Is there a better way of doing this?

Peter Otten __peter__ at web.de
Fri Mar 6 22:28:00 CET 2009


mattia wrote:

> Il Fri, 06 Mar 2009 14:06:14 +0100, Peter Otten ha scritto:
> 
>> mattia wrote:
>> 
>>> Hi, I'm new to python, and as the title says, can I improve this
>>> snippet (readability, speed, tricks):
>>> 
>>> def get_fitness_and_population(fitness, population):
>>>     return [(fitness(x), x) for x in population]
>>> 
>>> def selection(fitness, population):
>>>     '''
>>>     Select the parent chromosomes from a population according to their
>>>     fitness (the better fitness, the bigger chance to be selected) '''
>>>     selected_population = []
>>>     fap = get_fitness_and_population(fitness, population) pop_len =
>>>     len(population)
>>>     # elitism (it prevents a loss of the best found solution) # take
>>>     the only 2 best solutions
>>>     elite_population = sorted(fap)
>>>     selected_population += [elite_population[pop_len-1][1]] +
>>> [elite_population[pop_len-2][1]]
>>>     # go on with the rest of the elements for i in range(pop_len-2):
>>>         # do something
>> 
>> def selection1(fitness, population, N=2):
>>     rest = sorted(population, key=fitness, reverse=True) best = rest[:N]
>>     del rest[:N]
>>     # work with best and rest
>> 
>> 
>> def selection2(fitness, population, N=2):
>>     decorated = [(-fitness(p), p) for p in population]
>>     heapq.heapify(decorated)
>>     
>>     best = [heapq.heappop(decorated)[1] for _ in range(N)] rest = [p for
>>     f, p in decorated]
>>     # work with best and rest
>> 
>> Both implementations assume that you are no longer interested in the
>> individuals' fitness once you have partitioned the population in two
>> groups.
>> 
>> In theory the second is more efficient for "small" N and "large"
>> populations.
>> 
>> Peter
> 
> Ok, but the fact is that I save the best individuals of the current
> population, than I'll have to choose the others elements of the new
> population (than will be N-2) in a random way. The common way is using a
> roulette wheel selection (based on the fitness of the individuals, if the
> total fitness is 200, and one individual has a fitness of 10, that this
> individual will have a 0.05 probability to be selected to form the new
> population). So in the selection of the best solution I have to use the
> fitness in order to get the best individual, the last individual use the
> fitness to have a chance to be selected. Obviously the old population anf
> the new population must have the same number of individuals.

You're right, it was a bad idea.

Peter



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