我正在尝试在 Julia 中实现一个简单的正则化逻辑回归算法。我想使用 Optim.jl 库来最小化我的成本函数,但我无法让它工作。
我的成本函数和梯度如下:
function cost(X, y, theta, lambda)
m = length(y)
h = sigmoid(X * theta)
reg = (lambda / (2*m)) * sum(theta[2:end].^2)
J = (1/m) * sum( (-y).*log(h) - (1-y).*log(1-h) ) + reg
return J
end
function grad(X, y, theta, lambda, gradient)
m = length(y)
h = sigmoid(X * theta)
# gradient = zeros(size(theta))
gradient = (1/m) * X' * (h - y)
gradient[2:end] = gradient[2:end] + (lambda/m) * theta[2:end]
return gradient
end
(其中
theta
是假设函数的参数向量,lambda
是正则化参数。)然后,根据此处给出的说明:https://github.com/JuliaOpt/Optim.jl 我尝试像这样调用优化函数:
# those are handle functions I define to pass them as arguments:
c(theta::Vector) = cost(X, y, theta, lambda)
g!(theta::Vector, gradient::Vector) = grad(X, y, theta, lambda, gradient)
# then I do
optimize(c,some_initial_theta)
# or maybe
optimize(c,g!,initial_theta,method = :l_bfgs) # try a different algorithm
在这两种情况下,它都说它无法收敛并且输出看起来有点尴尬:
julia> optimize(c,initial_theta)
Results of Optimization Algorithm
* Algorithm: Nelder-Mead
* Starting Point: [0.0,0.0,0.0,0.0,0.0]
* Minimum: [1.7787162051775145,3.4584135105727145,-6.659680628594007,4.776952006060713,1.5034743945407143]
* Value of Function at Minimum: -Inf
* Iterations: 1000
* Convergence: false
* |x - x'| < NaN: false
* |f(x) - f(x')| / |f(x)| < 1.0e-08: false
* |g(x)| < NaN: false
* Exceeded Maximum Number of Iterations: true
* Objective Function Calls: 1013
* Gradient Call: 0
julia> optimize(c,g!,initial_theta,method = :l_bfgs)
Results of Optimization Algorithm
* Algorithm: L-BFGS
* Starting Point: [0.0,0.0,0.0,0.0,0.0]
* Minimum: [-6.7055e-320,-2.235e-320,-6.7055e-320,-2.244e-320,-6.339759952602652e-7]
* Value of Function at Minimum: 0.693148
* Iterations: 1
* Convergence: false
* |x - x'| < 1.0e-32: false
* |f(x) - f(x')| / |f(x)| < 1.0e-08: false
* |g(x)| < 1.0e-08: false
* Exceeded Maximum Number of Iterations: false
* Objective Function Calls: 75
* Gradient Call: 75
问题
我的方法(来自我的第一个代码 list )不正确吗?还是我滥用了 Optim.jl 函数?无论哪种方式,在这里定义和最小化成本函数的正确方法是什么?
这是我第一次和 Julia 在一起,可能我做错了什么,但我不知道到底是什么。任何帮助将不胜感激!
编辑
X
和 y
是训练集,X
是一个 90x5 的矩阵,y
是一个 90x1 的向量(也就是说,我的训练集取自 Iris - 我认为这无关紧要)。 最佳答案
下面是我使用闭包和柯里化进行逻辑回归的成本和梯度计算函数(适用于那些已经习惯了返回成本和梯度的函数的人的版本):
function cost_gradient(θ, X, y, λ)
m = length(y)
return (θ::Array) -> begin
h = sigmoid(X * θ)
J = (1 / m) * sum(-y .* log(h) .- (1 - y) .* log(1 - h)) + λ / (2 * m) * sum(θ[2:end] .^ 2)
end, (θ::Array, storage::Array) -> begin
h = sigmoid(X * θ)
storage[:] = (1 / m) * (X' * (h .- y)) + (λ / m) * [0; θ[2:end]]
end
end
Sigmoid函数实现:
sigmoid(z) = 1.0 ./ (1.0 + exp(-z))
要在 Optim.jl 中应用
cost_gradient
,请执行以下操作:using Optim
#...
# Prerequisites:
# X size is (m,d), where d is the number of training set features
# y size is (m,1)
# λ as the regularization parameter, e.g 1.5
# ITERATIONS number of iterations, e.g. 1000
X=[ones(size(X,1)) X] #add x_0=1.0 column; now X size is (m,d+1)
initialθ = zeros(size(X,2),1) #initialTheta size is (d+1, 1)
cost, gradient! = cost_gradient(initialθ, X, y, λ)
res = optimize(cost, gradient!, initialθ, method = ConjugateGradient(), iterations = ITERATIONS);
θ = Optim.minimizer(res);
现在,您可以轻松预测(例如训练集验证):
predictions = sigmoid(X * θ) #X size is (m,d+1)
要么尝试我的方法,要么将其与您的实现进行比较。
关于regression - 使用 Optim.jl 在 Julia 中进行逻辑回归,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/32703119/