Question

我正在寻找一种节省内存的方法。正如您在下面看到的，我创建了很多列表来设计我自己的概率分布（F）。当T = 2时，这意味着我创建了2个离散概率分布。一个仅用于第一个时期，一个用于两个时期。直到11个周期（T = 11），它工作正常，但如果我选择T = 12，我在sc_prob上有一个memoryError。我使用64位版本，所以这不是问题。有没有办法避免这种memoryError？

#number of periods
T=2

# problem data
k=100
b=9
h=1

def multiply(n): 
    temp = 1
    for i in range(len(n)):
        temp *= n[i]
    return temp

minimum=[2, 11, 13, 20, 1, 0, 11, 9, 18, 20, 18, 20, 6, 20, 10, 17, 18, 18, 11, 9, 3, 16, 0, 20, 9]
mean=[23.5, 16.5, 43.0, 43.5, 8.0, 5.0, 40.0, 18.0, 41.5, 51.5, 46.0, 31.5, 53.5, 26.5, 21.5, 23.0, 49.5, 54.0, 20.5, 37.0, 20.5, 23.5, 17.0, 39.5, 55.5]
maximum=[45, 22, 73, 67, 15, 10, 69, 27, 65, 83, 74, 43, 101, 33, 33, 29, 81, 90, 30, 65, 38, 31, 34, 59, 102]
mad=[18, 5, 30, 23, 6, 3, 29, 9, 23, 31, 28, 11, 42, 6, 11, 6, 31, 36, 9, 28, 15, 7, 13, 19, 43]


#scenarios 
scenario=list(product([0,1,2],repeat=T))
#print "scenarios are:",scenario

#Assigning probability to every 0,1,and 2
probability= [[[] for n in range(3)] for s in range(T)]
for n in range(T):
    probability[n][0]=mad[n]/float(2*(mean[n]-minimum[n]))
    probability[n][1]=1-(mad[n]/float(2*(mean[n]-minimum[n])))-(mad[n]/float(2*(maximum[n]-mean[n])))
    probability[n][2]=mad[n]/float(2*(maximum[n]-mean[n]))
#print "probabilities are:", probability

# Create same list as 'scenarios', but now 0,1,2 are replaced with its probability
sc_prob=[[[] for n in range(len(scenario[s]))] for s in range(len(scenario))]
for s in range(len(scenario)):
    for n in range(len(scenario[s])):
        if scenario[s][n]==0:
             sc_prob[s][n]=probability[n][0]
        if scenario[s][n]==1:
             sc_prob[s][n]=probability[n][1]
        if scenario[s][n]==2:
             sc_prob[s][n]=probability[n][2]
#print "sc_prob per period per scenario are:",sc_prob

# all probabilities within 1 scenario are multiplied to calculate probability of whole scenario
sc_prob_2=range(len(scenario))
for s in range(len(scenario)):
    sc_prob_2[s]=multiply(sc_prob[s])
#print "sc_prob_2", sc_prob_2

# create same list as scenarios, but now 0,1,2 are replaced with its demand
demand=[[[] for n in range(len(scenario[s]))] for s in range(len(scenario))]
for s in range(len(scenario)):
    for n in range(len(scenario[s])):
        if scenario[s][n]==0:
            demand[s][n]=minimum[n]
        if scenario[s][n]==1:
            demand[s][n]=mean[n]
        if scenario[s][n]==2:
            demand[s][n]=maximum[n]
#print "demands are: ", demand

# creating list where demand and probability are merged
# to get sum probability equal to 1, probability is divided by (3**(T-(t+1)))
prob_C_Demand=range(T)
for t in range(T):
        prob_C_Demand[t]=list((sum(demand[s][:t+1]),multiply(sc_prob[s][:t+1])/(3**(T-(t+1)))) for s in range(len(scenario)))
#print "prob_C_Demand:", prob_C_Demand


#merge demand with probability when demand is the same
#create sorted dict
F=range(T)
for t in range(T):
    F[t]= {x:0 for x,_ in prob_C_Demand[t]}
    for name,num in prob_C_Demand[t]: 
        F[t][name] += num
    F[t]=sorted(F[t].iteritems())
#print "F:",F

#to calculate estimated demand 'D'
probxDemand=[[[] for s in range(len(scenario))] for t in range(T)]
for t in range(T):
        for s in range(len(scenario)):
                probxDemand[t][s]=multiply(prob_C_Demand[t][s])
print "probxDemand:", probxDemand

D=range(T)
for t in range(T):
        D[t]=sum(probxDemand[t])
print "D:", D

创建了很多长列表，如何避免MemoryError？

0 个答案: