Julia CUDA - 减少矩阵列

Question

考虑以下内核，它沿二维矩阵行减少

function row_sum!(x, ncol, out)
    """out = sum(x, dims=2)"""
    row_idx = (blockIdx().x-1) * blockDim().x + threadIdx().x
    for i = 1:ncol
        @inbounds out[row_idx] += x[row_idx, i]
    end
    return
end

N = 1024
x = CUDA.rand(Float64, N, 2*N)
out = CUDA.zeros(Float64, N)
@cuda threads=256 blocks=4 row_sum!(x, size(x)[2], out)
isapprox(out, sum(x, dims=2))  # true

除了沿着列（二维矩阵）进行缩减外，我该如何编写类似的内核？特别是，我如何获取每一列的索引，类似于我们如何使用 row_idx?

获取每一行的索引

Answer 1

代码如下：

function col_sum!(x, nrow, out)
    """out = sum(x, dims=1)"""
    col_idx = (blockIdx().x-1) * blockDim().x + threadIdx().x
    for i = 1:nrow
        @inbounds out[col_idx] += x[i, col_idx]
    end
    return
end

N = 1024
x = CUDA.rand(Float64, N, 2N)
out = CUDA.zeros(Float64, 2N)
@cuda threads=256 blocks=8 col_sum!(x, size(x, 1), out)

这是测试：

julia> isapprox(out, vec(sum(x, dims=1)))
true

如您所见，结果向量的大小现在是 2N 而不是 N，因此我们必须相应地调整 blocks 的数量（即乘以 2 现在我们有 8 而不是 4)

可以在这里找到更多资料：https://juliagpu.gitlab.io/CUDA.jl/tutorials/introduction/

Julia CUDA - 减少矩阵列

Julia CUDA - Reduce matrix columns

cuda

julia

gpu