match.arg(p.adjust.method)中的错误：“arg”必须为NULL或字符向量

Question

这里是我的数据

mydat=structure(list(id = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), group = c(1L, 
1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 1L, 1L, 1L, 1L, 2L, 
2L, 2L, 2L, 3L, 3L, 3L, 3L), var = c(23L, 24L, 24L, 23L, 23L, 
24L, 24L, 23L, 23L, 24L, 24L, 23L, 23L, 24L, 24L, 23L, 23L, 24L, 
24L, 23L, 23L, 24L, 24L, 23L)), .Names = c("id", "group", "var"
), class = "data.frame", row.names = c(NA, -24L))

我想加入两张桌子。id是标识符。

library(tidyverse)
mdyat %>% 
  with(.,pairwise.wilcox.test(var,id, group, exact =F)) %>% 
  broom::tidy() %>% 
 complete(id,group) %>% 
  left_join(mydat %>% 
              group_by(id,group)) %>% 
              summarise_all(c("mean", "sd", "median")) 
            by=c("id,group")

并得到错误

Error in match.arg(p.adjust.method) : 
  'arg' must be NULL or a character vector

如何分别为每个标识符执行此脚本，即所需的输出

id      mean    sd      median  p.value
1   1   23,5    0.5773503   23,5    NA
1   2   23,5    0.5773503   23,5    1
1   3   23,5    0.5773503   23,5    1
2   1   23,5    0.5773503   23,5    NA
2   2   23,5    0.5773503   23,5    1
2   3   23,5    0.5773503   23,5    1

Answer 1

第一部分可以使用group_by和do进行修复，如下所示。

mydat %>% 
  group_by(id) %>%
  do({
    with(., pairwise.wilcox.test(var, group, exact =F)) %>% broom::tidy()
  }) 

 ## # A tibble: 6 x 4
 ## # Groups:   id [2]
 ##      id group1 group2 p.value
 ##   <int> <fctr>  <chr>   <dbl>
 ## 1     1      2      1       1
 ## 2     1      3      1       1
 ## 3     1      3      2       1
 ## 4     2      2      1       1
 ## 5     2      3      1       1
 ## 6     2      3      2       1

为了将其与摘要统计相结合，您需要决定要加入哪个组(group1还是group2)。在下面的文章中，我加入了group1，所以mean、sd和median指的是group1，p.value是指group1和group2之间的区别。

mydat %>% 
  group_by(id) %>%
  do({
    with(., pairwise.wilcox.test(var, group, exact =F)) %>% broom::tidy()
  }) %>% 
  mutate(group1 = as.numeric(as.character(group1)), 
         group2 = as.numeric(as.character(group2))) %>%
  complete(group1 = mydat$group) %>%
  left_join(mydat %>% group_by(id,group) %>% summarise_all(c("mean", "sd", "median")), 
            by=c('id', 'group1'='group'))

Answer 2

函数参数是错误的：

pairwise.wilcox.test(var,id, group, exact =F)

?pairwise.wilcox.test将正确的语法声明为：

pairwise.wilcox.test(x, g, p.adjust.method = p.adjust.methods,
                      paired = FALSE, ...)

这意味着第三个函数参数应该是p.adjust.method，而不是group。