问题描述
您能告诉我 glm $ residuals 和 resid(glm)返回什么，其中glm是拟泊松对象.例如我将如何使用glm $ y和glm $ linear.predictors创建它们.
Can you tell me what is returned by glm$residuals and resid(glm) where glm is a quasipoisson object. e.g. How would I create them using glm$y and glm$linear.predictors.
glm $残余
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
残基(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
推荐答案
调用resid(model)将默认为偏差残差，而model $ resid将为您提供工作残差.由于具有链接功能，因此对模型残差没有唯一的定义.存在偏差，工作残差，部分皮尔逊残差和响应残差.因为这些仅依赖于均值结构(而不是方差)，所以拟泊松和泊松的残差具有相同的形式.您可以查看residuals.glm函数以了解详细信息，但这是一个示例:
Calling resid(model) will default to the deviance residuals, whereas model$resid will give you the working residuals. Because of the link function, there is no single definition of what a model residual is. There are the deviance, working, partial, Pearson, and response residuals. Because these only rely on the mean structure (not the variance), the residuals for the quasipoisson and poisson have the same form. You can take a look at the residuals.glm function for details, but here is an example:
counts <- c(18,17,15,20,10,20,25,13,12)
outcome <- gl(3,1,9)
treatment <- gl(3,3)
glm.D93 <- glm(counts ~ outcome + treatment, family=quasipoisson())
glm.D93$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)
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