Julia跳跃多元ML估计
我试图在Julia中使用跳跃和NLopt解算器在线性回归设置中执行正态分布变量的ML估计 有一个很好的工作示例,但是如果我尝试估计回归参数(斜率),代码编写起来会变得非常乏味,特别是当参数空间增加时 也许有人知道如何写得更简洁。这是我的密码:Julia跳跃多元ML估计,julia,julia-jump,Julia,Julia Jump,我试图在Julia中使用跳跃和NLopt解算器在线性回归设置中执行正态分布变量的ML估计 有一个很好的工作示例,但是如果我尝试估计回归参数(斜率),代码编写起来会变得非常乏味,特别是当参数空间增加时 也许有人知道如何写得更简洁。这是我的密码: #type definition to store data type data n::Int A::Matrix β::Vector y::Vector ls::Vector err::Vector end
#type definition to store data
type data
n::Int
A::Matrix
β::Vector
y::Vector
ls::Vector
err::Vector
end
#generate regression data
function Data( n = 1000 )
A = [ones(n) rand(n, 2)]
β = [2.1, 12.9, 3.7]
y = A*β + rand(Normal(), n)
ls = inv(A'A)A'y
err = y - A * ls
data(n, A, β, y, ls, err)
end
#initialize data
d = Data()
println( var(d.y) )
function ml( )
m = Model( solver = NLoptSolver( algorithm = :LD_LBFGS ) )
@defVar( m, b[1:3] )
@defVar( m, σ >= 0, start = 1.0 )
#this is the working example.
#As you can see it's quite tedious to write
#and becomes rather infeasible if there are more then,
#let's say 10, slope parameters to estimate
@setNLObjective( m, Max,-(d.n/2)*log(2π*σ^2) \\cont. next line
-sum{(d.y[i]-d.A[i,1]*b[1] \\
-d.A[i,2]*b[2] \\
-d.A[i,3]*b[3])^2, i=1:d.n}/(2σ^2) )
#julia returns:
> slope: [2.14,12.85,3.65], variance: 1.04
#which is what is to be expected
#however:
#this is what I would like the code to look like:
@setNLObjective( m, Max,-(d.n/2)*log(2π*σ^2) \\
-sum{(d.y[i]-(d.A[i,j]*b[j]))^2, \\
i=1:d.n, j=1:3}/(2σ^2) )
#I also tried:
@setNLObjective( m, Max,-(d.n/2)*log(2π*σ^2) \\
-sum{sum{(d.y[i]-(d.A[i,j]*b[j]))^2, \\
i=1:d.n}, j=1:3}/(2σ^2) )
#but unfortunately it returns:
> slope: [10.21,18.89,15.88], variance: 54.78
solve(m)
println( getValue(b), " ", getValue(σ^2) )
end
ml()
有什么想法吗
编辑
如Reza所述,工作示例如下:
@setNLObjective( m, Max,-(d.n/2)*log(2π*σ^2) \\
-sum{(d.y[i]-sum{d.A[i,j]*b[j],j=1:3})^2,
i=1:d.n}/(2σ^2) )
我没有跟踪您的代码,但在任何地方,我希望以下内容适用于您:
sum([(d.y[i]-sum([d.A[i,j]*b[j] for j=1:3]))^2 for i=1:d.n])
正如@iaindanning所提到的,跳转包在其宏中有一种特殊的求和语法,因此更有效和抽象的方法是:
sum{sum{(d.y[i]-d.A[i,j]*b[j], j=1:3}^2,i=1:d.n}
sum{}
语法是一种特殊的语法,仅在跳转宏中有效,是sums的首选语法
因此,您的示例可以写成:
function ml( )
m = Model( solver = NLoptSolver( algorithm = :LD_LBFGS ) )
@variable( m, b[1:3] )
@variable( m, σ >= 0, start = 1.0 )
@NLobjective(m, Max,
-(d.n/2)*log(2π*σ^2)
- sum{
sum{(d.y[i]-d.A[i,j]*b[j], j=1:3}^2,
i=1:d.n}/(2σ^2) )
我将其扩展到多行,使其尽可能清晰
从技术上讲,雷扎的回答没有错,但不是惯用的跳跃,对于大型车型来说也没有那么有效。的确如此。非常感谢,谢谢你的回答。我知道这在技术上是错误的(由于括号,正如你正确指出的那样),但他完全按照我的需要写出了问题。