我正在使用Rstudio运行一个广义的估算方程式,其公式如下:
new <- gee(deaths~ subtypes,
data = hi, family = "binomial", id = iso,
corstr = "exchangeable", scale.fix = TRUE, scale.value = 1)
然后,我想通过运行来获得强大的标准化错误:
summary(new)
但是控制台中的答案从中间开始就像这样
[71,] 0.01557088 0.01557088
[72,] 0.01557088 0.01557088
[73,] 0.01557088 0.01557088
[74,] 0.01557088 0.01557088
[ reached getOption("max.print") -- omitted 60 rows ]
我希望它看起来像这样(通过R而不是Rstudio获得)。
GEE: GENERALIZED LINEAR MODELS FOR DEPENDENT DATA
gee S-function, version 4.13 modified 98/01/27 (1998)
Model:
Link: Logit
Variance to Mean Relation: Binomial
Correlation Structure: Exchangeable
Call:
gee(formula = deaths ~ subtypes, id = iso, data = hi, family = "binomial",
corstr = "exchangeable", scale.fix = TRUE, scale.value = 1)
Summary of Residuals:
Min 1Q Median 3Q Max
-0.5882720 -0.5824096 0.4175904 0.4175904 0.4909421
Coefficients:
Estimate Naive S.E. Naive z Robust S.E. Robust z
(Intercept) 0.33267295 0.1214047 2.74019759 0.1604175 2.07379434
subtypes2 -0.29643730 0.3437622 -0.86233233 0.4620348 -0.64159083
subtypes3 0.02415334 0.2680558 0.09010566 0.2650652 0.09112226
Estimated Scale Parameter: 1
Number of Iterations: 3
如果没有省略答案的顶部,我怎样才能在Rstudio中实现这一点。
答案 0 :(得分:0)
如果代码已实现,summary.gee
有一个标志可能会隐藏工作关联的显示。
> gee:::summary.gee
function (object, correlation = TRUE, ...)
{
...
if (correlation) {
}
...
}
如果实施此功能,summary(new, correlation=FALSE)
可能有所帮助。
在此之前,您可以操纵summary.gee
对象。
> g <- gee(breaks ~ tension, id=wool, data=warpbreaks, corstr="exchangeable")
> s <- summary(g)
> s
...
Working Correlation
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1.00000000 0.02518404 0.02518404 0.02518404 0.02518404 0.02518404
[2,] 0.02518404 1.00000000 0.02518404 0.02518404 0.02518404 0.02518404
[3,] 0.02518404 0.02518404 1.00000000 0.02518404 0.02518404 0.02518404
...
> s$working.correlation <- NULL
> s
...
Working Correlation
NULL
(您可以在实际的working.correlation
对象中将gee
设置为NULL,但在摘要中执行此操作会保留原始输出。)