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Julia 中 for 循环的性能

转载 作者:行者123 更新时间:2023-12-04 15:10:21 24 4
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我尝试了以下代码,但性能却大不相同
代码之间。听说顶层的代码不适合数值计算,但性能似乎也取决于
顶级变量(此处为 N)出现在 for 循环的范围内。
避免这样的顶级变量总是更好吗?

N = 10000000

# case1 (slow)
x = 0.0
@time for k = 1:N
x += float( k )
end

# case2 (slow)
@time let
y = 0.0
for j = 1:N
y += float( j )
end
end

# case3 (very fast)
@time let
n::Int64
n = N
z = 0.0
for m = 1:n
z += float( m )
end
end

# case 4 (slow)
function func1()
c = 0.0
for i = 1:N
c += float( i )
end
end

# case 5 (fast)
function func2( n )
c = 0.0
for i = 1:n
c += float( i )
end
end

# case 6 (fast)
function func3()
n::Int
n = N
c = 0.0
for i = 1:n
c += float( i )
end
end

# case 7 (slow)
function func4()
n = N # n = int( N ) is also slow
c = 0.0
for i = 1:n
c += float( i )
end
end

@time func1()
@time func2( N )
@time func3()
@time func4()

使用 Julia 0.3.7(在 Linux x86_64 上)获得的结果是
elapsed time: 2.595440598 seconds (959985496 bytes allocated, 10.70% gc time)
elapsed time: 2.469471127 seconds (959983688 bytes allocated, 11.49% gc time)
elapsed time: 1.608e-6 seconds (16 bytes allocated)
elapsed time: 2.535243279 seconds (960021976 bytes allocated, 11.21% gc time)
elapsed time: 0.002601149 seconds (75592 bytes allocated)
elapsed time: 0.003471583 seconds (84456 bytes allocated)
elapsed time: 2.480343146 seconds (960020752 bytes allocated, 11.48% gc time)

最佳答案

答案是“避免此类顶级变量总是更好吗?”文字当然是“不,这取决于”,但有用的评论是将全局变量声明为常量

const N = 10000000

使案例 2 与案例 3 一样快。

编辑:我应该补充一点,案例 2 的问题是顶级 N使范围的类型 1:N和迭代器变量 j不稳定,即使累加器变量 y是本地的。这个问题的一个更灵活的解决方法是
   let
y = 0.0
for j = 1:(N::Int)
y += float( j )
end
y
end

关于Julia 中 for 循环的性能,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29877563/

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