Unsure why I'm getting errors with LoopVectorization

New to Julia
1

Julia v1.8.5
LoopVectorization v0.12.159

Working on trying to maximize performance of some functions and I’m getting errors. Here is a minimum code example of one:

function test(a,n)
    rows = size(a,1)
    cols = size(a,2)

    println(rows,":",cols)
    # a[n,:] = mean.(eachcol(a[1:n,:]))
    @turbo for j in 1:cols
        _s = 0.0
        for i in 1:n
            _s += a[i,j]
        end
        a[n,j] = _s / n
    end

end

N = 100_000
x = rand(N,2)
test(x,10)

y = rand(N)
test(y,10)

The second call gives: ERROR: MethodError: reducing over an empty collection is not allowed; consider supplying init to the reducer

I realize the first passes a Matrix and the second passes a Vector.

Function works as expected without the @turbo.

Why is this error being generated and how can I modify the code for this to work with both Matrix and Vector types?

2

At first glance it seems this code is explicitly designed for a two-dimensional array, for instance because it mentions rows and cols. What do you expect to achieve when giving it a vector?

3

Redefine the existing method as test(a::AbstractaMatrix, n), and add a dispatch for test(a::AbstractVector, n), which would have only a single loop.

You can define different methods for the different combinations of input parameter types: https://docs.julialang.org/en/v1/manual/methods/

1 Like
4

If you don’t want, as others suggest, to define a vector-specific method, try moving @turbo to the inner loop, which is the more demanding one.

function test(a,n)
    rows = size(a,1)
    cols = size(a,2)

    println(rows,":",cols)
    # a[n,:] = mean.(eachcol(a[1:n,:]))
    for j in 1:cols
        _s = 0.0
        @turbo     for i in 1:n
            _s += a[i,j]
        end
        a[n,j] = _s / n
    end

end

Or reshape the vector as a column matrix. (although, it doesn’t seem like a nice solution to me)


function test1(a,n)
    rows = size(a,1)
    cols = size(a,2)
    a=reshape(a,:,cols)
    println(rows,":",cols)
    # a[n,:] = mean.(eachcol(a[1:n,:]))
    @turbo for j in 1:cols
        _s = 0.0
        for i in 1:n
            _s += a[i,j]
        end
        a[n,j] = _s / n
    end
end

PS
I don’t know the terms and applicability limits of the macro, but the highlighted problem deserves further investigation by those who know the macro well