以有效的方式转换和最小化 data.frame答案

【问题标题】：Converting and minimizing a data.frame in an efficient way以有效的方式转换和最小化 data.frame
【发布时间】：2022-01-13 04:22:48
【问题描述】：

将我的数据转换为所需输出的最快方法是什么？

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"

【问题讨论】：

您想要的输出是如何计算的？你尝试了什么？哪里失败了？
@Sotos，亲爱的 Sotos，我是 R 新手，想知道是否有办法实现这一目标。

标签： r dataframe dplyr tidyverse

【解决方案1】：

这是一个基本的 R 解决方案，

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

【讨论】：

【解决方案2】：

有点乱，不过你可以试试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

或者使用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

【讨论】：

【解决方案3】：

新答案，dplyr 唯一解决方案：

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

【讨论】：