如何自动化 R 中嵌套变量的 t 检验？

Question

我想自动收集来自 t 检验的汇总统计数据。在下面的示例中，我有嵌套变量Age、Location 和Treatment。对于每个Age 和Location，我想运行一个基于Treatment 的t 检验，它有两个分类名称Control 和Treatment。换句话说，我想知道每个Location 对每个Age 的控制和处理手段之间的区别。

我想使用 matrixTests 中的 col_t_welch 函数运行 t 检验，因为输出已经包含我正在寻找的几个汇总统计信息（即 mean.diff、stderr 和 @ 987654335@)。我如何设置我的数据框 (df1) 以便能够为嵌套 t 检验使用 for 循环？

可重现的例子：

library(matrixTests)
library(ggplot2)

set.seed(123)
df1 <- data.frame(matrix(ncol = 4, nrow = 36))
x <- c("Age","Location","Treatment","Value")
colnames(df1) <- x
df1$Age <- as.factor(rep(c(1,2,3), each = 12))
df1$Location <- as.factor(rep(c("Central","North"), each = 6))
df1$Treatment <- as.factor(rep(c("Control","Treatment"), each = 3))
df1$Value <- round(rnorm(36,200,25),0)

# I can't get the for-loop below to work because I'm not sure how to set up the data frame, but I was thinking something along these lines.

i <- 1
p <- numeric(length = 3*2)
mean_diff <- numeric(length = 3*2)
SE_diff <- numeric(length = 3*2)

for(j in c("1", "2", "3")){
  for(k in c("Control", "Treatment")){
    ttest <- col_t_welch(Value, data = df1, subset = Age == j & Treatment == k))
    p[i] <- a$pvalue
    mean_diff[i] <- ttest$mean.diff
    SE_diff[i] <- ttest$stderr
    i <- i + 1
  }
}

理想的最终数据框应如下面的d2 所示。

d2 <- expand.grid(Age = rep(c(1,2,3), 1),
                  Location = rep(c("Central","North"), 1), 
                  mean_diff = NA,
                  SE_diff = NA,
                  pvalue = NA)

C1 <- df1[c(1:6),3:4]
N1 <- df1[c(7:12),3:4]
C2 <- df1[c(13:18),3:4]
N2 <- df1[c(19:24),3:4]
C3 <- df1[c(25:30),3:4]
N3 <- df1[c(31:36),3:4]

c1_mod <- col_t_welch(x=C1[1:3,2], y=C1[4:6,2])
n1_mod <- col_t_welch(x=N1[1:3,2], y=N1[4:6,2])
c2_mod <- col_t_welch(x=C2[1:3,2], y=C2[4:6,2])
n2_mod <- col_t_welch(x=N2[1:3,2], y=N2[4:6,2])
c3_mod <- col_t_welch(x=C3[1:3,2], y=C3[4:6,2])
n3_mod <- col_t_welch(x=N3[1:3,2], y=N3[4:6,2])

d2[1,3] <- c1_mod$mean.diff
d2[1,4] <- c1_mod$stderr
d2[1,5] <- c1_mod$pvalue
d2[2,3] <- c2_mod$mean.diff
d2[2,4] <- c2_mod$stderr
d2[2,5] <- c2_mod$pvalue
d2[3,3] <- c3_mod$mean.diff
d2[3,4] <- c3_mod$stderr
d2[3,5] <- c3_mod$pvalue
d2[4,3] <- n1_mod$mean.diff
d2[4,4] <- n1_mod$stderr
d2[4,5] <- n1_mod$pvalue
d2[5,3] <- n2_mod$mean.diff
d2[5,4] <- n2_mod$stderr
d2[5,5] <- n2_mod$pvalue
d2[6,3] <- n3_mod$mean.diff
d2[6,4] <- n3_mod$stderr
d2[6,5] <- n3_mod$pvalue

d2

Answer 1

我想这可能对你有帮助

库

library(matrixTests)
library(tidyverse)

数据

set.seed(123)
df1 <- data.frame(matrix(ncol = 4, nrow = 36))
x <- c("Age","Location","Treatment","Value")
colnames(df1) <- x
df1$Age <- as.factor(rep(c(1,2,3), each = 12))
df1$Location <- as.factor(rep(c("Central","North"), each = 6))
df1$Treatment <- as.factor(rep(c("Control","Treatment"), each = 3))
df1$Value <- round(rnorm(36,200,25),0)

如何

df1 %>% 
  group_nest(Age,Location,Treatment) %>% 
  pivot_wider(names_from = Treatment,values_from = data) %>% 
  mutate(
    test = map2(
      .x = Control,
      .y = Treatment,
      .f = ~col_t_welch(.x,.y)
      )
  ) %>% 
  unnest(test) %>% 
  select(Age,Location,pvalue,mean.diff,stderr)

结果

# A tibble: 6 x 5
  Age   Location pvalue mean.diff stderr
  <fct> <fct>     <dbl>     <dbl>  <dbl>
1 1     Central   0.675     -9.67  21.3 
2 1     North     0.282    -22     17.7 
3 2     Central   0.925     -3     28.4 
4 2     North     0.570      9.33  14.6 
5 3     Central   0.589    -14.7   25.0 
6 3     North     0.311    -11.3    8.59