如何将wilcox.test应用于R中的整个数据帧？

Question

我有一个数据框,其中一个分组因子(第一列)具有多个级别(多于两个)和几个包含数据的列.我想将wilcox.test应用于整个日期框架,以将每个组变量与其他变量进行比较.我怎样才能做到这一点？

更新:我知道wilcox.test只会测试两组之间的差异,而我的数据框包含三组.但我更感兴趣的是如何做到这一点,而不是使用什么测试.最有可能的一个组将被删除,但我还没有决定,所以我想测试所有变种.

这是一个示例:

structure(list(group = c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 3L, 3L), var1 = c(9.3, 
9.05, 7.78, 7.11, 7.14, 8.12, 7.5, 7.84, 7.8, 7.52, 8.84, 6.98, 
6.1, 6.89, 6.5, 7.5, 7.8, 5.5, 6.61, 7.65, 7.68), var2 = c(11L, 
11L, 10L, 1L, 3L, 7L, 11L, 11L, 11L, 11L, 4L, 1L, 1L, 1L, 2L, 
2L, 1L, 4L, 8L, 8L, 1L), var3 = c(7L, 11L, 3L, 7L, 11L, 2L, 11L, 
5L, 11L, 11L, 5L, 11L, 11L, 2L, 9L, 9L, 3L, 8L, 11L, 11L, 2L), 
    var4 = c(11L, 11L, 11L, 11L, 6L, 11L, 11L, 11L, 10L, 7L, 
    11L, 2L, 11L, 3L, 11L, 11L, 6L, 11L, 1L, 11L, 11L), var5 = c(11L, 
    1L, 2L, 2L, 11L, 11L, 1L, 10L, 2L, 11L, 1L, 3L, 11L, 11L, 
    8L, 8L, 11L, 11L, 11L, 2L, 9L)), .Names = c("group", "var1", 
"var2", "var3", "var4", "var5"), class = "data.frame", row.names = c(NA, 
-21L))

UPDATE

感谢大家的所有答案!

Answer 1

这个pairwise.wilcox.test函数似乎在这里很有用; 也许是这样的？

out <- lapply(2:6, function(x) pairwise.wilcox.test(d[[x]], d$group))
names(out) <- names(d)[2:6]
out

如果你只想要p值,你可以通过并提取它们并制作一个矩阵.

sapply(out, function(x) {
    p <- x$p.value
    n <- outer(rownames(p), colnames(p), paste, sep='v')
    p <- as.vector(p)
    names(p) <- n
    p
})
##         var1      var2      var3 var4      var5
## 2v1 0.5414627 0.8205958 0.4851572    1 1.0000000
## 3v1 0.1778222 0.3479835 1.0000000    1 1.0000000
## 2v2        NA        NA        NA   NA        NA
## 3v2 0.5414627 0.3479835 0.3784941    1 0.6919826

另请注意,pairwise.wilcox.test使用Holm方法调整多次比较; 如果您想做一些不同的事情,请查看p.adjust参数.

Answer 2

您可以使用循环遍历列apply,然后使用匿名函数将列传递给您要使用的任何测试,如下所示(假设数据框已命名df):

apply(df[-1],2,function(x) kruskal.test(x,df$group))

注意:我使用了Kruskal-Wallis测试,因为它适用于多个组.如果只有两组,使用Wilcoxon测试也可以正常工作.

如果你做想做的所有变量配对Wilcoxon检验的,这里有一个两班轮,将遍历所有列,所有对并返回结果的列表:

group.pairs <- combn(unique(df$group),2,simplify=FALSE)
# this loops over the 2nd margin - the columns - of df and makes each column
# available as x
apply(df[-1], 2, function(x)
             # this loops over the list of group pairs and makes each such pair
             # available as an integer vector y
             lapply(group.pairs, function(y)
                    wilcox.test(x[df$group %in% y],df$group[df$group %in% y])))

Answer 3

更新我的答案以跨列工作

test.fun <- function(dat, col) { 

 c1 <- combn(unique(dat$group),2)
 sigs <- list()
 for(i in 1:ncol(c1)) {
    sigs[[i]] <- wilcox.test(
                   dat[dat$group == c1[1,i],col],
                   dat[dat$group == c1[2,i],col]
                 )
    }
    names(sigs) <- paste("Group",c1[1,],"by Group",c1[2,])

 tests <- data.frame(Test=names(sigs),
                    W=unlist(lapply(sigs,function(x) x$statistic)),
                    p=unlist(lapply(sigs,function(x) x$p.value)),row.names=NULL)

 return(tests)
}


tests <- lapply(colnames(dat)[-1],function(x) test.fun(dat,x))
names(tests) <- colnames(dat)[-1]
# tests <- do.call(rbind, tests) reprints as data.frame

# This solution is not "slow" and outperforms the other answers significantly: 
system.time(
  rep(
   tests <- lapply(colnames(dat)[-1],function(x) test.fun(dat,x)),10000
  )
)

#   user  system elapsed 
#  0.056   0.000   0.053 

结果：

tests

$var1
                Test  W          p
1 Group 1 by Group 2 28 0.36596737
2 Group 1 by Group 3 39 0.05927406
3 Group 2 by Group 3 38 0.27073136

$var2
                Test    W         p
1 Group 1 by Group 2 19.0 0.8205958
2 Group 1 by Group 3 36.5 0.1159945
3 Group 2 by Group 3 40.5 0.1522726

$var3
                Test    W         p
1 Group 1 by Group 2 13.0 0.2425786
2 Group 1 by Group 3 23.5 1.0000000
3 Group 2 by Group 3 41.0 0.1261647

$var4
                Test  W         p
1 Group 1 by Group 2 26 0.4323470
2 Group 1 by Group 3 30 0.3729664
3 Group 2 by Group 3 29 0.9479518

$var5
                Test    W         p
1 Group 1 by Group 2 24.0 0.7100968
2 Group 1 by Group 3 19.0 0.5324295
3 Group 2 by Group 3 17.5 0.2306609