R 是否有一种方法可以运行多个t.test，从而生成可以轻松存储为表格式的结果？
我正在处理一组数据，我想对这些数据进行子集和比较t检验。最终目标是拥有一个易于阅读的表，作为可以呈现给读者的输出 目前，我使用的是一次给出一个结果的单个t测试，如下面的代码R 是否有一种方法可以运行多个t.test，从而生成可以轻松存储为表格式的结果？,r,t-test,R,T Test,我正在处理一组数据，我想对这些数据进行子集和比较t检验。最终目标是拥有一个易于阅读的表，作为可以呈现给读者的输出 目前，我使用的是一次给出一个结果的单个t测试，如下面的代码 t.test(survey$numericDV[survey$condition == 0 & survey$partisan_guess == "Republican"], survey$numericDV[survey$condition == 1 & survey$partisan_
t.test(survey$numericDV[survey$condition == 0 & survey$partisan_guess == "Republican"], survey$numericDV[survey$condition == 1 & survey$partisan_guess == "Republican"])
survey <- structure(list(numericDVedu = c(0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 0, 0,
1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1,
0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1,
0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0,
1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1,
0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0,
0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0,
0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1,
1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0,
0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0,
0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0,
0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1,
1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0,
0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0), condition =
structure(c(5L, 4L, 2L, 5L, 1L, 1L, 2L, 3L, 2L, 2L, 5L, 2L, 3L, 1L,
5L, 4L, 5L, 2L, 4L, 4L, 1L, 2L, 3L, 5L, 2L, 4L, 5L, 4L, 5L, 5L, 5L,
2L, 1L, 4L, 3L, 5L, 2L, 5L, 1L, 4L, 2L, 3L, 2L, 5L, 1L, 2L, 1L, 2L,
3L, 1L, 2L, 4L, 3L, 5L, 3L, 4L, 1L, 5L, 1L, 2L, 4L, 2L, 2L, 3L, 4L,
3L, 1L, 2L, 3L, 2L, 4L, 2L, 1L, 5L, 4L, 1L, 3L, 5L, 4L, 3L, 2L, 4L,
5L, 3L, 4L, 2L, 4L, 2L, 4L, 3L, 5L, 2L, 3L, 1L, 1L, 1L, 3L, 5L, 5L,
3L, 1L, 3L, 2L, 3L, 4L, 5L, 2L, 2L, 1L, 1L, 5L, 5L, 2L, 4L, 5L, 3L,
1L, 4L, 5L, 3L, 4L, 1L, 5L, 3L, 1L, 2L, 1L, 3L, 5L, 3L, 1L, 2L, 4L,
4L, 1L, 3L, 4L, 5L, 3L, 3L, 5L, 4L, 2L, 3L, 5L, 4L, 1L, 5L, 3L, 4L,
2L, 4L, 5L, 3L, 4L, 2L, 4L, 5L, 3L, 2L, 1L, 2L, 4L, 1L, 3L, 5L, 2L,
1L, 3L, 4L, 1L, 2L, 4L, 5L, 2L, 2L, 3L, 3L, 5L, 1L, 2L, 5L, 2L, 3L,
4L, 2L, 4L, 1L, 3L, 4L, 1L, 4L, 1L, 5L, 4L, 2L, 2L, 5L, 1L, 4L, 5L,
3L, 1L, 1L, 4L, 5L, 3L, 2L, 1L, 1L, 5L, 1L, 4L, 5L, 3L, 4L, 5L, 3L,
1L, 5L, 2L, 4L, 5L, 1L, 4L, 1L, 3L, 2L, 4L, 3L, 5L, 5L, 1L, 4L, 1L,
3L, 4L, 5L, 1L, 3L, 1L, 3L, 4L, 2L, 3L, 3L, 2L, 3L, 2L, 4L, 5L, 2L,
4L, 5L, 1L, 2L, 5L, 3L, 2L, 3L, 5L, 4L, 1L, 3L, 4L, 5L, 1L, 2L, 5L,
5L, 3L, 1L, 4L, 5L, 3L, 2L, 1L, 1L, 4L, 5L, 1L, 2L, 1L, 3L, 1L, 5L,
2L, 2L, 5L, 1L, 3L, 4L, 3L, 1L, 3L, 2L, 1L, 2L, 5L, 3L, 1L, 4L, 2L,
3L, 1L, 2L, 3L, 4L, 1L, 3L, 2L, 5L, 1L, 4L, 5L, 1L, 2L, 1L, 2L, 4L,
5L, 5L, 3L, 5L, 4L, 2L, 4L, 3L, 5L, 2L), .Label = c("0", "1", "2",
"3", "4"), class = "factor"),
partisan_guess = structure(c(2L, 1L, 2L, 2L, 2L, 2L, 2L,
1L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 1L, 2L, 2L, 2L,
