来自 glmer 的反向变换系数与用于预测的缩放自变量

Question

我使用 lme4 包安装了一个混合模型。在拟合模型之前，我使用 scale() 函数转换了自变量。我现在想使用 predict() 在图表上显示我的结果，因此我需要将预测数据恢复到原始比例。我该怎么做？

简化示例：

database <- mtcars

# Scale data
database$wt <- scale(mtcars$wt)
database$am <- scale(mtcars$am)

# Make model
model.1 <- glmer(vs ~ scale(wt) + scale(am) + (1|carb), database, family = binomial, na.action = "na.fail")

# make new data frame with all values set to their mean
xweight <- as.data.frame(lapply(lapply(database[, -1], mean), rep, 100))

# make new values for wt
xweight$wt <- (wt = seq(min(database$wt), max(database$wt), length = 100))

#  predict from new values
a <- predict(model.1, newdata = xweight, type="response", re.form=NA)

# returns scaled prediction

我已经尝试使用 this example 来反向转换预测：

# save scale and center values
scaleList <- list(scale = attr(database$wt, "scaled:scale"),
              center = attr(database$wt, "scaled:center"))

# back-transform predictions
a.unscaled <- a * scaleList$scale + scaleList$center

# Make model with unscaled data to compare
un.model.1 <- glmer(vs ~ wt + am + (1|carb), mtcars, family = binomial, na.action = "na.fail")

# make new data frame with all values set to their mean
un.xweight <- as.data.frame(lapply(lapply(mtcars[, -1], mean), rep, 100))

# make new values for wt
un.xweight$wt <- (wt = seq(min(mtcars$wt), max(mtcars$wt), length = 100))

#  predict from new values
b <- predict(un.model.1, newdata = xweight, type="response", re.form=NA)

all.equal(a.unscaled,b)
# [1] "Mean relative difference: 0.7223061"

这不起作用 - 应该没有任何区别。 我做错了什么？

我也看过一些类似的问题，但没有设法将任何问题应用到我的案例中 (How to unscale the coefficients from an lmer()-model fitted with a scaled response, unscale and uncenter glmer parameters, Scale back linear regression coefficients in R from scaled and centered data, https://stats.stackexchange.com/questions/302448/back-transform-mixed-effects-models-regression-coefficients-for-fixed-effects-f)。

Answer 1

您的方法的问题在于它仅 "unscales" 基于 wt 变量，而您缩放了回归模型中的所有变量。一种有效的方法是使用原始数据框上使用的 centering/scaling 值调整新（预测）数据框中的所有变量：

## scale variable x using center/scale attributes
## of variable y
scfun <- function(x,y) {
    scale(x, 
          center=attr(y,"scaled:center"), 
          scale=attr(y,"scaled:scale"))
}
## scale prediction frame
xweight_sc <- transform(xweight,
                        wt = scfun(wt, database$wt),
                        am = scfun(am, database$am))
## predict
p_unsc <- predict(model.1, 
                  newdata=xweight_sc, 
                  type="response", re.form=NA)

将此 p_unsc 与未缩放模型（代码中的 b）的预测进行比较，即 all.equal(b,p_unsc)，得出 TRUE。

另一种合理的方法是

unscale/uncenter 使用链接问题之一（例如 this one）中提出的 "unscaling" 方法的所有参数，生成系数向量 beta_unsc
根据您的预测框架构建合适的模型矩阵：

X <- model.matrix(formula(model,fixed.only=TRUE), 
         newdata=pred_frame)

计算线性预测器和反变换：

pred <- plogis(X %*% beta_unsc)

来自 glmer 的反向变换系数与用于预测的缩放自变量

Back-transform coefficients from glmer with scaled independent variables for prediction

regression

r

scale

lme4