将 t 检验结果列表转换为数据框
Transform a list of t-tests results into a data frame
我有一些来自 perm.t.test 的结果,我想将其转换为数据框。
这是我的数据集的简化版本:
treat = c("C","C","C","C","C","C","C","C","C","C","C","C","C",
"C","C","C","C","C","C","C","T","T","T","T","T","T",
"T","T","T","T","T","T","T","T","T","T","T","T","T","T")
subj = c("B16","B17","B18","B19","B20","B16","B17","B18","B19",
"B20","B16","B17","B18","B19","B20","B16","B17","B18",
"B19","B20","B1","B2","B3","B4","B5","B1","B2","B3","B4"
,"B5","B1","B2","B3","B4","B5","B1","B2","B3","B4","B5")
t = c("T0","T0","T0","T0","T0","T1","T1","T1","T1","T1","T2",
"T2","T2","T2","T2","T3","T3","T3","T3","T3","T0","T0",
"T0","T0","T0","T1","T1","T1","T1","T1","T2","T2","T2",
"T2","T2","T3","T3","T3","T3","T3")
exparat = c(0.11,0.27,0.04,0.47,-0.11,-0.05,-0.05,0.33,-0.11,
0.47,-0.01,0.43,0.47,0.33,-0.11,-0.09,0.20,-0.11,
0.47,0.33,0.19,0.02,0.33,0.47,-0.11,0.42,0.13,0.47,
-0.11,0.33,0.42,0.19,-0.11,0.33,0.47,0.42,0.17,
0.33,0.47,-0.11)
data = data.frame(treat, subj, t, exparat)
data$treat <- factor(data$treat)
data$t <- factor(data$t,levels=unique(data$t))
head(data)
treat subj t exparat
1 C B16 T0 0.11
2 C B17 T0 0.27
3 C B18 T0 0.04
4 C B19 T0 0.47
5 C B20 T0 -0.11
6 C B16 T1 -0.05
I 运行 多个 MKinfer::perm.t.test 时间组合之间 (t) 独立于每个治疗 (treat),使用此函数 ():
library(MKinfer)
combn(levels(data$t), 2, function(x) {
perm.t.test(exparat~t,data = subset(data, t %in% x), nperm=999, paired = T)
}, simplify = FALSE) -> result
但是现在我有两个问题:
1- 我的结果是 class 'c("perm.htest", "these") 对象的列表,但我需要将它转换成一个数据框,其中每一行都是一个测试,每一列都是测试的输出(即统计、参数、p.value)。所以我可以轻松地检查我的结果,更正多重比较的 p 值并导出它们。
2- 我不知道哪个检验对应于哪个因子 t 水平组合。在函数的输出中没有提到这一点。但我想我可以使用函数 combn(levels(data$t), 2)
检索这些,然后使用 t:
的所有级别组合创建一个向量(combt)
combt = combn(levels(data$t), 2)
combt = t(combt)
combt = data.frame(combt)
combt = paste(combt$X1, combt$X2, sep=" vs ")
我已经尝试搜索,但我可以找到解决方案。有谁可以帮助我吗?
