如何在Hmisc[r]中将成对参数传递给summaryM函数测试

我这里有一个可复制的例子

library(Hmisc)
set.seed(173)
sex <- factor(sample(c("m","f"), 500, rep=TRUE))
country <- factor(sample(c('US', 'Canada'), 500, rep=TRUE))
age <- rnorm(500, 50, 5)
sbp <- rnorm(500, 120, 12)
label(sbp) <- 'Systolic BP'
units(sbp) <- 'mmHg'
treatment <- factor(rep(c("PreTretment","PostTretment"), 250))

f <- summaryM(age + sex + sbp ~ treatment, test=TRUE)

库（Hmisc）
种子（173）
性别我目前正在用Hmisc总结方法进行配对Wilcoxon试验：

 conTestWP <- 
function (group, x) 
{
    st <- wilcox.test( x[as.numeric(group)==1], x[as.numeric(group)==2], paired=TRUE)
    list(P = st["p.value"], stat = st["statistic"], df = st[c("df1", "df2")], 
        testname = "Wilcoxon-paired",
        statname = "V", latexstat = "V", plotmathstat = "F[df]")}

尝试打印
f
会引发错误（错误地声称p值不是数字），但您可以查看内部，查看配对的
wilcox.test
结果是否已传递到对象中，并且与“手动”完成的结果相同：

通过为“df”值输入硬编码数字来修复抛出的错误的努力正在失败。我没有成功地跟踪我仅粘贴在顶部的回溯：

> f
Error in log10(ifelse(pv > 0, pv, 1e-50)) : 
  non-numeric argument to mathematical function
> traceback()
5: format.pval(pval, digits = pdig, eps = eps)
4: formatTestStats(tr, prtest = prtest, latex = latex, testUsed = testUsed, 
       pdig = pdig, eps = eps, footnoteTest = footnoteTest)
3: formatCons(stats[[i]], nam, tr, x$group.freq, prmsd, sep, formatArgs, 
       round, prtest, pdig = pdig, eps = eps)
2: print.summary.formula.reverse(list(stats = list(sbp = c(97.9191028465814, 
   94.9839938500064, 100.014783572809, 97.2881910017715, 107.288034416825, 
   105.746587052709, 111.864782483651, 112.689945667021, 116.229748640414, 
   115.604190135259, 119.743427097173, 119.276780441804, 123.380695706571, 
   122.111672516175, 128.138778071723, 126.592782661592, 133.726823015259, 
   132.141219449201, 140.847941698775, 136.762891898923, 145.175812916341, 
   141.635692905295, 120.013464038065, 118.994407752318, 12.1494617994813, 
   11.9252958974706)), type = 2, group.name = "treatment", group.label = "treatment", 

这对我很有用：

conTestWP <- 
function (group, x) 
{
    st <- wilcox.test( x[as.numeric(group)==1], x[as.numeric(group)==2], paired=TRUE)
    list(P=st$p.value, stat=st$statistic, df = "NA", 
    testname = "Wilcoxon-paired",
    statname = "V", namefun = "fstat", latexstat = "V", plotmathstat = "F[df]")}

But，But，But。。。数据不成对。怎么可能呢？它们有不同的计数。一旦你解决了这个问题，为什么不预先计算成对案例的相关变量的差异，然后（也许）使用（注意：未测试的概念）
summaryM
给你配对测试（这实际上只是针对H0=0的差异进行的测试。请参阅
？wilcox.test
@bondedust中的详细信息部分。我添加了简化版本。你能对此发表意见吗？我尝试了一个
conTest
对象，该对象以conTestkw为模型，但我不认为应该将成对比较传递给summaryM。我认为
summary.formula
是进行配对测试的正确方法，我有一个工作示例。不幸的是，尽管构造了一个提供正确统计信息的方法，但由于我无法跟踪的
pval
错误，它无法打印。@BondedDust感谢您的努力。SummaryM是旧summary.formul的新版本一个但唯一相反的方法。在这两个函数中，所有剩余的都是相等的。我认为Harrell使用了某种内部Wilcoxon方法，它不是wilcox.test统计包的通用方法。嗯，这可能解释了summaryM文档中的一些明显错误。
> f
Error in log10(ifelse(pv > 0, pv, 1e-50)) : 
  non-numeric argument to mathematical function
> traceback()
5: format.pval(pval, digits = pdig, eps = eps)
4: formatTestStats(tr, prtest = prtest, latex = latex, testUsed = testUsed, 
       pdig = pdig, eps = eps, footnoteTest = footnoteTest)
3: formatCons(stats[[i]], nam, tr, x$group.freq, prmsd, sep, formatArgs, 
       round, prtest, pdig = pdig, eps = eps)
2: print.summary.formula.reverse(list(stats = list(sbp = c(97.9191028465814, 
   94.9839938500064, 100.014783572809, 97.2881910017715, 107.288034416825, 
   105.746587052709, 111.864782483651, 112.689945667021, 116.229748640414, 
   115.604190135259, 119.743427097173, 119.276780441804, 123.380695706571, 
   122.111672516175, 128.138778071723, 126.592782661592, 133.726823015259, 
   132.141219449201, 140.847941698775, 136.762891898923, 145.175812916341, 
   141.635692905295, 120.013464038065, 118.994407752318, 12.1494617994813, 
   11.9252958974706)), type = 2, group.name = "treatment", group.label = "treatment", 

conTestWP <- 
function (group, x) 
{
    st <- wilcox.test( x[as.numeric(group)==1], x[as.numeric(group)==2], paired=TRUE)
    list(P=st$p.value, stat=st$statistic, df = "NA", 
    testname = "Wilcoxon-paired",
    statname = "V", namefun = "fstat", latexstat = "V", plotmathstat = "F[df]")}