如何预测merMod对象(lme4)的术语? [英] How to predict terms of merMod objects (lme4)?

问题描述

对于简单的glm对象，我可以使用predict(fit, type = "terms")检索具有每个术语的拟合值的矩阵.

For simple glm objects, I can use predict(fit, type = "terms") to retrieve a matrix with fitted values for each term.

lmer的等效项是什么. glmer合适的型号?据我所知，predict.merMod函数不支持type = terms.

What is the equivalent for lmer resp. glmer fitted models? As far as I can see, the predict.merMod function does not support type = terms.

推荐答案

lmer的等效项是什么. glmer合适的型号?

What is the equivalent for lmer resp. glmer fitted models?

我不认为有一个.不过，您可以轻松地制作以下内容之一

I do not think there is one. Though, you can easily make one as follows

#####
# fit model with one terms which is a matrix
library(lme4)
fit <- lmer(Reaction ~ cbind(Days, (Days > 3) * Days) + (Days | Subject), 
            sleepstudy)

#####
# very similar code to `predict.lm`
pred_terms_merMod <- function(fit, newdata){
  tt <- terms(fit)
  beta <- fixef(fit)

  mm <- model.matrix(tt, newdata)
  aa <- attr(mm, "assign")
  ll <- attr(tt, "term.labels")
  hasintercept <- attr(tt, "intercept") > 0L
  if (hasintercept) 
    ll <- c("(Intercept)", ll)
  aaa <- factor(aa, labels = ll)
  asgn <- split(order(aa), aaa)
  if (hasintercept) {
    asgn$"(Intercept)" <- NULL
    avx <- colMeans(mm)
    termsconst <- sum(avx * beta)
  }
  nterms <- length(asgn)
  if (nterms > 0) {
    predictor <- matrix(ncol = nterms, nrow = NROW(mm))
    dimnames(predictor) <- list(rownames(mm), names(asgn))
    if (hasintercept) 
      mm <- sweep(mm, 2L, avx, check.margin = FALSE)
    for (i in seq.int(1L, nterms, length.out = nterms)) {
      idx <- asgn[[i]]
      predictor[, i] <- mm[, idx, drop = FALSE] %*% beta[idx]
    }
  } else {
    predictor <- ip <- matrix(0, n, 0L)
  }
  attr(predictor, "constant") <- if (hasintercept) termsconst else 0
  predictor
}

# use the function
newdata <- data.frame(Days = c(1, 5), Reaction = c(0, 0))
(out <- pred_terms_merMod(fit, newdata))
#R>   cbind(Days, (Days > 3) * Days)
#R> 1                        -21.173
#R> 2                         21.173
#R> attr(,"constant")
#R> [1] 283.24

#####
# confirm results
beta. <-  fixef(fit)
beta.[1] + beta.[2]
#R> (Intercept) 
#R>      262.07 
out[1] + attr(out, "constant")
#R> [1] 262.07

beta.[1] + (beta.[2] + beta.[3]) * 5
#R> (Intercept) 
#R>      304.41 
out[2] + attr(out, "constant")
#R> [1] 304.41

据我所知，将上述内容扩展为还包括标准错误应该很简单.

Extending the above to also include standard errors should be straightforward as far as I gather.

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