当 id 变量在列 header 中编码时,将数据从宽格式转换为长格式

Converting data from wide to long format when id variables are encoded in column header

我对 R 比较陌生,有如下宽格式数据

subject_id   age    sex  treat1.1.param1    treat1.1.param2   treat1.2.param1   treat1.2.param2
-----------------------------------------------------------------------------------------------
1             23     M         1                  2                  3                   4
2             25     W         5                  6                  7                   8

这是我们针对给定治疗(此处为 treat1)的多个受试者的数据,在多轮重复测量(此处为第 1 轮和第 2 轮)中测量多个参数(此处为 param1 和 param2)。该受试者条目所属的处理、轮次和参数的信息编码在列header中,如上例所示。

我想把长格式的数据举例如下:

subject_id  age sex treatment   round       param1      param2
------------------------------------------------------------------------------------------
1           23   M   treat1      1           1          2
1           23   M   treat1      2           3          4
2           25   W   treat1      1           5          6
2           25   W   treat1      2           7          8

即id变量识别单个观察值分别是subject_id、treatment、round。但是由于后两个变量是使用点作为分隔符在 header 列中编码的,所以我不知道如何从上面的宽格式转换为长格式。所有使用标准示例(使用 reshape2tidyr)的尝试都失败了。因为在现实中,我每 30 轮有 12 次治疗,每轮大约有 50 个参数,相对手动的方法对我帮助不大。

我们可以使用 tidyr 中的 pivot_longer 指定 names_tonames_pattern 参数。

tidyr::pivot_longer(df, 
                    cols = starts_with("treat"), 
                    names_to = c("treatmeant", "round", ".value"), 
                    names_pattern =  "(\w+)\.(\d+)\.(\w+)")

#  subject_id   age sex   treatmeant round param1 param2
#       <int> <int> <fct> <chr>      <chr>  <int>  <int>
#1          1    23 M     treat1     1          1      2
#2          1    23 M     treat1     2          3      4
#3          2    25 W     treat1     1          5      6
#4          2    25 W     treat1     2          7      8

数据

df <- structure(list(subject_id = 1:2, age = c(23L, 25L), sex = structure(1:2, 
.Label = c("M", "W"), class = "factor"), 
treat1.1.param1 = c(1L, 5L), treat1.1.param2 = c(2L, 6L), 
treat1.2.param1 = c(3L, 7L), treat1.2.param2 = c(4L, 8L)), 
class = "data.frame", row.names = c(NA, -2L))

您可以使用 tidyr gatherseparatespread:

tibble::tibble(subject_id = 1:2,
               age = c(23,25),
               sex = c("M", "W"),
               round_1_param_1 = c(1,5),
               round_1_param_2 = c(2,6),
               round_2_param_1 = c(3,7),
               round_2_param_2 = c(4,8)) %>% 
  tidyr::gather("key", "value", -subject_id, -age, -sex) %>% 
  tidyr::separate(key, c("round", "param"), sep = "param") %>%
  dplyr::mutate_at(vars("round", "param"), ~ tidyr::extract_numeric(.)) %>% 
  tidyr::spread(param, value)

# A tibble: 4 x 6
  subject_id   age sex   round   `1`   `2`
       <int> <dbl> <chr> <dbl> <dbl> <dbl>
1          1    23 M         1     1     2
2          1    23 M         2     3     4
3          2    25 W         1     5     6
4          2    25 W         2     7     8

这里有一个可能的data.table方法,

library(data.table)

dcast(melt(dd, id.vars = c("subject_id", "age", 'sex'))
      [, .(subject_id, age, sex, gsub('(\w+)\.\d+\.\w+', '\1', variable),
                                 gsub('\w+\.(\d+)\.\w+', '\1', variable),
                                 gsub('\w+\.\d+\.(\w+)', '\1', variable), value)],
      subject_id + age + sex + V4 + V5 ~ V6)

这给出了,

   subject_id age sex     V4 V5 param1 param2
1:          1  23   M treat1  1      1      2
2:          1  23   M treat1  2      3      4
3:          2  25   W treat1  1      5      6
4:          2  25   W treat1  2      7      8