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r - 从 gamlss R 对象预测二元响应概率

转载 作者:行者123 更新时间:2023-12-05 05:17:44 27 4
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我想从 gamlss R 函数预测二元类概率/类标签,如何使用预测函数来获取它们?

我有以下示例代码

library(gamlss)
X1 <- rnorm(500)
X2 <- sample(c("A","C","D","E"),500, replace = TRUE)
Y <- ifelse(X1>0.2& X2=="A",1,0)

n <- 500
training <- sample(1:n, 400)
testing <- (1:n)[-training]

fit <- gamlss(Y[training]~pcat(X2[training],Lp=1)+ri(X1[training],Lp=1),family=BI())

pred <- predict(fit,newdata = data.frame(X1,X2)[testing,],type = "response")

Error in predict.gamlss(fit, newdata = data.frame(X1, X2)[testing, ], : define the original data using the option data

有什么想法吗?

最佳答案

您需要使用gamlssdata 选项定义原始数据:

library(gamlss)
set.seed(1)
n <- 500
X1 <- rnorm(n)
X2 <- sample(c("A","C","D","E"), n, replace = TRUE)
Y <- ifelse(X1>0.2 & X2=="A", 1, 0)
dtset <- data.frame(X1, X2, Y)

training <- sample(1:n, 400)
XYtrain <- dtset[training,]
XYtest <- dtset[-training,]

fit <- gamlss(Y ~ pcat(X2, Lp=1) + ri(X1, Lp=1), family=BI(), data=XYtrain)
pred <- predict(fit, type="response", newdata=XYtest)

不幸的是,predict 现在会生成一条新的错误消息:

Error in if (p != ap) stop("the dimensions of the penalty matrix and of the design matrix are incompatible") : argument is of length zero

这个问题可以通过修改predict.gamlss使用的gamlss.ri函数来解决:

