Setup
set.seed(0)
y <- matrix(rnorm(20), 10, 2)
x <- 1:10
fit <- lm(y ~ x) ## class: "mlm", "lm"
fit1 <- lm(y[, 1] ~ x) ## class: "lm"
fit2 <- lm(y[, 2] ~ x) ## class: "lm"
rstudent(fit)
# [,1] [,2]
#1 0.74417620 0.89121744
#2 -0.67506054 -0.50275275
#3 0.76297805 -0.74363941
#4 0.71164461 0.01075898
#5 0.03337192 0.03355209
#6 -1.75099724 -0.02701558
#7 -1.05594284 0.56993056
#8 -0.48486883 -0.35286612
#9 -0.23468552 0.79610101
#10 2.90701182 -0.93665406
cbind(rstudent(fit1), rstudent(fit2))
# [,1] [,2]
#1 0.74417620 1.90280959
#2 -0.67506054 -0.92973971
#3 0.76297805 -1.47237918
#4 0.71164461 0.01870820
#5 0.03337192 0.06042497
#6 -1.75099724 -0.04056992
#7 -1.05594284 1.02171222
#8 -0.48486883 -0.64316472
#9 -0.23468552 1.69605079
#10 2.90701182 -1.25676088
正如您所观察到的，只有第一个响应的结果才能正确返回rstandard(fit).
Why rstudent“传销”失败
问题是，没有“传销”方法rstudent.
methods(rstudent)
#[1] rstudent.glm* rstudent.lm*
你打电话时rstudent(fit)，S3方法调度机制发现rstudent.lm相反，因为inherits(fit, "lm") is TRUE。很遗憾，stats:::rstudent.lm没有为“传销”模型进行正确的计算。
stats:::rstudent.lm
#function (model, infl = lm.influence(model, do.coef = FALSE),
# res = infl$wt.res, ...)
#{
# res <- res/(infl$sigma * sqrt(1 - infl$hat))
# res[is.infinite(res)] <- NaN
# res
#}
lm.influence没有给出正确的sigma对于“传销”。底层 C 例程C_influence只计算sigma对于一个“lm”。如果你给lm.influence“mlm”，仅返回第一个响应变量的结果。
## pass in "mlm"
.Call(stats:::C_influence, fit$qr, FALSE, residuals(fit), 10 * .Machine$double.eps)$sigma
# [1] 1.3130265 1.3216357 1.3105706 1.3171621 1.3638689 1.1374385 1.2668101
# [8] 1.3416338 1.3586428 0.9180828
## pass in "lm"
.Call(stats:::C_influence, fit1$qr, FALSE, residuals(fit1), 10 * .Machine$double.eps)$sigma
# [1] 1.3130265 1.3216357 1.3105706 1.3171621 1.3638689 1.1374385 1.2668101
# [8] 1.3416338 1.3586428 0.9180828
显然对于“传销”来说sigma应该是一个矩阵。现在给出这个不正确的sigma, 《回收规则》 https://cran.r-project.org/doc/manuals/r-devel/R-intro.html#The-recycling-rule被应用在后面"/"在下面的行中stats:::rstudent.lm，因为res它左边的（残差）是一个矩阵，但它右边的东西是一个向量。
res <- res / (infl$sigma * sqrt(1 - infl$hat))
实际上，计算结果仅对于第一个响应变量是正确的；所有其余的响应变量都会使用错误的sigma.
R核心团队需要修补一些诊断功能
请注意，文档页面中列出的几乎所有功能?influence.measures对于“传销”来说是有问题的。他们应该发出警告，说“传销”方法尚未实施。
lm.influnce需要在 C 级进行修补。一旦这件事完成，rstudent.lm将在“传销”上正常工作。
其他功能可以在 R 级别轻松修补，例如，stats:::cooks.distance.lm and stats:::rstandard。目前（R 3.5.1）它们是：
stats:::cooks.distance.lm
#function (model, infl = lm.influence(model, do.coef = FALSE),
# res = weighted.residuals(model),
# sd = sqrt(deviance(model)/df.residual(model)),
# hat = infl$hat, ...)
#{
# p <- model$rank
# res <- ((res/(sd * (1 - hat)))^2 * hat)/p
# res[is.infinite(res)] <- NaN
# res
#}
stats:::rstandard.lm
#function (model, infl = lm.influence(model, do.coef = FALSE),
# sd = sqrt(deviance(model)/df.residual(model)), type = c("sd.1",
# "predictive"), ...)
#{
# type <- match.arg(type)
# res <- infl$wt.res/switch(type, sd.1 = sd * sqrt(1 - infl$hat),
# predictive = 1 - infl$hat)
# res[is.infinite(res)] <- NaN
# res
#}
并且可以对它们进行修补（通过使用outer) with
patched_cooks.distance.lm <-
function (model, infl = lm.influence(model, do.coef = FALSE),
res = weighted.residuals(model),
sd = sqrt(deviance(model)/df.residual(model)),
hat = infl$hat, ...)
{
p <- model$rank
res <- ((res / c(outer(1 - hat, sd))) ^ 2 * hat) / p
res[is.infinite(res)] <- NaN
res
}
patched_rstandard.lm <-
function (model, infl = lm.influence(model, do.coef = FALSE),
sd = sqrt(deviance(model)/df.residual(model)), type = c("sd.1",
"predictive"), ...)
{
type <- match.arg(type)
res <- infl$wt.res/switch(type, sd.1 = c(outer(sqrt(1 - infl$hat), sd)),
predictive = 1 - infl$hat)
res[is.infinite(res)] <- NaN
res
}
快速测试：
oo <- cbind(cooks.distance(fit1), cooks.distance(fit2)) ## correct result
all.equal(cooks.distance(fit), oo)
#[1] "Mean relative difference: 0.8211302"
all.equal(patched_cooks.distance.lm(fit), oo)
#[1] TRUE
rr <- cbind(rstandard(fit1), rstandard(fit2)) ## correct result
all.equal(rstandard(fit), rr)
#[1] "Mean relative difference: 0.5290036"
all.equal(patched_rstandard.lm(fit), rr)
#[1] TRUE
