r - 如果在 R 的病例对照研究中一组为“0”，如何计算 OR（奇数比）？

Question

我是 RI 的新手，我正在使用 Epicalc 包计算优势比。我有这个数据

真实价值

values <- matrix(c(5,0,35,2,30,38,40,8,11,7,10,9,35,8,66,72,34,4,31,14), nrow = 2)
values
colnames(values) <- c("Laikipia", "West Pokot", "Garissa", "Isiolo", "Mandera", "Marsabit", "Samburu", "Tana River", "Turkana","Wajir")
rownames(values) <- c("0","1")
values <- as.table(values)
values
cc(outcome=NULL, exposure=NULL, cctable=values)}

但是我的赔率比是无限的，这是可以理解的，因为我在莱基皮亚有一个零值。所以我尝试了 Haldane-Anscombe 校正，并在下面给出的每个单元格代码中添加了 0.5，但我仍然得到无限值

Haldane-Anscombe 校正

values1 <- matrix(c(5.5,0.5,35.5,2.5,30.5,38.5,40.5,8.5,11.5,7.5,10.5,9.5,35.5,8.5,66.5,72.5,34.5,4.5,31.5,14.5), nrow = 2)
values1
colnames(values1) <- c("Laikipia", "West Pokot", "Garissa", "Isiolo", "Mandera", "Marsabit", "Samburu", "Tana River", "Turkana","Wajir")
rownames(values1) <- c("O","1")
values <- as.table(values1)
values1
cc(outcome=NULL, exposure=NULL, cctable=values1)

"cc {epicalc}" 优势比计算和绘图

Description

Odds ratio calculation and graphing

Usage

cc(outcome, exposure, decimal = 2, cctable = NULL, graph = TRUE, 
    original = TRUE, design = "cohort", main, xlab = "auto", ylab,
  alpha = .05, fisher.or = FALSE, exact.ci.or = TRUE)

如何解决这个问题呢。谢谢

标签： rdataframe