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这是运行单个固定效果方法的三种不同方法,它们给出或多或少相同的结果(见下文)。我的主要问题是如何使用第二个模型(model_plm)或第三个模型(model_felm)获得预测概率或平均边际效应。我知道如何使用第一个模型model_lmggeffects

由于我有超过一百万个人,我的模型只能使用model_plmand model_felm。如果我使用model_lm,因为他们在模型中受到控制,所以需要花费大量时间来运行一百万个人。我还收到以下错误:Error: vector memory exhausted (limit reached?). 我检查了 StackOverflow 上的许多线程以解决该错误,但似乎没有任何解决方法。

我想知道是否有一种有效的方法来解决这个问题。我的主要兴趣是提取交互的预测概率residence*union。我通常使用以下软件包之一提取预测概率或平均边际效应ggeffectsemmeansmargins.

library(lfe)
library(plm)
library(ggeffects)
data("Males")  

model_lm = lm(wage ~ exper + residence+health + residence*union +factor(nr)-1, data=Males)
model_plm = plm(wage ~ exper + residence + health + residence*union,model = "within", index=c("nr", "year"), data=Males)
model_felm = felm(wage ~ exper + residence + health + residence*union | nr, data= Males)

pred_ggeffects <- ggpredict(model_lm, c("residence","union"), 
                    vcov.fun = "vcovCL", 
                    vcov.type = "HC1",
                    vcov.args = list(cluster = Males$nr))
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3 回答 3

2

我尝试调整公式/数据集以使 emmeans 和 plm 发挥出色。让我知道这里是否有什么东西。经过一些测试,我意识到 biglm 的答案不会为一百万个人削减它。

library(emmeans)
library(plm)
data("Males")  

## this runs but we need to get an equivalent result with expanded formula
## and expanded dataset
model_plm = plm(wage ~ exper + residence + health + residence*union,model = "within", index=c("nr"), data=Males)

## expanded dataset
Males2 <- data.frame(wage=Males[complete.cases(Males),"wage"],
                     model.matrix(wage ~ exper + residence + health + residence*union, Males), 
                     nr=Males[complete.cases(Males),"nr"])


(fmla2 <- as.formula(paste("wage ~ ", paste(names(coef(model_plm)), collapse= "+"))))

## expanded formula
model_plm2 <- plm(fmla2,
                  model = "within",
                  index=c("nr"), 
                  data=Males2)

(fmla2_rg <- as.formula(paste("wage ~ -1 +", paste(names(coef(model_plm)), collapse= "+"))))

plm2_rg <- qdrg(fmla2_rg,
                data = Males2,
                coef = coef(model_plm2),
                vcov = vcov(model_plm2),
                df = model_plm2$df.residual)

plm2_rg

### when all 3 residences are 0, that's `rural area`
### then just pick the rows when one of the residences are 1
emmeans(plm2_rg, c("residencenorth_east","residencenothern_central","residencesouth", "unionyes"))

在删除了一些行之后,这给出了:

> ### when all 3 residences are 0, that's `rural area`
> ### then just pick the rows when one of the residences are 1
> emmeans(plm2_rg, c("residencenorth_east","residencenothern_central","residencesouth", "unionyes"))
 residencenorth_east residencenothern_central residencesouth unionyes emmean     SE   df lower.CL upper.CL
                   0                        0              0        0 0.3777 0.0335 2677  0.31201    0.443
                   1                        0              0        0 0.3301 0.1636 2677  0.00929    0.651
                   0                        1              0        0 0.1924 0.1483 2677 -0.09834    0.483
                   0                        0              1        0 0.2596 0.1514 2677 -0.03732    0.557
                   0                        0              0        1 0.2875 0.1473 2677 -0.00144    0.576
                   1                        0              0        1 0.3845 0.1647 2677  0.06155    0.708
                   0                        1              0        1 0.3326 0.1539 2677  0.03091    0.634
                   0                        0              1        1 0.3411 0.1534 2677  0.04024    0.642

Results are averaged over the levels of: healthyes 
Confidence level used: 0.95
于 2021-10-23T03:51:21.367 回答
2

问题似乎是当我们添加-1到公式中时,会在模型矩阵中创建一个额外的列,该列不包含在回归系数中。(这是 R 创建因子编码方式的副产品。)所以我可以通过添加一个策略性放置的零系数来解决这个问题。我们还必须以相同的方式修复协方差矩阵:

library(emmeans)
library(plm)
data("Males")

mod <- plm(wage ~ exper + residence + health + residence*union,
           model = "within", 
           index = "nr", 
           data = Males)

BB <- c(coef(mod)[1], 0, coef(mod)[-1])
k <- length(BB)
VV <- matrix(0, nrow = k, ncol = k)
VV[c(1, 3:k), c(1, 3:k)] <- vcov(mod)

RG <- qdrg(~ -1 + exper + residence + health + residence*union, 
           data = Males, coef = BB, vcov = VV, df = df.residual(mod))

验证事情是否对齐:

> names(RG@bhat)
 [1] "exper"                             ""                                 
 [3] "residencenorth_east"               "residencenothern_central"         
 [5] "residencesouth"                    "healthyes"                        
 [7] "unionyes"                          "residencenorth_east:unionyes"     
 [9] "residencenothern_central:unionyes" "residencesouth:unionyes"
> colnames(RG@linfct)
 [1] "exper"                             "residencerural_area"              
 [3] "residencenorth_east"               "residencenothern_central"         
 [5] "residencesouth"                    "healthyes"                        
 [7] "unionyes"                          "residencenorth_east:unionyes"     
 [9] "residencenothern_central:unionyes" "residencesouth:unionyes"

