朱莉娅比Java慢得多

Question

我是朱莉娅的新手,我写了一个计算RMSE(均方根误差)的简单函数.ratings是一个评级矩阵,每一行都是[user, film, rating].有1500万个评级.该rmse()方法需要12.0秒,但Java实现速度快约188倍:0.064秒.为什么Julia实施会变慢？在Java中,我正在使用一个Rating对象数组,如果它是一个多维int数组,它会更快.

ratings = readdlm("ratings.dat", Int32)

function predict(user, film)
    return 3.462
end

function rmse()
    total = 0.0
    for i in 1:size(ratings, 1)
        r = ratings[i,:]
        diff = predict(r[1], r[2]) - r[3]
        total += diff * diff
    end
    return sqrt(total / size(ratings)[1])
end

编辑:避免全局变量后,它在1.99秒内完成(比Java慢31倍).删除后r = ratings[i,:],它是0.856秒(慢13倍).

Answer 1

一些建议:

• 不要使用全局变量.出于恼人的技术原因,它们很慢.相反,传入ratings作为参数.
• 该r = ratings[i,:]行制作副本,速度很慢.相反,使用predict(r[i,1], r[i,2]) - r[i,3].
• square()可能比x*x- 尝试更快.
• 如果您从源头使用最前沿的Julia,请查看全新的NumericExtensions.jl软件包,该软件包具有针对许多常见数值操作的疯狂优化功能.(参见julia-dev列表)
• Julia必须在第一次执行代码时编译代码.在朱莉娅进行基准测试的正确方法是多次进行计时并忽略第一次.

Answer 2

对我来说,以下代码在0.024秒内运行(我怀疑我的笔记本电脑比你的机器快得多).我用注释掉的行初始化了评级,因为我没有你提到的文件.

function predict(user, film)
    return 3.462
end

function rmse(r)
    total = 0.0
    for i = 1:size(r,1)
        diff = predict(r[i,1],r[i,2]) - r[i,3]
        total += diff * diff
    end
    return sqrt(total / size(r,1))
end

# ratings = rand(1:20, 5000000, 3)

Answer 3

在我的系统上,问题似乎是你的常量值predict函数没有得到优化.更换多余的调用predict使代码在0.01秒内运行.

function time()
    ratings = ones(15_000_000, 3)
    predict(user, film) = 3.462
    function rmse(ratings)
        total = 0.0
        for i in 1:size(ratings, 1)
            diff = predict(ratings[i, 1], ratings[i, 2]) - ratings[3]
            total += diff * diff
        end
        return sqrt(total / size(ratings, 1))
    end
    rmse(ratings)
    @elapsed rmse(ratings)
end

time()

function time2()
    ratings = ones(15_000_000, 3)
    predict(user, film) = 3.462
    function rmse(ratings)
        total = 0.0
        for i in 1:size(ratings, 1)
            diff = 3.462 - ratings[3]
            total += diff * diff
        end
        return sqrt(total / size(ratings, 1))
    end
    rmse(ratings)
    @elapsed rmse(ratings)
end

time2()