提前致谢。
您可以在 result
-
的列表中应用 broom::tidy
library(dplyr)
library(purrr)
map_df(result, broom::tidy) %>%
mutate(combination = combn(levels(data$t), 2, paste0, collapse = ' vs '), .before = 1)
# combination estimate statistic p.value parameter conf.low conf.high method alternative
# <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <chr> <chr>
#1 T0 vs T1 -0.015 -0.116 0.910 9 -0.307 0.277 Permutation Paired t-test two.sided
#2 T0 vs T2 -0.073 -0.764 0.465 9 -0.289 0.143 Permutation Paired t-test two.sided
#3 T0 vs T3 -0.04 -0.670 0.519 9 -0.175 0.0950 Permutation Paired t-test two.sided
#4 T1 vs T2 -0.058 -0.482 0.641 9 -0.330 0.214 Permutation Paired t-test two.sided
#5 T1 vs T3 -0.025 -0.220 0.831 9 -0.282 0.232 Permutation Paired t-test two.sided
#6 T2 vs T3 0.033 0.292 0.777 9 -0.222 0.288 Permutation Paired t-test two.sided
我有一些来自 perm.t.test 的结果,我想将其转换为数据框。
这是我的数据集的简化版本:
treat = c("C","C","C","C","C","C","C","C","C","C","C","C","C",
"C","C","C","C","C","C","C","T","T","T","T","T","T",
"T","T","T","T","T","T","T","T","T","T","T","T","T","T")
subj = c("B16","B17","B18","B19","B20","B16","B17","B18","B19",
"B20","B16","B17","B18","B19","B20","B16","B17","B18",
"B19","B20","B1","B2","B3","B4","B5","B1","B2","B3","B4"
,"B5","B1","B2","B3","B4","B5","B1","B2","B3","B4","B5")
t = c("T0","T0","T0","T0","T0","T1","T1","T1","T1","T1","T2",
"T2","T2","T2","T2","T3","T3","T3","T3","T3","T0","T0",
"T0","T0","T0","T1","T1","T1","T1","T1","T2","T2","T2",
"T2","T2","T3","T3","T3","T3","T3")
exparat = c(0.11,0.27,0.04,0.47,-0.11,-0.05,-0.05,0.33,-0.11,
0.47,-0.01,0.43,0.47,0.33,-0.11,-0.09,0.20,-0.11,
0.47,0.33,0.19,0.02,0.33,0.47,-0.11,0.42,0.13,0.47,
-0.11,0.33,0.42,0.19,-0.11,0.33,0.47,0.42,0.17,
0.33,0.47,-0.11)
data = data.frame(treat, subj, t, exparat)
data$treat <- factor(data$treat)
data$t <- factor(data$t,levels=unique(data$t))
head(data)
treat subj t exparat
1 C B16 T0 0.11
2 C B17 T0 0.27
3 C B18 T0 0.04
4 C B19 T0 0.47
5 C B20 T0 -0.11
6 C B16 T1 -0.05
I 运行 多个 MKinfer::perm.t.test 时间组合之间 (t) 独立于每个治疗 (treat),使用此函数 (
library(MKinfer)
combn(levels(data$t), 2, function(x) {
perm.t.test(exparat~t,data = subset(data, t %in% x), nperm=999, paired = T)
}, simplify = FALSE) -> result
但是现在我有两个问题:
1- 我的结果是 class 'c("perm.htest", "these") 对象的列表,但我需要将它转换成一个数据框,其中每一行都是一个测试,每一列都是测试的输出(即统计、参数、p.value)。所以我可以轻松地检查我的结果,更正多重比较的 p 值并导出它们。
2- 我不知道哪个检验对应于哪个因子 t 水平组合。在函数的输出中没有提到这一点。但我想我可以使用函数 combn(levels(data$t), 2)
检索这些,然后使用 t:
combt = combn(levels(data$t), 2)
combt = t(combt)
combt = data.frame(combt)
combt = paste(combt$X1, combt$X2, sep=" vs ")
我已经尝试搜索,但我可以找到解决方案。有谁可以帮助我吗?
提前致谢。
您可以在 result
-
broom::tidy
library(dplyr)
library(purrr)
map_df(result, broom::tidy) %>%
mutate(combination = combn(levels(data$t), 2, paste0, collapse = ' vs '), .before = 1)
# combination estimate statistic p.value parameter conf.low conf.high method alternative
# <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <chr> <chr>
#1 T0 vs T1 -0.015 -0.116 0.910 9 -0.307 0.277 Permutation Paired t-test two.sided
#2 T0 vs T2 -0.073 -0.764 0.465 9 -0.289 0.143 Permutation Paired t-test two.sided
#3 T0 vs T3 -0.04 -0.670 0.519 9 -0.175 0.0950 Permutation Paired t-test two.sided
#4 T1 vs T2 -0.058 -0.482 0.641 9 -0.330 0.214 Permutation Paired t-test two.sided
#5 T1 vs T3 -0.025 -0.220 0.831 9 -0.282 0.232 Permutation Paired t-test two.sided
#6 T2 vs T3 0.033 0.292 0.777 9 -0.222 0.288 Permutation Paired t-test two.sided