R 中 fisher.test 中的差异-解网

问：

我虽然运行了两次相同的测试，但结果却不同。

首先，我从我的数据中创建了一个列联表：

> SAtab1 <- table(dt2$ACE_SA_BIN, dt2$IHD_other_healthy)
> SAtab1
   
      0   1   2
  0 345 577  29
  1  10  25   2

然后我运行 Fisher 测试：

> fisher.test(SAtab1, workspace = 2e8, simulate.p.value=TRUE)

    Fisher's Exact Test for Count Data with simulated p-value (based on 2000 replicates)

data:  SAtab1
p-value = 0.2969
alternative hypothesis: two.sided

然后我在某处看到了代码，他们在其中传递了一个矩阵而不是列联表。所以我创建了一个矩阵：

> SAtab2 <- 
+   matrix(c(345,10,577,25,29,2),
+          nrow = 2,
+          dimnames = list(Gender = c("no", "yes"),
+                          Illness =c("no illness","other than IHD", "HD")))
> SAtab2
      Illness
Gender no illness other than IHD HD
   no         345            577 29
   yes         10             25  2

并再次运行 fisher.test：

> fisher.test(SAtab2, workspace = 2e8, simulate.p.value=TRUE)

    Fisher's Exact Test for Count Data with simulated p-value (based on 2000 replicates)

data:  SAtab2
p-value = 0.2829
alternative hypothesis: two.sided

测试怎么可能得不到相同的结果？轻微的偏差从何而来？

R 比较差分析卡方

> fisher.test(SAtab2, workspace = 2e8, simulate.p.value=TRUE)

    Fisher's Exact Test for Count Data with simulated p-value (based on 2000 replicates)

data:  SAtab2
p-value = 0.3088
alternative hypothesis: two.sided

> fisher.test(SAtab2, workspace = 2e8, simulate.p.value=TRUE)

    Fisher's Exact Test for Count Data with simulated p-value (based on 2000 replicates)

data:  SAtab2
p-value = 0.3018
alternative hypothesis: two.sided

而否则它将给出相同的输出

> lst1 <- replicate(5, fisher.test(SAtab2, workspace = 2e8, simulate.p.value=FALSE), simplify = FALSE)

> all.equal(lst1[[1]], lst1[[2]])
[1] TRUE
> all.equal(lst1[[1]], lst1[[3]])
[1] TRUE
> all.equal(lst1[[1]], lst1[[5]])
[1] TRUE

上一个：测试两个矩阵重叠的角度区域

下一个：emmeans：：emmeans（ mod， ~ group） %>% multcomp：：could（Letters=letters）

R 中 fisher.test 中的差异

discrepancies in fisher.test in R

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