Question

我试图在R中运行Mantel-Haenszel分析，以确定在考虑组内的“诊断”比率时，比例测试的比较是否仍然很重要。此测试可在统计数据包中找到。

library(stats)
mantelhaen.test(x)

完成了一些阅读后，我发现这个测试可以在 n x n x k 的列联表上执行优势比测试，而不是简单地 n x n 。但是，我无法以正确的方式安排我的数据，因为我对R很新。我创建了一些示例数据......

ex.label <- c("A","A","A","A","A","A","A","B","B","B")
ex.status  <- c("+","+","-","+","-","-","-","+","+","-")
ex.diag <- c("X","X","Z","Y","Y","Y","X","Y","Z","Z")
ex.data <- data.frame(ex.label,ex.diag,ex.status)

看起来像这样......

  ex.label ex.diag ex.status
1         A       X         +
2         A       X         +
3         A       Z         -
4         A       Y         +
5         A       Y         -
6         A       Y         -
7         A       X         -
8         B       Y         +
9         B       Z         +
10        B       Z         -

我最初能够使用一个简单的N-1卡方来运行比较测试+到 - 仅针对A和B，但现在我希望能够考虑{{1}也是。我将在这里显示我想要查看的图表，这基本上是比较每列中比率的显着性。我能够做到这一点，但我现在想要能够解释ex.diag。

我尝试使用ex.diag函数以可行的方式排列数据。

ftable()

看起来像这样......

ex.ftable <- ftable(ex.data)

但是，当我运行ex.status - + ex.label ex.diag A X 1 2 Y 2 1 Z 1 0 B X 0 0 Y 0 1 Z 1 1时，我收到错误mantelhaen.test(ex.ftable)。如何以可以实际运行此测试的方式排列数据？

Answer 1

在mantelhaen.test中，三维列联表x的最后一个维度需要是分层变量（ex.diag）。该矩阵可以如下生成：

ex.label <- c("A","A","A","A","A","A","A","B","B","B")
ex.status  <- c("+","+","-","+","-","-","-","+","+","-")
ex.diag <- c("X","X","Z","Y","Y","Y","X","Y","Z","Z")

# Now ex.diag is in the first column
ex.data <- data.frame(ex.diag, ex.label, ex.status)

# The flat table
( ex.ftable <- ftable(ex.data) )

#                  ex.status - +
# ex.diag ex.label              
# X       A                  1 2
#         B                  0 0
# Y       A                  2 1
#         B                  0 1
# Z       A                  1 0
#         B                  1 1

可以使用aperm生成3D矩阵。

# Trasform the ftable into a 2 x 2 x 3 array
# First dimension: ex.label
# Second dimension: ex.status
# Third dimension: ex.diag
( mtx3D <- aperm(array(t(as.matrix(ex.ftable)),c(2,2,3)),c(2,1,3)) )

#     , , 1
# 
#      [,1] [,2]
# [1,]    1    2
# [2,]    0    0
# 
# , , 2
# 
#      [,1] [,2]
# [1,]    2    1
# [2,]    0    1
# 
# , , 3
# 
#      [,1] [,2]
# [1,]    1    0
# [2,]    1    1

现在可以进行Cochran-Mantel-Haenszel卡方检验。

# Cochran-Mantel-Haenszel chi-squared test of the null that 
# two nominal variables are conditionally independent in each stratum
#
mantelhaen.test(mtx3D, exact=FALSE)

测试结果是

        Mantel-Haenszel chi-squared test with continuity correction

data:  mtx3D
Mantel-Haenszel X-squared = 0.23529, df = 1, p-value = 0.6276
alternative hypothesis: true common odds ratio is not equal to 1
95 percent confidence interval:
 NaN NaN
sample estimates:
common odds ratio 
              Inf

鉴于案例数量较少，最好计算一个精确的条件测试（选项exact=TRUE）。

mantelhaen.test(mtx3D, exact=T)

#         Exact conditional test of independence in 2 x 2 x k tables
# 
# data:  mtx3D
# S = 4, p-value = 0.5
# alternative hypothesis: true common odds ratio is not equal to 1
# 95 percent confidence interval:
#  0.1340796       Inf
# sample estimates:
# common odds ratio 
#               Inf

在R中安排一个三维列联表以运行Cochran-Mantel-Haenszel分析？

1 个答案: