如何为 `pivot_longer()` 创建 `id`

Question

我在尝试从下面的dat 制作长格式数据方面已经完成了一半。

两个小问题：在我下面的pivot_longer() 代码中，

(1)如何为times 的数量添加id 变量，如我的预期输出所示？

(2)如何将列time1, . . .,time8 转为0, . . ., 7？

dat <- read.csv('https://raw.githubusercontent.com/rnorouzian/e/master/wi.csv')

# Top 3 rows of current data:

      time1    time2     time3     time4     time5     time6    time7    time8       ses
1  1.203999 2.278898  3.716495  3.550721  3.375575  4.029231 5.292819 4.117426 -0.428465
2  0.291965 1.882300  0.958540  0.793806  0.021239  1.709134 3.127197 1.713560 -0.831093
3 -0.634382 0.847460 -0.801319 -0.126182 -0.496423 -1.009533 1.067997 0.131556  0.936131

# Top 3 rows of EXPECTED output:

  id       ses   Reading  time 
1  1 -0.428465  1.203999     0      
2  1 -0.428465  2.278898     1     
3  1 -0.428465  3.716495     2     
 
# What I tried:-----------------------------------------------------------------
pivot_longer(dat, time1:time8, names_to = "time", values_to = "Reading")

        ses  time  Reading
1 -0.428465 time1 1.203999
2 -0.428465 time2 2.278898
3 -0.428465 time3 3.716495

Answer 1

您可以在获取长格式数据之前创建一个id 变量，并为每个id 创建一个time 列，即当前行号-1。

library(dplyr)
library(tidyr)

dat %>%
  mutate(id = row_number()) %>%
  pivot_longer(cols = time1:time8, names_to = "time", values_to = "Reading") %>%
  group_by(id) %>%
  mutate(time = row_number() - 1)

#     ses    id  time Reading
#    <dbl> <int> <dbl>   <dbl>
# 1 -0.428     1     0   1.20 
# 2 -0.428     1     1   2.28 
# 3 -0.428     1     2   3.72 
# 4 -0.428     1     3   3.55 
# 5 -0.428     1     4   3.38 
# 6 -0.428     1     5   4.03 
# 7 -0.428     1     6   5.29 
# 8 -0.428     1     7   4.12 
# 9 -0.831     2     0   0.292
#10 -0.831     2     1   1.88 
# … with 1,590 more rows

Answer 2

在基础 R 中你可以这样做：

cbind(dat[9],stack(dat,-9))

甚至

 reshape(dat,1:8, dir="long",sep="")