gamlss.ri <- function (x, y, w, xeval = NULL, ...) 
{
regpen <- function(sm, D, P0, lambda) {
for (it in 1:iter) {
RD <- rbind(R, sqrt(lambda) * sqrt(omega.) * D)
svdRD <- svd(RD)
rank <- sum(svdRD$d > max(svdRD$d) * .Machine$double.eps^0.8)
np <- min(p, N)
U1 <- svdRD$u[1:np, 1:rank]
y1 <- t(U1) %*% Qy
beta <- svdRD$v[, 1:rank] %*% (y1/svdRD$d[1:rank])
dm <- max(abs(sm - beta))
sm <- beta
omega. <- c(1/(abs(sm)^(2 - Lp) + kappa^2))
if (dm < c.crit)
break
}
HH <- (svdRD$u)[1:p, 1:rank] %*% t(svdRD$u[1:p, 1:rank])
edf <- sum(diag(HH))
fv <- X %*% beta
row.names(beta) <- namesX
out <- list(fv = fv, beta = beta, edf = edf, omega = omega.)
}
fnGAIC <- function(lambda, k) {
fit <- regpen(sm, D, P0, lambda)
fv <- fit$fv
GAIC <- sum(w * (y - fv)^2) + k * fit$edf
GAIC
}
X <- if (is.null(xeval))
as.matrix(attr(x, "X"))
else as.matrix(attr(x, "X"))[seq(1, length(y)), , drop=FALSE] # Added drop=FALSE
namesX <- as.character(attr(x, "namesX"))
D <- as.matrix(attr(x, "D"))
order <- as.vector(attr(x, "order"))
lambda <- as.vector(attr(x, "lambda"))
df <- as.vector(attr(x, "df"))
Lp <- as.vector(attr(x, "Lp"))
kappa <- as.vector(attr(x, "kappa"))
iter <- as.vector(attr(x, "iter"))
k <- as.vector(attr(x, "k"))
c.crit <- as.vector(attr(x, "c.crit"))
method <- as.character(attr(x, "method"))
gamlss.env <- as.environment(attr(x, "gamlss.env"))
startLambdaName <- as.character(attr(x, "NameForLambda"))
N <- sum(w != 0)
n <- nrow(X)
p <- ncol(X)
aN <- nrow(D)
ap <- ncol(D)
qrX <- qr(sqrt(w) * X, tol = .Machine$double.eps^0.8)
R <- qr.R(qrX)
Q <- qr.Q(qrX)
Qy <- t(Q) %*% (sqrt(w) * y)
if (p != ap)
stop("the dimensions of the penalty matrix and of the design matrix are incompatible")
P0 <- diag(p) * 1e-06
sm <- rep(0, p)
omega. <- rep(1, p)
tau2 <- sig2 <- NULL
lambdaS <- get(startLambdaName, envir = gamlss.env)
if (lambdaS >= 1e+07)
lambda <- 1e+07
if (lambdaS <= 1e-07)
lambda <- 1e-07
if (is.null(df) && !is.null(lambda) || !is.null(df) && !is.null(lambda)) {
fit <- regpen(sm, D, P0, lambda)
fv <- fit$fv
}
else if (is.null(df) && is.null(lambda)) {
lambda <- lambdaS
switch(method, ML = {
for (it in 1:20) {
fit <- regpen(sm, D, P0, lambda)
gamma. <- D %*% as.vector(fit$beta) * sqrt(fit$omega)
fv <- X %*% fit$beta
sig2 <- sum(w * (y - fv)^2)/(N - fit$edf)
tau2 <- sum(gamma.^2)/(fit$edf - order)
lambda.old <- lambda
lambda <- sig2/tau2
if (abs(lambda - lambda.old) < 1e-04 || lambda >
1e+05) break
}
}, GAIC = {
lambda <- nlminb(lambda, fnGAIC, lower = 1e-07, upper = 1e+07,
k = k)$par
fit <- regpen(sm, D, P0, lambda)
fv <- fit$fv
assign(startLambdaName, lambda, envir = gamlss.env)
}, )
}
else {
edf1_df <- function(lambda) {
edf <- sum(1/(1 + lambda * UDU$values))
(edf - df)
}
Rinv <- solve(R)
S <- t(D) %*% D
UDU <- eigen(t(Rinv) %*% S %*% Rinv)
lambda <- if (sign(edf1_df(0)) == sign(edf1_df(1e+05)))
1e+05
else uniroot(edf1_df, c(0, 1e+05))$root
fit <- regpen(sm, D, P0, lambda)
fv <- fit$fv
}
waug <- as.vector(c(w, rep(1, nrow(D))))
xaug <- as.matrix(rbind(X, sqrt(lambda) * D))
lev <- hat(sqrt(waug) * xaug, intercept = FALSE)[1:n]
var <- lev/w
coefSmo <- list(coef = fit$beta, lambda = lambda, edf = fit$edf,
sigb2 = tau2, sige2 = sig2, sigb = if (is.null(tau2)) NA else sqrt(tau2),
sige = if (is.null(sig2)) NA else sqrt(sig2), fv = as.vector(fv),
se = sqrt(var), Lp = Lp)
class(coefSmo) <- "ri"
if (is.null(xeval)) {
list(fitted.values = as.vector(fv), residuals = y - fv,
var = var, nl.df = fit$edf - 1, lambda = lambda,
coefSmo = coefSmo)
}
else {
ll <- dim(as.matrix(attr(x, "X")))[1]
nx <- as.matrix(attr(x, "X"))[seq(length(y) + 1, ll),
]
pred <- drop(nx %*% fit$beta)
pred
}
}
# Replace "gamlss.ri" in the package "gamlss"
assignInNamespace("gamlss.ri", gamlss.ri, pos="package:gamlss")

pred <- predict(fit, type="response", newdata=XYtest)

print(head(pred))
# [1] 2.220446e-16 2.220446e-16 2.220446e-16 4.142198e-12 2.220446e-16 2.220446e-16

关于r - 从 gamlss R 对象预测二元响应概率,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48855164/

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