lapply:拟合数千个混合模型并能够提取 lsmeans

lapply: Fitting thousands of mixed models and being able to extract lsmeans

我有一个适用于数据集的线性混合模型 (lme4) 的公式列表 (> 10,000)。我成功地使用了 lapply() 和一个包含 tryCatch() 的自定义函数来适应这些模型。现在我想提取所有这些模型的 P 值和 lsmeans。我已成功提取 P 值,但 lsmeans 函数遇到错误。

library(lme4)
library(lmerTest)
library(pbkrtest)
library(lsmeans)

formulaS <- list() #Not going to detail generation of list, generically: 'Yvar~X1*X2+(1|subject)'
dataSET <- X #dataframe with first 3 columns containing fixed and random factors, 
             # as well as >10,000 columns of variables of interest

modelSeq <- function (x, dat) {
  return(tryCatch(lmer(x, data = dat), error=function(e) NULL))
}

modelsOutput <- lapply(formulaS, function(x) modelSeq(x, dat = dataSET))

lsmeans(modelsOutput[[1]], pairwise ~ X1:X2) #recieves error

solve.default(L %% V0 %% t(L), L) 错误: Lapack例程dgesv: system is exactly singular: U[1,1] = 0

我认为这不是模型问题的原因是,如果我单独拟合模型,我可以很好地提取 lsmeans。是否有关于 1) 为什么我不能提取 lsmeans,2) 如何有效地提取均值,或 3) 另一种有效方法的评论。

谢谢!

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更新和编辑:这是 RNAseq 数据,随着时间的推移,我正在使用重复的受试者样本,所以 >10,000 个模型具有相同的固定和随机效应,描述了实验设计。响应(基因)是唯一变化的变量。我试图在下面的代码中更明确地说明这一点。认识到具有身份 link 的混合模型可能不适合数据,我在下面使用了新的包装器。我在提取方法时仍然遇到问题。此外,欢迎任何关于更合适、更省时的 P 值计算方法的评论。

library(lme4)
library(blmeco)
library(ggeffects)

formulaS <- list() #Not going to detail generation of list, generically: 'GeneI~TRT*TIME+(1|subject)'
dataSET <- X #dataframe with first 3 columns containing fixed and random factors, 
             # as well as >10,000 columns of variables of interest (gene TPM)

wrap.glmer.nb <- function (modelForm, dat) {
  m <- tryCatch(glmer.nb(formula = modelForm, data = dat), error = function(e) NULL)
  if (!is.null(m)) {
    m.disp <- tryCatch(dispersion_glmer(m), error = function(e) NULL)
    m.wald <- tryCatch(anova(m), error = function(e) NULL)
    m.means.c <- tryCatch(ggemmeans(model = m, terms = c('TRT')), error = function(e) NULL)
    m.means.e <- tryCatch(ggemmeans(model = m, terms = c('TIME')), error = function(e) NULL)
    m.means.cxe <- tryCatch(ggemmeans(model = m, terms = c('TRT', 'TIME')), error = function(e) NULL)
    x <- list(m.disp, m.wald, m.means.c, m.means.e, m.means.cxe)
    print(paste0('Done with a model at ', Sys.time()))
    return(x)
  } else{
    x <- m
    return(x)
  }
}

startTime <- Sys.time()
modelOUTPUTS <- lapply(formulaS, function(modelForm) wrap.glmer.nb(modelForm, dat = dataSET))
endTime <- Sys.time()
print(paste('Victory! The analysis took:', endTime - startTime))

如果您在 modelSeq():

中添加一行,您的原始设置就会起作用
modelSeq <- function (x, dat) {
  environment(x) <- environment()
  return(tryCatch(lmer(x, data = dat), error=function(e) NULL))
}

这会将公式的环境设置为函数体的环境,从而可以找到名为 dat.

的数据集

类似的例子:

fitm <- function(formula, data, ...) {
    environment(formula) <- environment()
    lm(formula, data = data, ...)
}

fl <- list(breaks ~ tension, breaks ~ wool + tension, breaks ~ wool*tension)

md <- lapply(fl, fitm, data = warpbreaks[c(1,2,3,5,8,13,21,34,54), ])

lapply(md, function(m) emmeans(m, "tension"))

产生:

NOTE: Results may be misleading due to involvement in interactions

[[1]]
 tension emmean    SE df lower.CL upper.CL
 L         41.2  6.64  6    24.91     57.4
 M         17.0 16.27  6   -22.82     56.8
 H         26.0 11.51  6    -2.16     54.2

Confidence level used: 0.95 

[[2]]
 tension emmean    SE df lower.CL upper.CL
 L         41.6  8.91  5    18.73     64.5
 M         17.7 19.41  5   -32.21     67.6
 H         26.0 12.59  5    -6.38     58.4

Results are averaged over the levels of: wool 
Confidence level used: 0.95 

[[3]]
 tension emmean   SE df lower.CL upper.CL
 L         41.1 10.9  4     10.9     71.3
 M       nonEst   NA NA       NA       NA
 H         26.0 14.1  4    -13.0     65.0

Results are averaged over the levels of: wool 
Confidence level used: 0.95 

顺便说一句,你不需要 lsmeans 包;它只是 emmeans 的前端。其实lsmeans函数本身就在emmeans;它只是运行 emmeans 并重新标记结果。