Question

我已经编写了一个用于解决拉普拉斯方程的串行代码，但是当我尝试在Julia中并行编写时，它需要比串行代码更多的时间和内存。我写了一个简单的例子。我该如何与这段代码并行？

有一个域t1。

t2将被计算，然后t1 = t2

@everywhere function left!(t1,t2,n,l_type,b_left,dx=1.0,k=50.0)
    if l_type==1
            for i=1:n
                t2[i,1]=(b_left*dx/k)+t1[i,2];
                t1[i,1]=t2[i,1];
            end
    else 
        for i=1:n
        t1[i,1]=b_left;
        end
    end
    return t1 end

# parallel left does not work.
@everywhere function pleft!(t1,t2,n,l_type,b_left,dx=1.0,k=50.0)
    if l_type==1
           @parallel for i=1:n
                t2[i,1]=(b_left*dx/k)+t1[i,2];
                t1[i,1]=t2[i,1];
            end
    else 
     @parallel for i=1:n
        t1[i,1]=b_left;
        end
    end
    return t1
end
n = 10;
t1 = SharedArray(Float64,(n,n));
t2=t1;
typ = 0;
value = 10;
dx = 1;
k=50;

@time t3 = pleft!(t1,t2,n,typ,value,dx,k)
@time t2 = left!(t1,t2,n,typ,value,dx,k)

答案是：

0.000872 seconds (665 allocations: 21.328 KB) # for parallel one 
0.000004 seconds (4 allocations: 160 bytes)   #for usual one

我该如何解决这个问题？

计算后我应该在while循环中计算。我需要在代码下面并行。

@everywhere function oneStepseri(t1,N)
    t2 = t1;
     for j = 2:(N-1) 
        for i = 2:(N-1)
         t2[i,j]=0.25*(t1[i-1,j]+t1[i+1,j]+t1[i,j-1]+t1[i,j+1]);
        end
               end
    return t2;
end

感谢...

Answer 1

我试了很多东西。 @parallel SharedArray，Distributed Array，域名划分和使用@spawn。没有加速。但最近Julia添加了“Threads”，您可以按导出JULIA_NUM_THREADS=4添加主题在命令窗口中。使用Threads.@threads，您可以并行代码。按Threads.nthreads()检查线程数这是我的代码它给了我一个很好的加速。

#to add threads export JULIA_NUM_THREADS=4

nth = Threads.nthreads(); #print number of threads

println(nth);

a = zeros(10);

Threads.@threads for i = 1:10
           a[i] = Threads.threadid()
       end

show(a)

b = zeros(100000);
c = zeros(100000);
b[1] = b[end] = 1;
c[1] = c[end] = 1;

function noth(A)
    B = A;
    for i=2:(length(A)-1)
        B[i] = (A[i-1] + A[i+1])*0.5;
    end 
    return B
end

function th(A)
    B = A;
    Threads.@threads for i=2:(length(A)-1)
        B[i] = (A[i-1] + A[i+1])*0.5;
    end
    return B
end


println("warmup noth , th")
@time bb = noth(b)
@time cc = th(c)
println("end ")
@time bb = noth(b)
@time cc = th(c)

@time bb = noth(b)
@time cc = th(c)

@time bb = noth(b)
@time cc = th(c)
@time bb = noth(b)
@time cc = th(c)
@time bb = noth(b)
@time cc = th(c)
@time bb = noth(b)
@time cc = th(c)
show(bb[10])
println("\nbb ------------------------------------------------------------------------------------------------------------------> cc")
show(cc[10])

答案是这样的

5                                                                                                                                                                    
[1.0,1.0,2.0,2.0,3.0,3.0,4.0,4.0,5.0,5.0]warmup noth , th                                                                                                            
  0.008661 seconds (2.53 k allocations: 113.180 KB)                                                                                                                  
  0.020738 seconds (7.94 k allocations: 336.981 KB)                                                                                                                  
end                                                                                                                                                                  
  0.000446 seconds (4 allocations: 160 bytes)                                                                                                                        
  0.000122 seconds (6 allocations: 224 bytes)                                                                                                                        
  0.000437 seconds (4 allocations: 160 bytes)                                                                                                                        
  0.000135 seconds (6 allocations: 224 bytes)                                                                                                                        
  0.000435 seconds (4 allocations: 160 bytes)                                                                                                                        
  0.000115 seconds (6 allocations: 224 bytes)                                                                                                                        
  0.000447 seconds (4 allocations: 160 bytes)                                                                                                                        
  0.000112 seconds (6 allocations: 224 bytes)                                                                                                                        
  0.000440 seconds (4 allocations: 160 bytes)                                                                                                                        
  0.000109 seconds (6 allocations: 224 bytes)                                                                                                                        
  0.000439 seconds (4 allocations: 160 bytes)                                                                                                                        
  0.000116 seconds (6 allocations: 224 bytes)                                                                                                                        
0.052478790283203125                                                                                                                                                 
bb ------------------------------------------------------------------------------------------------------------------> cc                                            
0.052478790283203125juser@juliabox:~/threads$

用于5个线程和100000个节点。

请注意，对于热身，没有加速。但在那之后有加速。

0.000446 seconds (4 allocations: 160 bytes)   # usual code run                    
0.000122 seconds (6 allocations: 224 bytes)   #parallel code run

如何在julia语言中并行简单的for循环？

1 个答案: