Reputation: 125
I have some results from a perm.t.test
that I would like to transform into a data frame. I've tried a lot and searched a lot but I don't manage to fix this, I'm still new to R.
This is a reduced version of my dataset:
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$t))
head(data)
treat subj t exparat
1 C B16 T0 0.11
2 C B17 T0 0.27
3 C B18 T0 0.04
4 C B19 T0 0.47
5 C B20 T0 -0.11
6 C B16 T1 -0.05
I run multiple MKinfer::perm.t.test
between combinations of times (t) independently for each treatment (treat), using this function (Thanks to Ronak Shah):
library(MKinfer)
combn(levels(data$t), 2, function(x) {
perm.t.test(exparat~t,data = subset(data, t %in% x), R=999, paired = T)
}, simplify = FALSE) -> result
But now I have two problems:
1- My result is a list of class "perm.htest" objects, but I need to transform it into a data frame where each row is a test and, each column is an output of the test (i.e. statistic, parameter, p.value). So I can easily check my results, correct the p-values for multiple comparisons and export them.
I've been advised to use:
map_df(result, broom::tidy) %>%
mutate(combination = combn(levels(data$t), 2, paste0, collapse = ' vs '), .before = 1)
The problem is that MKinfer::perm.t.test perform both a permutation-t-test and a parametric t-test. So in the results, there is both a p.value
and a perm.p.value
, a there are both conf.int
and perm.conf.int
. So if I use broom::tidy
, it will recognise it an htest
and I will just get the results of the parametric t-tests.
Maybe something similar would work:
dtlist <- combn(levels(data$t), 2, function(x) {
dt <- MKinfer::perm.t.test(respvar~t,data = subset(data, t %in% x), R=999, paired = T)
df <- data.frame(
perm.p = dt$perm.p.value,
stat = dt$statistic)
return(df)
})
But I need a data frame as an output that I can then export in csv.
Does anyone have a solution?
Thank you in advance.
Upvotes: 2
Views: 116
Reputation: 389135
You can use -
do.call(rbind, combn(levels(data$t), 2, function(x) {
dt <- MKinfer::perm.t.test(exparat~t,
data = subset(data, t %in% x), R=999, paired = T)
data.frame(perm.p = dt$perm.p.value,stat = dt$statistic)
}, simplify = FALSE)) -> result
result$combination <- combn(levels(data$t), 2, paste0, collapse = ' vs ')
Upvotes: 2
Reputation: 1474
library(tidyverse)
tidy_result <- function(result) {
tibble(
estimate = result$estimate,
statistic = result$statistic,
conf.low = result$conf.int[1],
conf.high = result$conf.int[2],
perm.p.value = result$perm.p.value
)
}
combn(levels(data$t), 2, function(x) {
perm.t.test(exparat~t,data = subset(data, t %in% x), R=999, paired = T)
}, simplify = FALSE) -> result
map_dfr(result, tidy_result)
# # A tibble: 6 x 5
# estimate statistic conf.low conf.high perm.p.value
# <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 -0.015 -0.116 -0.307 0.277 0.918
# 2 -0.073 -0.764 -0.289 0.143 0.470
# 3 -0.04 -0.670 -0.175 0.0950 0.528
# 4 -0.058 -0.482 -0.330 0.214 0.593
# 5 -0.025 -0.220 -0.282 0.232 0.824
# 6 0.033 0.292 -0.222 0.288 0.748
Upvotes: 2