无序列表中相对概率的累积和
Cumulative sum of relative probabilities in an non-ordered list
population_d = {'0,0,1,0,1,1,0,1,1,1,1,0,0,0,0,1': 6,
'0,0,1,1,1,0,0,1,1,0,1,1,0,0,0,1': 3,
'0,1,1,0,1,1,0,0,1,1,1,0,0,1,0,0': 5,
'1,0,0,1,1,1,0,0,1,1,0,1,1,0,0,0': 1}
def ProbabilityList(population_d):
fitness = population_d.values()
total_fit = (sum(fitness))
relative_fitness = [f/total_fit for f in fitness]
probabilities = [sum(relative_fitness[:i+1]) for i in range(len(relative_fitness))]
return (probabilities)
我试图计算这种数据结构的累积概率,但是,我需要保持值的顺序以便将它们索引到另一个列表中相同位置的各个个体。
程序按顺序执行操作,为最后一个位置赋予更高的权重,在这种情况下,这将是最低的适应性。
有谁知道是否有一种方法可以在不改变它们在输出列表中的位置的情况下以正确的方式(适应度值的新月顺序)执行累积和?
非常感谢!
population_d = {'0,0,1,0,1,1,0,1,1,1,1,0,0,0,0,1': 6,
'0,0,1,1,1,0,0,1,1,0,1,1,0,0,0,1': 3,
'0,1,1,0,1,1,0,0,1,1,1,0,0,1,0,0': 5,
'1,0,0,1,1,1,0,0,1,1,0,1,1,0,0,0': 1}
在您的字典中,您将适合度 (?) 值与唯一标识符相关联。大概这些标识符来自您的程序和数据集中的其他地方。我没有试图依赖字典的构造顺序来保持这种关系,而是维护了关联并构造了一个新字典,其值是将适应度从低到高排序后获得的累积和。
import operator
def ProbabilityList(population_d):
fitness = population_d.values()
total_fit = (sum(fitness))
#create list of (individual, fitness) tuples
items = population_d.items()
#sort by fitness value
items = sorted(items, key = operator.itemgetter(1))
#some people prefer
#items = sorted(items, key = lambda item: item[1])
#print(items)
#maintain association and calculate relative fitness
relative_fitness = [(ind,fit/total_fit) for (ind,fit) in items]
#print(relative_fitness)
cumsum = 0
probabilities = {}
for ind, fit in relative_fitness:
cumsum += fit
probabilities[ind] = cumsum
return (probabilities)
d = ProbabilityList(population_d)
for k, v in d.items():
print('key:{}, fitness:{}, cumsum:{}'.format(k, population_d[k], v))
>>>
key:1,0,0,1,1,1,0,0,1,1,0,1,1,0,0,0, fitness:1, cumsum:0.06666666666666667
key:0,0,1,1,1,0,0,1,1,0,1,1,0,0,0,1, fitness:3, cumsum:0.26666666666666666
key:0,1,1,0,1,1,0,0,1,1,1,0,0,1,0,0, fitness:5, cumsum:0.6
key:0,0,1,0,1,1,0,1,1,1,1,0,0,0,0,1, fitness:6, cumsum:1.0
>>>
希望通过字典,您能够将累积总和与代码其他部分中的原始 个人 联系起来。
我看到您一直在询问与此数据集和项目相关的其他问题。您可能想花一些时间学习 Pandas,甚至考虑将数据保存在数据库中,而不是分散在整个项目中的各个容器中。
population_d = {'0,0,1,0,1,1,0,1,1,1,1,0,0,0,0,1': 6,
'0,0,1,1,1,0,0,1,1,0,1,1,0,0,0,1': 3,
'0,1,1,0,1,1,0,0,1,1,1,0,0,1,0,0': 5,
'1,0,0,1,1,1,0,0,1,1,0,1,1,0,0,0': 1}
def ProbabilityList(population_d):
fitness = population_d.values()
total_fit = (sum(fitness))
relative_fitness = [f/total_fit for f in fitness]
probabilities = [sum(relative_fitness[:i+1]) for i in range(len(relative_fitness))]
return (probabilities)
我试图计算这种数据结构的累积概率,但是,我需要保持值的顺序以便将它们索引到另一个列表中相同位置的各个个体。
程序按顺序执行操作,为最后一个位置赋予更高的权重,在这种情况下,这将是最低的适应性。
有谁知道是否有一种方法可以在不改变它们在输出列表中的位置的情况下以正确的方式(适应度值的新月顺序)执行累积和?
非常感谢!
population_d = {'0,0,1,0,1,1,0,1,1,1,1,0,0,0,0,1': 6,
'0,0,1,1,1,0,0,1,1,0,1,1,0,0,0,1': 3,
'0,1,1,0,1,1,0,0,1,1,1,0,0,1,0,0': 5,
'1,0,0,1,1,1,0,0,1,1,0,1,1,0,0,0': 1}
在您的字典中,您将适合度 (?) 值与唯一标识符相关联。大概这些标识符来自您的程序和数据集中的其他地方。我没有试图依赖字典的构造顺序来保持这种关系,而是维护了关联并构造了一个新字典,其值是将适应度从低到高排序后获得的累积和。
import operator
def ProbabilityList(population_d):
fitness = population_d.values()
total_fit = (sum(fitness))
#create list of (individual, fitness) tuples
items = population_d.items()
#sort by fitness value
items = sorted(items, key = operator.itemgetter(1))
#some people prefer
#items = sorted(items, key = lambda item: item[1])
#print(items)
#maintain association and calculate relative fitness
relative_fitness = [(ind,fit/total_fit) for (ind,fit) in items]
#print(relative_fitness)
cumsum = 0
probabilities = {}
for ind, fit in relative_fitness:
cumsum += fit
probabilities[ind] = cumsum
return (probabilities)
d = ProbabilityList(population_d)
for k, v in d.items():
print('key:{}, fitness:{}, cumsum:{}'.format(k, population_d[k], v))
>>>
key:1,0,0,1,1,1,0,0,1,1,0,1,1,0,0,0, fitness:1, cumsum:0.06666666666666667
key:0,0,1,1,1,0,0,1,1,0,1,1,0,0,0,1, fitness:3, cumsum:0.26666666666666666
key:0,1,1,0,1,1,0,0,1,1,1,0,0,1,0,0, fitness:5, cumsum:0.6
key:0,0,1,0,1,1,0,1,1,1,1,0,0,0,0,1, fitness:6, cumsum:1.0
>>>
希望通过字典,您能够将累积总和与代码其他部分中的原始 个人 联系起来。
我看到您一直在询问与此数据集和项目相关的其他问题。您可能想花一些时间学习 Pandas,甚至考虑将数据保存在数据库中,而不是分散在整个项目中的各个容器中。