1L, 2L, 2L, 1L, 2L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 2L, 1L,
1L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 1L, 2L,
2L, 2L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 2L, 2L, 1L, 2L, 1L, 2L,
2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 1L,
2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 2L, 2L, 1L, 2L,
2L, 1L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 1L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 2L, 1L,
2L, 2L, 1L, 2L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 1L, 1L, 2L, 2L,
2L, 1L, 2L, 2L, 2L, 1L, 2L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 2L,
2L, 2L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 2L, 2L, 1L, 2L, 2L, 2L,
2L, 1L, 1L, 2L, 1L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 1L, 2L, 1L,
2L, 2L, 1L, 1L, 1L, 1L, 1L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 1L,
2L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 1L,
2L, 2L, 1L, 1L, 2L, 2L, 2L, 2L, 1L, 2L, 1L, 1L, 2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L,
2L, 2L, 1L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 2L, 2L, 1L, 1L,
1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 2L, 1L, 2L, 1L,
2L, 1L, 2L, 2L, 1L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 2L, 1L, 2L,
2L, 2L, 1L, 2L, 1L, 2L, 2L, 1L, 2L, 1L, 2L, 2L, 2L, 1L, 1L,
2L, 2L, 1L, 2L, 1L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 2L,
1L, 1L, 2L, 2L, 2L, 2L, 1L, 2L), .Label = c("Democrat", "Republican"
), class = "factor")), class = "data.frame", row.names = c(NA, -330L))
survey我们可以编写一个函数，针对条件0对每个条件应用t.test
run_t_test <- function(data, partisan_guess) {
data <- subset(data, partisan_guess == partisan_guess)
zero_data <- data$numericDV[survey$condition == 0]
purrr::map_df(1:4, function(x) broom::tidy(t.test(
zero_data, data$numericDV[survey$condition == x])), .id = 'condition')
}
run_t_test(survey, 'Republican')
# A tibble: 4 x 11
# condition estimate estimate1 estimate2 statistic p.value parameter conf.low
# <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 1 -0.0490 0.333 0.382 -0.588 0.557 132. -0.214
#2 2 -0.113 0.333 0.446 -1.32 0.188 128. -0.282
#3 3 0.0635 0.333 0.270 0.782 0.436 127. -0.0972
#4 4 0.0980 0.333 0.235 1.25 0.212 130. -0.0565
# … with 3 more variables: conf.high <dbl>, method <chr>, alternative <chr>
run_t_test(survey, 'Democrat')
# A tibble: 4 x 11
# condition estimate estimate1 estimate2 statistic p.value parameter conf.low
# <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
#1 1 -0.0490 0.333 0.382 -0.588 0.557 132. -0.214
#2 2 -0.113 0.333 0.446 -1.32 0.188 128. -0.282
#3 3 0.0635 0.333 0.270 0.782 0.436 127. -0.0972
#4 4 0.0980 0.333 0.235 1.25 0.212 130. -0.0565
# … with 3 more variables: conf.high <dbl>, method <chr>, alternative <chr>
run_t_testpairwise.t.test（survey$numericDV，survey$condition）
您希望条件0与1:4相对还是每个条件彼此相对？@RonakShah条件0与1:4相对。例如，不需要将1与2进行比较。