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| import random | |
| def createChild(individual1, individual2): | |
| child = "" | |
| for i in range(len(individual1)): | |
| if (int(100 * random.random()) < 50): | |
| child += individual1[i] | |
| else: | |
| child += individual2[i] | |
| return child | |
| def createChildren(breeders, number_of_child): | |
| nextPopulation = [] | |
| for i in range(len(breeders)/2): | |
| for j in range(number_of_child): | |
| nextPopulation.append(createChild(breeders[i], breeders[len(breeders) -1 -i])) | |
| return nextPopulation |
If you follow the code in passwordTuto from his profile then by design all individuals in the population will have the same length as the desired password. Also, all children will have the same length so there is no need to account for differences in length as there simply will not be.
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Does createChild() account for the possibility that the length of individual1 is greater than that of individual2? It looks like that scenario would produce an out of range error.