1-12个核心的itertools.product并行化

Question

我想考虑自然数的几种组合。现在我花了很多时间甚至使用N = 3，我想介绍代码的并行化。这段代码在这里。对不起，大字符串！

from itertools import product

n1=[]
n2=[] 
n3=[]
n4=[]
N=2
for i in range(N+1):
    n1.append(i)
    n2.append(i)    
    n3.append(i)
    n4.append(i) 

c=list(product(n1,n2,n3,n4))
for i in range(len(c)):
    d=list(itertools.islice(c[i],4))    
    d=np.array(d)
    n1=d[0]
    n2=d[1]
    n3=d[2]
    n4=d[3]
    modes=[]
    ymain=[]
    for i in range(n1+n2):
        modes.append(1400)
    for i in range(n3+n4):
        modes.append(600)
    for i in range(n1):
        ymain.append(X)    
    for i in range(n2):
        ymain.append(Y)    
    for i in range(n3):
        ymain.append(X)    
    for i in range(n4):
        ymain.append(Y)    
    for p in range(1,30):
        E=500*p
        numer=int(round(E/600.0))+1
        M=np.size((np.array(modes)))
        A=[]
        for i in range(M):
            A.append(range(numer))    
        for i in range (M):
            for j in range(numer):
                A[i][j]=(A[i][j])*modes[i]
        modes=np.array(modes)
        ymain=np.array(ymain)

        sums = ((vs,sum(vs)) for vs in product(*A))
        for k,v in sums:
            print k,v, E

如何为1-12个内核并行化itertools.product？