我是 R 的新手，我想知道將我的轉換data為Desired_Output下面的最有效的方法是什么？
m="
n min max median q1 q3 group m SD
1 30 30 55 44.5 43.25 49.75 treat 45.38524 5.593169
2 30 31 55 47.0 44.00 49.00 treat 46.11951 4.886821
3 30 40 55 48.0 45.00 51.00 treat 47.92676 4.173242
4 15 30 51 44.0 42.50 45.50 treat 43.21604 4.245150
5 15 31 54 46.0 42.50 48.50 treat 44.94723 5.759449
6 15 44 55 48.0 45.00 48.50 treat 47.66393 3.012334
7 15 39 55 49.0 44.00 52.00 treat 48.01439 5.571240
8 15 41 55 48.0 44.50 50.00 treat 47.59677 4.261415
9 15 40 55 47.0 45.00 50.00 treat 47.38081 4.200670
10 18 42 55 46.0 44.00 49.50 cont 46.91764 3.996259
11 18 40 55 44.0 43.00 47.00 cont 45.25704 3.667377
12 18 41 55 44.5 44.00 50.00 cont 46.58674 4.334604
13 9 42 49 46.0 43.00 48.00 cont 45.60879 3.357931
14 9 42 48 44.0 43.00 45.00 cont 44.29745 1.878592
15 9 41 55 44.0 44.00 45.00 cont 45.43229 2.779801
16 9 43 55 50.0 44.00 52.00 cont 48.73261 5.506545
17 9 43 55 46.0 44.00 51.00 cont 47.61981 5.069204
18 9 41 55 50.0 44.00 51.00 cont 48.19267 5.403842"
data <- read.table(text=m, h=T)
Desired_Output="
nT mT sdT nC mC sdC
30 45.38524 5.593169 18 46.91764 3.996259
30 46.11951 4.886821 18 45.25704 3.667377
30 47.92676 4.173242 18 46.58674 4.334604
. . . . . .
. . . . . .
. . . . . .
15 47.38081 4.200670 9 48.19267 5.403842"
uj5u.com熱心網友回復：
它有點亂，但你可以嘗試使用 dplyr
library(dplyr)
data <- data %>%
group_by(group) %>%
mutate(idx = 1:n()) %>%
select(group, n, idx, m, SD) %>%
ungroup
df1 <- data %>%
filter(group == "treat") %>%
select(-group)
df2 <- data %>%
filter(group == "cont") %>%
select(-group)
df <- df1 %>%
left_join(df2, by = "idx") %>%
select(-idx)
names(df) <- c("nT", "mT", "sdT", "nC", "mC", "sdC")
df
nT mT sdT nC mC sdC
<int> <dbl> <dbl> <int> <dbl> <dbl>
1 30 45.4 5.59 18 46.9 4.00
2 30 46.1 4.89 18 45.3 3.67
3 30 47.9 4.17 18 46.6 4.33
4 15 43.2 4.25 9 45.6 3.36
5 15 44.9 5.76 9 44.3 1.88
6 15 47.7 3.01 9 45.4 2.78
7 15 48.0 5.57 9 48.7 5.51
8 15 47.6 4.26 9 47.6 5.07
9 15 47.4 4.20 9 48.2 5.40
或使用tidyr::pivot_wider，
library(tidyr)
data %>%
group_by(group) %>%
mutate(idx = 1:n()) %>%
select(group, n, idx, m, SD) %>%
ungroup %>%
pivot_wider(id_cols = idx, values_from = c(n, m, SD), names_from = group) %>%
select(-idx)
n_treat n_cont m_treat m_cont SD_treat SD_cont
<int> <int> <dbl> <dbl> <dbl> <dbl>
1 30 18 45.4 46.9 5.59 4.00
2 30 18 46.1 45.3 4.89 3.67
3 30 18 47.9 46.6 4.17 4.33
4 15 9 43.2 45.6 4.25 3.36
5 15 9 44.9 44.3 5.76 1.88
6 15 9 47.7 45.4 3.01 2.78
7 15 9 48.0 48.7 5.57 5.51
8 15 9 47.6 47.6 4.26 5.07
9 15 9 47.4 48.2 4.20 5.40
uj5u.com熱心網友回復：
這是一個基本的R解決方案，
l2 <- split(data[c('n', 'm', 'SD')], data$group)
do.call(cbind, Map(function(x, y){names(x) <- paste0(names(x), '_', y); x}, l2, names(l2)))
# cont.n_cont cont.m_cont cont.SD_cont treat.n_treat treat.m_treat treat.SD_treat
#10 18 46.91764 3.996259 30 45.38524 5.593169
#11 18 45.25704 3.667377 30 46.11951 4.886821
#12 18 46.58674 4.334604 30 47.92676 4.173242
#13 9 45.60879 3.357931 15 43.21604 4.245150
#14 9 44.29745 1.878592 15 44.94723 5.759449
#15 9 45.43229 2.779801 15 47.66393 3.012334
#16 9 48.73261 5.506545 15 48.01439 5.571240
#17 9 47.61981 5.069204 15 47.59677 4.261415
#18 9 48.19267 5.403842 15 47.38081 4.200670
uj5u.com熱心網友回復：
新答案，dplyr唯一解決方案：
library(dplyr)
df %>%
select(n, m, SD, group) %>%
slice(10:18) %>%
rename_with(~ paste0(.x, "C")) %>%
bind_cols(df[1:9, ]) %>%
rename_with( ~ paste0(.x, "T"), c(n, m, SD)) %>%
select(nT, mT, sdT=SDT, nC, mC, sdC=SDC)
nT mT sdT nC mC sdC
10 30 45.38524 5.593169 18 46.91764 3.996259
11 30 46.11951 4.886821 18 45.25704 3.667377
12 30 47.92676 4.173242 18 46.58674 4.334604
13 15 43.21604 4.245150 9 45.60879 3.357931
14 15 44.94723 5.759449 9 44.29745 1.878592
15 15 47.66393 3.012334 9 45.43229 2.779801
16 15 48.01439 5.571240 9 48.73261 5.506545
17 15 47.59677 4.261415 9 47.61981 5.069204
18 15 47.38081 4.200670 9 48.19267 5.403842
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