背景:nlm函数在R是一个通用的优化程序,使用牛顿的方法。为了优化一个函数,牛顿方法需要函数,以及函数的第一和第二导数(梯度向量和Hessian矩阵)。在R中,nlm函数允许您指定与梯度和Hessian计算相对应的R函数,或者可以保留这些未指定的函数,并根据数值导数(通过deriv函数)提供数值解。通过提供计算梯度和Hessian的函数可以找到更精确的解,因此它是一个有用的特性。
我的问题是:当提供解析的Hessian函数时,nlm函数速度较慢,并且常常无法在合理的时间内收敛。我猜这是底层代码中的某种错误,但我很高兴错了。有没有一种方法可以使nlm更好地使用解析的Hessian矩阵?
示例:下面的R代码使用逻辑回归示例演示了这个问题,其中
log(Pr(Y=1)/Pr(Y=0)) = b0 + Xb
其中X是维数N的多元正规的p,b是长度p的系数的向量。
library(mvtnorm)
# example demonstrating a problem with NLM
expit <- function(mu) {1/(1+exp(-mu))}
mk.logit.data <- function(N,p){
set.seed(1232)
U = matrix(runif(p*p), nrow=p, ncol=p)
S = 0.5*(U+t(U)) + p*diag(rep(1,p))
X = rmvnorm(N, mean = runif(p, -1, 1), sigma = S)
Design = cbind(rep(1, N), X)
beta = sort(sample(c(rep(0,p), runif(1))))
y = rbinom(N, 1, expit(Design%*%beta))
list(X=X,y=as.numeric(y),N=N,p=p)
}
# function to calculate gradient vector at given coefficient values
logistic_gr <- function(beta, y, x, min=TRUE){
mu = beta[1] + x %*% beta[-1]
p = length(beta)
n = length(y)
D = cbind(rep(1,n), x)
gri = matrix(nrow=n, ncol=p)
for(j in 1:p){
gri[,j] = D[,j]*(exp(-mu)*y-1+y)/(1+exp(-mu))
}
gr = apply(gri, 2, sum)
if(min) gr = -gr
gr
}
# function to calculate Hessian matrix at given coefficient values
logistic_hess <- function(beta, y, x, min=TRUE){
# allow to fail with NA, NaN, Inf values
mu = beta[1] + x %*% beta[-1]
p = length(beta)
n = length(y)
D = cbind(rep(1,n), x)
h = matrix(nrow=p, ncol=p)
for(j in 1:p){
for(k in 1:p){
h[j,k] = -sum(D[,j]*D[,k]*(exp(-mu))/(1+exp(-mu))^2)
}
}
if(min) h = -h
h
}
# function to calculate likelihood (up to a constant) at given coefficient values
logistic_ll <- function(beta, y,x, gr=FALSE, he=FALSE, min=TRUE){
mu = beta[1] + x %*% beta[-1]
lli = log(expit(mu))*y + log(1-expit(mu))*(1-y)
ll = sum(lli)
if(is.na(ll) | is.infinite(ll)) ll = -1e16
if(min) ll=-ll
# the below specification is required for using analytic gradient/Hessian in nlm function
if(gr) attr(ll, "gradient") <- logistic_gr(beta, y=y, x=x, min=min)
if(he) attr(ll, "hessian") <- logistic_hess(beta, y=y, x=x, min=min)
ll
}第一个例子,使用p=3:
dat = mk.logit.data(N=100, p=3)glm函数估计可供参考。nlm应该给出同样的答案,考虑到由于近似而产生的小误差。
(glm.sol <- glm(dat$y~dat$X, family=binomial()))$coefficients
> (Intercept) dat$X1 dat$X2 dat$X3
> 0.00981465 0.01068939 0.04417671 0.01625381
# works when correct analytic gradient is specified
(nlm.sol1 <- nlm(p=runif(dat$p+1), f=logistic_ll, gr=TRUE, y=dat$y, x=dat$X))$estimate
> [1] 0.009814547 0.010689396 0.044176627 0.016253966
# works, but less accurate when correct analytic hessian is specified (even though the routine notes convergence is probable)
(nlm.sol2 <- nlm(p=runif(dat$p+1), f=logistic_ll, gr=TRUE, he=TRUE, y=dat$y, x=dat$X, hessian = TRUE, check.analyticals=TRUE))$estimate
> [1] 0.009827701 0.010687278 0.044178416 0.016255630但是当p更大时,问题变得很明显,这里是10。
dat = mk.logit.data(N=100, p=10)同样,glm解决方案可供参考。nlm应该给出同样的答案,考虑到由于近似而产生的小误差。
(glm.sol <- glm(dat$y~dat$X, family=binomial()))$coefficients
> (Intercept) dat$X1 dat$X2 dat$X3 dat$X4 dat$X5 dat$X6 dat$X7
> -0.07071882 -0.08670003 0.16436630 0.01130549 0.17302058 0.03821008 0.08836471 -0.16578959
> dat$X8 dat$X9 dat$X10
> -0.07515477 -0.08555075 0.29119963
# works when correct analytic gradient is specified
(nlm.sol1 <- nlm(p=runif(dat$p+1), f=logistic_ll, gr=TRUE, y=dat$y, x=dat$X))$estimate
> [1] -0.07071879 -0.08670005 0.16436632 0.01130550 0.17302057 0.03821009 0.08836472
> [8] -0.16578958 -0.07515478 -0.08555076 0.29119967
