如何编写循环来运行数据帧的 Mann whitney U 测试？答案

【问题标题】：How to write a loop to run a Mann whitney U test of a data frame?如何编写循环来运行数据帧的 Mann whitney U 测试？
【发布时间】：2020-12-01 13:49:27
【问题描述】：

我正在尝试在 R 中进行一些分析。我需要对我的数据框执行非配对 Wilcox 测试。首先我为分析做了一个脚本：

wilcox.test(df$Obs1[df$Var1 == "1" & df$Var2 == "1"], 
            df$Obs1[df$Var1 == "1" & df$Var2 == "2"])$p.value

但是，我想通过将列 Obs1 更改为 Obs2 然后更改为 Obs3 来执行此分析的 for 循环，但不知道如何开始。

Sample    Var1    Var2    Obs1    Obs2    Obs3
A         1       2       3.12    4.13    5.35
A         2       2       6.54    2.67    1.11
B         1       3       4.14    1.96    8.98
B         2       3       5.47    9.42    6.72
C         1       1       8.56    4.95    2.62
C         2       1       4.92    3.21    3.97
D         1       1       3.36    9.99    2.62
D         2       1       8.57    1.25    6.74
etc.

输入(df)

结构（列表（样本 = c（A，A，B，B，C，C，D，D，E，E，F，F，G，G，H，H，I，I），Var1 = c（“1”，“2”，“1”，“2”，“1”，“2”，“1”，“2”，“1”，“2”，“1”，“2”， “1”，“2”，“1”，“2”，“1”，“2”），Var2 = c（“2”，“2”，“3”，“3”，“1”，“ 1”、“1”、“1”、“2”、“2”、“2”、“2”、“1”、“1”、“3”、“3”、“3”、“3” ), Obs1 = c(3.12, 6.54, 4.14, 5.40, 8.56, 4.92, 3.36, 8.57, 7.56, 7.89. 2.34, 5.43, 6.21, 7.43, 5.83, 9.28, 10.30, 3.42, 26), Obs2 = c(4.16 , 1.96, 9.42, 4.95, 3.21, 9.99, 1.25, 9.78, 9.54, 7.23, 6.24, 1.23, 2.35, 4.73, 4.12, 5.30, 0.12), Obs3 = c(5.35, 1.11, 8.98, 3..7, 2..7, 6.76 2.62, 6.74, 9.41, 5.37, 7.43, 9.62, 4.69, 4.27, 2.74, 3.53, 5.87, 7.98), Obs4 = c(8.52, 9.59, 3.45, 9.40, 8.77, 4.26, 9.3,5, 5.12., 2.3,5, 5.12., 2.3,5, 5.12., , 4.33, 7.61, 5.37, 6.84, 9.98, 8.65, 0.43), row.names = c(NA, -6L), class= c("tbl_df", "tbl", "data.frame"))

有人可以帮我解决循环吗？提前致谢。

【问题讨论】：

显示您的数据的dput。小例子
已添加。然而，这只是一个随机的例子

标签： r loops pairwise.wilcox.test

【解决方案1】：

使用tidyverse

library(tidyverse)
df <- structure(
  list(Sample = c("A", "A", "B", "B", "C", "C", "D", "D", "E", "E", "F", "F", "G", "G", "H", "H", "I", "I"), 
       Var1 = c("1", "2", "1", "2", "1", "2", "1", "2", "1", "2", "1", "2", "1", "2", "1", "2", "1", "2"), 
       Var2 = c("2", "2", "3", "3", "1", "1", "1", "1", "2", "2", "2", "2", "1", "1", "3", "3", "3", "3"),
       Obs1 = c(3.12, 6.54, 4.14, 5.40, 8.56, 4.92, 3.36, 8.57, 7.56, 7.89, 2.34, 5.43, 6.21, 7.43, 5.83, 9.28, 10.30, 3.42), 
       Obs2 = c(4.13, 2.67, 1.96, 9.42, 4.95, 3.21, 9.99, 1.25, 9.78, 9.54, 7.23, 6.24, 1.23, 2.35, 4.73, 4.12, 5.30, 0.12),
       Obs3 = c(5.35, 1.11, 8.98, 6.72, 2.62, 3.97, 2.62, 6.74, 9.41, 5.37, 7.43, 9.62, 4.69, 4.27, 2.74, 3.53, 5.87, 7.98),
       Obs4 = c(8.52, 9.59, 3.45, 9.40, 8.77, 4.26, 9.78, 5.55, 2.31, 5.12, 2.35, 4.33, 7.61, 5.37, 6.84, 9.98, 8.65, 0.43)))

df <- as_tibble(df)

WT_data <- df %>% 
  mutate(Var12 = paste0(Var1, Var2)) %>% 
  select(Var12, starts_with("Obs")) %>% 
  filter(Var12 == "11" | Var12 == "12")

map_df(WT_data[-1], ~ wilcox.test(.x ~ Var12, data = WT_data)$p.value)
#> Warning in wilcox.test.default(x = c(2.62, 2.62, 4.69), y = c(5.35, 9.41, :
#> cannot compute exact p-value with ties
#> # A tibble: 1 x 4
#>    Obs1  Obs2   Obs3  Obs4
#>   <dbl> <dbl>  <dbl> <dbl>
#> 1   0.4     1 0.0765   0.2

^{由reprex package (v0.3.0) 于 2020-12-01 创建}

【讨论】：