他们确实排队,所以我们可以得到我们需要的结果:

(EMM <- emmeans(RG, ~ residence * union))
 residence       union emmean     SE   df lower.CL upper.CL
 rural_area      no     0.378 0.0335 2677  0.31201    0.443
 north_east      no     0.330 0.1636 2677  0.00929    0.651
 nothern_central no     0.192 0.1483 2677 -0.09834    0.483
 south           no     0.260 0.1514 2677 -0.03732    0.557
 rural_area      yes    0.287 0.1473 2677 -0.00144    0.576
 north_east      yes    0.385 0.1647 2677  0.06155    0.708
 nothern_central yes    0.333 0.1539 2677  0.03091    0.634
 south           yes    0.341 0.1534 2677  0.04024    0.642

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

一般来说,关键是确定添加的列出现在哪里。它将是模型公式中第一个因素的第一级的位置。您可以通过查看删除截距的模型公式names(coef(mod))以及colnames(model.matrix(formula), data = data)在哪里来检查它。formula

更新:一般功能

这是一个可用于为任何plm对象创建参考网格的函数。事实证明,有时这些对象确实有截距(例如,随机效应模型),所以我们必须检查。对于缺少截距的模型,您真的应该仅将其用于对比。

plmrg = function(object, ...) {
    form = formula(formula(object))
    if (!("(Intercept)" %in% names(coef(object))))
        form = update(form, ~ . - 1)
    data = eval(object$call$data, environment(form))
    mmat = model.matrix(form, data)
    sel = which(colnames(mmat) %in% names(coef(object)))
    k = ncol(mmat)
    b = rep(0, k)
    b[sel] = coef(object)
    v = matrix(0, nrow = k, ncol = k)
    v[sel, sel] = vcov(object)
    emmeans::qdrg(formula = form, data = data, 
        coef = b, vcov = v, df = df.residual(object), ...)
}

测试运行:

> (rg = plmrg(mod, at = list(exper = c(3,6,9))))
'emmGrid' object with variables:
    exper = 3, 6, 9
    residence = rural_area, north_east, nothern_central, south
    health = no, yes
    union = no, yes

> emmeans(rg, "residence")
NOTE: Results may be misleading due to involvement in interactions
 residence       emmean     SE   df lower.CL upper.CL
 rural_area       0.313 0.0791 2677   0.1579    0.468
 north_east       0.338 0.1625 2677   0.0190    0.656
 nothern_central  0.243 0.1494 2677  -0.0501    0.536
 south            0.281 0.1514 2677  -0.0161    0.578

Results are averaged over the levels of: exper, health, union 
Confidence level used: 0.95 
于 2021-10-26T15:35:09.400 回答
1

这个潜在biglm::biglm()的解决方案用于拟合 lm 模型,然后使用emmeans::qdrg()指定的麻烦。这种方法对您的情况有帮助吗?

library(biglm)
library(emmeans)
## the biglm coefficients using factor() with all the `nr` levels has NAs.
## so restrict data to complete cases in the `biglm()` call
model_biglm <- biglm(wage ~ -1 +exper + residence+health + residence*union + factor(nr), data=Males[!is.na(Males$residence),])
summary(model_biglm)

## double check that biglm and lm give same/similar model
## summary(model_biglm)
## summary(model_lm)
summary(model_biglm)$rsq
summary(model_lm)$r.squared
identical(coef(model_biglm), coef(model_lm)) ## not identical!  but plot the coefficients...
head(cbind(coef(model_biglm), coef(model_lm)))
tail(cbind(coef(model_biglm), coef(model_lm)))
plot(cbind(coef(model_biglm), coef(model_lm))); abline(0,1,col="blue")


## do a "[q]uick and [d]irty [r]eference [g]rid and follow examples 
### from ?qdrg and https://cran.r-project.org/web/packages/emmeans/vignettes/FAQs.html 
  rg1 <- qdrg(wage ~ -1 + exper + residence+health + residence*union + factor(nr), 
              data = Males,
              coef = coef(model_biglm),
              vcov = vcov(model_biglm), 
              df = model_biglm$df.resid,
              nuisance="nr")
  
## Since we already specified nuisance in qdrg() we don't in emmeans():
  emmeans(rg1, c("residence","union"))

这使:

>   emmeans(rg1, c("residence","union"))
 residence       union emmean     SE   df lower.CL upper.CL
 rural_area      no      1.72 0.1417 2677     1.44     2.00
 north_east      no      1.67 0.0616 2677     1.55     1.79
 nothern_central no      1.53 0.0397 2677     1.45     1.61
 south           no      1.60 0.0386 2677     1.52     1.68
 rural_area      yes     1.63 0.2011 2677     1.23     2.02
 north_east      yes     1.72 0.0651 2677     1.60     1.85
 nothern_central yes     1.67 0.0503 2677     1.57     1.77
 south           yes     1.68 0.0460 2677     1.59     1.77

Results are averaged over the levels of: 1 nuisance factors, health 
Confidence level used: 0.95 
于 2021-10-22T17:46:28.967 回答