Question

dplyr中是否有办法比较各组？这里有一个具体的例子：我想对以下组合应用t检验：a vs b，a vs c和b vs c

set.seed(1)
tibble(value = c(rnorm(1000, 1, 1), rnorm(1000, 5, 1), rnorm(1000, 10,1)),
       group=c(rep("a", 1000), rep("b", 1000), rep("c", 1000))) %>%
   nest(value)

# A tibble: 3 x 2
  group data                
  <chr> <list>              
1 a     <tibble [1,000 × 1]>
2 b     <tibble [1,000 × 1]>
3 c     <tibble [1,000 × 1]>

如果dplyr没有提供解决方案，我也会对其他方法感到高兴...也许是data.table？

Answer 1

这里有一个基础R / tidyverse方法（有点手动，但是这个任务还不行）：

combn(df$group, 2, FUN = function(g) 
  t.test(filter(df, group == g[1]) %>% unnest %$% value , 
         filter(df, group == g[2]) %>% unnest %$% value ), 
  simplify = FALSE)

# [[1]]
# 
# Welch Two Sample t-test
# 
# data:  filter(df, group == g[1]) %>% unnest %$% value and filter(df, group == g[2]) %>% unnest %$% value
# t = -86.114, df = 1998, p-value < 2.2e-16
# alternative hypothesis: true difference in means is not equal to 0
# 95 percent confidence interval:
#   -4.086376 -3.904396
# sample estimates:
#   mean of x mean of y 
# 0.9883519 4.9837381 
# 
# 
# [[2]]
# 
# Welch Two Sample t-test
# 
# data:  filter(df, group == g[1]) %>% unnest %$% value and filter(df, group == g[2]) %>% unnest %$% value
# t = -195.4, df = 1998, p-value < 2.2e-16
# alternative hypothesis: true difference in means is not equal to 0
# 95 percent confidence interval:
#   -9.117558 -8.936356
# sample estimates:
#   mean of x  mean of y 
# 0.9883519 10.0153090 
# 
# 
# [[3]]
# 
# Welch Two Sample t-test
# 
# data:  filter(df, group == g[1]) %>% unnest %$% value and filter(df, group == g[2]) %>% unnest %$% value
# t = -108.65, df = 1997.9, p-value < 2.2e-16
# alternative hypothesis: true difference in means is not equal to 0
# 95 percent confidence interval:
#   -5.122395 -4.940747
# sample estimates:
#   mean of x mean of y 
# 4.983738 10.015309

比较组彼此

1 个答案: