Is there a better way of doing this?

[mattia]

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.