了解glm $ residuals和resid(glm)

Question

你能告诉我glm $ residuals和resid (glm)返回的是什么,其中glm是一个quasipoisson对象.例如,如何使用glm $ y和glm $ linear.predictors创建它们.

GLM $残差

     n missing  unique    Mean     .05     .10   .25  .50     .75     .90     .95

 37715   10042    2174 -0.2574 -2.7538 -2.2661 -1.4480 -0.4381  0.7542  1.9845  2.7749



lowest : -4.243 -3.552 -3.509 -3.481 -3.464
highest:  8.195  8.319  8.592  9.089  9.416

渣油(GLM)

        n    missing     unique       Mean        .05        .10        .25
    37715          0       2048 -2.727e-10    -1.0000    -1.0000    -0.6276
      .50        .75        .90        .95
  -0.2080     0.4106     1.1766     1.7333

lowest : -1.0000 -0.8415 -0.8350 -0.8333 -0.8288
highest:  7.2491  7.6110  7.6486  7.9574 10.1932

Answer 1

调用resid(模型)将默认为偏差残差,而模型$ resid将为您提供工作残差.由于链接功能,没有单一的模型残差定义.有偏差,工作,部分,皮尔逊和反应残差.因为这些只依赖于平均结构(而不​​是方差),所以quasipoisson和poisson的残差具有相同的形式.您可以查看residuals.glm函数以获取详细信息,但这是一个示例:

counts <- c(18,17,15,20,10,20,25,13,12)
outcome <- gl(3,1,9)
treatment <- gl(3,3)
glm.D93 <- glm(counts ~ outcome + treatment, family=quasipoisson())
glm.D93$resid


#working
resid(glm.D93,type="working")
(counts - glm.D93$fitted.values)/exp(glm.D93$linear)

#deviance
resid(glm.D93,type="dev")
fit <- exp(glm.D93$linear)
poisson.dev <- function (y, mu) 
    sqrt(2 * (y * log(ifelse(y == 0, 1, y/mu)) - (y - mu)))
poisson.dev(counts,fit) * ifelse(counts > fit,1,-1)

#response
resid(glm.D93,type="resp")
counts - fit

#pearson
resid(glm.D93,type="pear")
(counts - fit)/sqrt(fit)