julia - 如何在 Julia 中沿 bool 数组的轴进行按位或归约？

Question

我正试图找到在 Julia 中将 3D bool 掩码数组按位或缩减为 2D 的最佳方法。

当然，我总是可以写一个 for 循环:

x = randbool(3,3,3)
out = copy(x[:,:,1])
for i = 1:3
    for j = 1:3
        for k = 2:3
            out[i,j] |= x[i,j,k]
        end
    end
end

但我想知道是否有更好的方法来减少。

Answer 1

一个简单的答案是

out = x[:,:,1] | x[:,:,2] | x[:,:,3]

但我做了一些基准测试:

function simple(n,x)
    out = x[:,:,1] | x[:,:,2]
    for k = 3:n
        @inbounds out |= x[:,:,k]
    end
    return out
end

function forloops(n,x)
    out = copy(x[:,:,1])
    for i = 1:n
        for j = 1:n
            for k = 2:n
                @inbounds out[i,j] |= x[i,j,k]
            end
        end
    end
    return out
end

function forloopscolfirst(n,x)
    out = copy(x[:,:,1])
    for j = 1:n
        for i = 1:n
            for k = 2:n
                @inbounds out[i,j] |= x[i,j,k]
            end
        end
    end
    return out
end

shorty(n,x) = |([x[:,:,i] for i in 1:n]...)

timholy(n,x) = any(x,3)

function runtest(n)
    x = randbool(n,n,n)

    @time out1 = simple(n,x)
    @time out2 = forloops(n,x)
    @time out3 = forloopscolfirst(n,x)
    @time out4 = shorty(n,x)
    @time out5 = timholy(n,x)

    println(all(out1 .== out2))
    println(all(out1 .== out3))
    println(all(out1 .== out4))
    println(all(out1 .== out5))
end

runtest(3)
runtest(500)

结果如下

# For 500
simple: 0.039403016 seconds (39716840 bytes allocated)
forloops: 6.259421683 seconds (77504 bytes allocated)
forloopscolfirst 1.809124505 seconds (77504 bytes allocated)
shorty: elapsed time: 0.050384062 seconds (39464608 bytes allocated)
timholy: 2.396887396 seconds (31784 bytes allocated)

所以我会选择 simple 或 shorty