Fortran 提高大型数组写入文件速度的最佳方法

如何解决Fortran 提高大型数组写入文件速度的最佳方法

我有一个非常大的数组，大小为 20,000,000，我想将它写入一个未格式化的文件。 这是一个自相关函数。

使用 -O4 优化编译标志非常快，无需写入文件。 但是一旦我写入文件，它似乎需要一天多的时间才能完成。

最后是 f90 程序。下面是没有写入文件和写入文件的输出。

很明显，写入数组的单个元素需要大约 10 毫秒。

20,000 x 0.01 = 200,000 秒 = 3,333 分钟 = 55 小时

当读取只需要 45 秒时，写入文件需要这么长时间怎么可能？我可以做些什么来提高速度？

注意事项

系统：Ubuntu 20.04

编译行：fortran -o acorr.exe -O4 acorr.f90

无文件写入

elapsed time for reading:    43.4389992    
 Size of Jx:     20000000
 Loop Start Time:    43.5009995    
 correlation time magnitude 1e0 elapsed time:    43.5009995    
 correlation time magnitude 1e1 elapsed time:    43.5009995    
 correlation time magnitude 1e2 elapsed time:    43.5009995    
 correlation time magnitude 1e3 elapsed time:    43.5009995    
 correlation time magnitude 1e4 elapsed time:    43.5009995    
 correlation time magnitude 1e5 elapsed time:    43.5009995    
 correlation time magnitude 1e6 elapsed time:    43.5029984    
 correlation time magnitude 1e7 elapsed time:    43.5190010    
 elapsed time:    43.5369987  

带文件写入

 elapsed time for reading:    43.6349983    
 Size of Jx:     20000000
 Loop Start Time:    43.6949997    
 correlation time magnitude 1e0 elapsed time:    43.7319984    
 correlation time magnitude 1e1 elapsed time:    43.8969994    
 correlation time magnitude 1e2 elapsed time:    45.4980011    
 correlation time magnitude 1e3 elapsed time:    61.5289993

acorr.f90

PROGRAM acorr
    
    real:: a,b,c,d,sum,mean,var
    integer:: i,j,jsize,beginning,rate,end,end1
    real,dimension(20000000):: Jx,Jxm,corr
    integer:: skip_lines = 4
    call system_clock(beginning,rate)
    
    !reading file
    open(10,file='DiamHeat.log',status='old')
    do i = 1,skip_lines
        read(10,*)
    end do
    do i = 1,20000000
        read(10,*) a,Jx(i),d
    end do
    call system_clock(end)
    print *,"elapsed time for reading: ",real(end - beginning) / real(rate)
    close(10)
    !finished reading

    open(20,file='acorr.txt',form='UNFORMATTED')
    jsize = size(Jx)
    print *,"Size of Jx: ",jsize
    !print *,dot_product(Jx(10:jsize),Jx(1:jsize-10))

    !calculate mean
    mean = sum(Jx)/jsize
    Jxm(:) = Jx(:)-mean

    !calculate variance
    var = dot_product(Jxm,Jxm)/jsize
    
    !begin autocorrelation calc
    call system_clock(end1)
    print *,"Loop Start Time: ",real(end1 - beginning) / real(rate)
    do i =0,jsize-1
        !calculation
        corr(i+1) = dot_product(Jxm(i+1:jsize),Jxm(1:jsize-i))/var/(jsize-i)

        !clock timing
        if(i == 1) then
            call system_clock(end)
            print *,"correlation time magnitude 1e0 elapsed time: ",real(end - beginning) / real(rate)
        else if(i == 10) then
            call system_clock(end)
            print *,"correlation time magnitude 1e1 elapsed time: ",real(end - beginning) / real(rate)
        else if(i == 100) then
            call system_clock(end)
            print *,"correlation time magnitude 1e2 elapsed time: ",real(end - beginning) / real(rate)
        else if(i == 1000) then
            call system_clock(end)
            print *,"correlation time magnitude 1e3 elapsed time: ",real(end - beginning) / real(rate)
        else if(i == 10000) then
            call system_clock(end)
            print *,"correlation time magnitude 1e4 elapsed time: ",real(end - beginning) / real(rate)
        else if(i == 100000) then
            call system_clock(end)
            print *,"correlation time magnitude 1e5 elapsed time: ",real(end - beginning) / real(rate)
        else if(i == 1000000) then
            call system_clock(end)
            print *,"correlation time magnitude 1e6 elapsed time: ",real(end - beginning) / real(rate)
        else if(i == 10000000) then
            call system_clock(end)
            print *,"correlation time magnitude 1e7 elapsed time: ",real(end - beginning) / real(rate)
        end if 

    end do
    
    write(20,*) corr
    close(20)
    call system_clock(end)
    print *,"elapsed time: ",real(end - beginning) / real(rate)
END PROGRAM

解决方法

正如@francescalus 所评论的，编译器似乎会跳过计算，除非它用于其他目的。 添加

print*,sum corr 

在循环似乎使程序计算循环中的点积之后。这只是需要很长时间，但它正在以最佳容量进行计算。

再次感谢@francescalus