julia1.0中循环速度的改进
我有一个长向量V和一个大矩阵M。我的目的是在下面的Julia代码中julia1.0中循环速度的改进,julia,Julia,我有一个长向量V和一个大矩阵M。我的目的是在下面的Julia代码中 using LinearAlgebra function myfunction(M,V) n = size(V,1) sum = 0 summ = 0 for i = 1:n-1 for j = i+1:n a= [i,j] Y = V[a] X = M[a,a] sum += Y'*i
using LinearAlgebra
function myfunction(M,V)
n = size(V,1)
sum = 0
summ = 0
for i = 1:n-1
for j = i+1:n
a= [i,j]
Y = V[a]
X = M[a,a]
sum += Y'*inv(X)*Y
summ += tr(X)*Y'*Y
end
end
return sum, summ
end
因为向量很长,矩阵很大,所以这个过程需要很长时间。我在这个问题上花了很长时间。我真的很感谢你的帮助 我只是手动展开计算以避免分配:
function myfunction2(M::AbstractMatrix{T},V::AbstractVector{T}) where {T}
n = size(V, 1)
sum = zero(T)
summ = zero(T)
for i = 2:n
for j = 1:i-1
@inbounds y1, y2 = V[i], V[j]
y11 = y1*y1
y12 = y1*y2
y22 = y2*y2
@inbounds a, b, c, d = M[i,i], M[i,j], M[j,i], M[j,j]
sum += (d*y11-(c+b)*y12+a*y22) / (a*d-b*c)
summ += (a+d)*(y11+y22)
end
end
return sum, summ
end
(注意,我对M
和V
做出了明确的假设)
编辑这是最快的
function myfunction3(M::AbstractMatrix{T},V::AbstractVector{T}) where {T}
n = size(V, 1)
sum = zero(T)
summ = zero(T)
for i = 2:n
@inbounds y1 = V[i]
@inbounds a = M[i,i]
y11 = y1*y1
for j = 1:i-1
@inbounds y2 = V[j]
y12 = y1*y2
y22 = y2*y2
@inbounds b, c, d = M[i,j], M[j,i], M[j,j]
sum += (d*y11-(c+b)*y12+a*y22) / (a*d-b*c)
summ += (a+d)*(y11+y22)
end
end
return sum, summ
end
谢谢你的回复!这很有帮助。我只是想知道并行计算或线程是否会在这方面有所帮助?由于矩阵非常大,你可以使用我在回答另一个问题时向你展示的类似方法。
function myfunction3(M::AbstractMatrix{T},V::AbstractVector{T}) where {T}
n = size(V, 1)
sum = zero(T)
summ = zero(T)
for i = 2:n
@inbounds y1 = V[i]
@inbounds a = M[i,i]
y11 = y1*y1
for j = 1:i-1
@inbounds y2 = V[j]
y12 = y1*y2
y22 = y2*y2
@inbounds b, c, d = M[i,j], M[j,i], M[j,j]
sum += (d*y11-(c+b)*y12+a*y22) / (a*d-b*c)
summ += (a+d)*(y11+y22)
end
end
return sum, summ
end