如何在R中执行多个wilcox.test？

Question

我有这个矩阵并且目的是在R中进行Wilcoxon测试（控制与案例），但我不确定如何正确地放入我的矩阵。

gene.name  cont1 cont2  cont3  case1  case2  case3
A           10    2      3      21     18      8
B           14    8      7      12     34      22
C           16    9      19     21     2       8
D           32    81     17     29     43      25
..

Answer 1

您可以尝试：

# load your data 
d <- read.table(text="gene.name  cont1 cont2  cont3  case1  case2  case3
A           10    2      3      21     18      8
B           14    8      7      12     34      22
C           16    9      19     21     2       8
B           32    81     17     29     43      25", header=T)

library(tidyverse)
# transform to long format using dplyr (included in tidyverse)
dlong <- as.tbl(d) %>% 
  gather(key, value,-gene.name) %>% 
  mutate(group=ifelse(grepl("cont",key), "control", "case"))
# plot the data
dlong %>% 
  ggplot(aes(x=group, y=value)) +
   geom_boxplot()

# run the test
dlong %>% 
  with(., wilcox.test(value ~ group))

Wilcoxon rank sum test with continuity correction

data:  value by group
W = 94.5, p-value = 0.2034
alternative hypothesis: true location shift is not equal to 0

编辑

# as you don't clarified how to handle the double occurence of B I assume 
# thats a typo and fixed the second B to D
library(ggpubr)
dlong <- as.tbl(d) %>%
  mutate(gene.name=LETTERS[1:4]) %>% 
  gather(key, value,-gene.name) %>% 
  mutate(group=ifelse(grepl("cont",key), "control", "case"))

# plot the boxplot with Wilcoxen p-values using ggpubr
dlong %>% 
  ggplot(aes(x=gene.name, y=value, fill=group)) +
  geom_boxplot() +
  stat_compare_means(method= "wilcox.test")

# get the pvalues
dlong %>% 
  group_by(gene.name) %>% 
  summarise(p=wilcox.test(value~group)$p.value)
# A tibble: 4 x 2
   gene.name     p
       <chr> <dbl>
1         A   0.2
2         B   0.2
3         C   0.7
4         D   1.0

或者使用apply尝试基础R.

res <- apply(d[,-1], 1, function(x){
  wilcox.test(x ~ c(1,1,1,2,2,2))$p.value
})
cbind.data.frame(Genes=as.character(d$gene.name), p=res, BH=p.adjust(res, method = "BH"))
     Genes   p        BH
[1,]     1 0.2 0.4000000
[2,]     2 0.2 0.4000000
[3,]     3 0.7 0.9333333
[4,]     2 1.0 1.0000000