# fails to converge in 5000 iterations when correct analytic hessian is specified
(nlm.sol2 <- nlm(p=runif(dat$p+1), f=logistic_ll, gr=TRUE, he=TRUE, y=dat$y, x=dat$X, hessian = TRUE, iterlim=5000, check.analyticals=TRUE))$estimate
> [1] 0.31602065 -0.06185190 0.10775381 -0.16748897 0.05032156 0.34176104 0.02118631
> [8] -0.01833671 -0.20364929 0.63713991 0.18390489编辑:我还应该补充说,我已经通过多种不同的方法确认了正确的Hessian矩阵。
发布于 2019-05-21 21:01:05
我尝试了这些代码,但一开始它似乎使用的是与我在CRAN上找到的不同的rmvnorm。我在dae软件包中找到了一个rmvnorm,然后在mvtnorm包中找到了一个。后者是可以使用的。
nlm()是关于上述发布时间的补丁。我目前正在尝试验证这些补丁,现在它似乎正常工作了。请注意,我是许多R的优化代码的作者,包括optim()中的3/5。
uottawa.ca的nashjc
密码在下面。
发布于 2019-05-21 21:33:30
经修订的守则:
# example demonstrating a problem with NLM
expit <- function(mu) {1/(1+exp(-mu))}
mk.logit.data <- function(N,p){
set.seed(1232)
U = matrix(runif(p*p), nrow=p, ncol=p)
S = 0.5*(U+t(U)) + p*diag(rep(1,p))
X = rmvnorm(N, mean = runif(p, -1, 1), sigma = S)
Design = cbind(rep(1, N), X)
beta = sort(sample(c(rep(0,p), runif(1))))
y = rbinom(N, 1, expit(Design%*%beta))
list(X=X,y=as.numeric(y),N=N,p=p)
}
# function to calculate gradient vector at given coefficient values
logistic_gr <- function(beta, y, x, min=TRUE){
mu = beta[1] + x %*% beta[-1]
p = length(beta)
n = length(y)
D = cbind(rep(1,n), x)
gri = matrix(nrow=n, ncol=p)
for(j in 1:p){
gri[,j] = D[,j]*(exp(-mu)*y-1+y)/(1+exp(-mu))
}
gr = apply(gri, 2, sum)
if(min) gr = -gr
gr
}
# function to calculate Hessian matrix at given coefficient values
logistic_hess <- function(beta, y, x, min=TRUE){
# allow to fail with NA, NaN, Inf values
mu = beta[1] + x %*% beta[-1]
p = length(beta)
n = length(y)
D = cbind(rep(1,n), x)
h = matrix(nrow=p, ncol=p)
for(j in 1:p){
for(k in 1:p){
h[j,k] = -sum(D[,j]*D[,k]*(exp(-mu))/(1+exp(-mu))^2)
}
}
if(min) h = -h
h
}
# function to calculate likelihood (up to a constant) at given coefficient values
logistic_ll <- function(beta, y,x, gr=FALSE, he=FALSE, min=TRUE){
mu = beta[1] + x %*% beta[-1]
lli = log(expit(mu))*y + log(1-expit(mu))*(1-y)
ll = sum(lli)
if(is.na(ll) | is.infinite(ll)) ll = -1e16
if(min) ll=-ll
# the below specification is required for using analytic gradient/Hessian in nlm function
if(gr) attr(ll, "gradient") <- logistic_gr(beta, y=y, x=x, min=min)
if(he) attr(ll, "hessian") <- logistic_hess(beta, y=y, x=x, min=min)
ll
}
##!!!! NOTE: Must have this library loaded
library(mvtnorm)
dat = mk.logit.data(N=100, p=3)
(glm.sol <- glm(dat$y~dat$X, family=binomial()))$coefficients
# works when correct analytic gradient is specified
(nlm.sol1 <- nlm(p=runif(dat$p+1), f=logistic_ll, gr=TRUE, y=dat$y, x=dat$X))$estimate
# works, but less accurate when correct analytic hessian is specified (even though the routine notes convergence is probable)
(nlm.sol2 <- nlm(p=runif(dat$p+1), f=logistic_ll, gr=TRUE, he=TRUE, y=dat$y, x=dat$X, hessian = TRUE, check.analyticals=TRUE))$estimate
dat = mk.logit.data(N=100, p=10)
# Again, glm solution for reference. nlm should give the same answer, allowing for small errors due to approximation.
(glm.sol <- glm(dat$y~dat$X, family=binomial()))$coefficients
# works when correct analytic gradient is specified
(nlm.sol1 <- nlm(p=runif(dat$p+1), f=logistic_ll, gr=TRUE, y=dat$y, x=dat$X))$estimate
# fails to converge in 5000 iterations when correct analytic hessian is specified
(nlm.sol2 <- nlm(p=runif(dat$p+1), f=logistic_ll, gr=TRUE, he=TRUE, y=dat$y, x=dat$X, hessian = TRUE, iterlim=5000, check.analyticals=TRUE))$estimatehttps://stackoverflow.com/questions/48332808
复制相似问题