasEttinger Yeah, was thinking about that as well. Is there really such a big difference in speed between the two? Because operator.attrgetter is implemented in C? \$\endgroup\$
    – Graipher
    Sep 8, 2016 at 10:22
  • \$\begingroup\$ There should be, yes. \$\endgroup\$ Sep 8, 2016 at 10:27
  • \$\begingroup\$ @MathiasEttinger Alright, I'm convinced :) \$\endgroup\$
    – Graipher
    Sep 8, 2016 at 12:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.