r - pivot_longer 并返回 NA

Question

简单的 pivot_longer 问题。鉴于:

library(tidyverse)
df <- data.frame(id = c(1),
                 test_date = as.Date("01/01/1998", "%d/%m/%Y"),
                 test_result = c("yes"),
                 exit_date = as.Date("01/02/1998", "%d/%m/%Y"),
                 dob_date =  as.Date("01/01/1997", "%d/%m/%Y"),
                 exit_result_1 = c(10),
                 exit_result_2 = c(4),
                 exit_result_3 = c(2))
df
#   id  test_date test_result  exit_date   dob_date exit_result_1 exit_result_2 exit_result_3
# 1  1 1998-01-01         yes 1998-02-01 1997-01-01            10             4             2

我要:

#   id test_result exit_result_1 exit_result_2 exit_result_3 date_name      dates
# 1  1         yes            NA            NA            NA test_date 1998-01-01
# 2  1        <NA>            10             4             2 exit_date 1998-02-01
# 3  1        <NA>            NA            NA            NA  dob_date 1997-01-01

类似的东西

pivot_longer(df, cols = matches('date'), 
             names_to = "date_name", values_to = "dates")

但我想返回 NA。有什么建议吗？谢谢

Answer 1

可以在names_to中设置".value"来表示定义输出列名称的列名称对应的组成部分。在这种情况下，必须提供 names_sep 或 names_pattern 之一来指定应如何拆分列名。

df %>%
  pivot_longer(
    -id,
    names_to = c("date_name", ".value"),
    names_pattern = "([^_]+)_(.+)"
  )

# # A tibble: 3 × 7
#      id date_name date       result result_1 result_2 result_3
#   <dbl> <chr>     <date>     <chr>     <dbl>    <dbl>    <dbl>
# 1     1 test      1998-01-01 yes          NA       NA       NA
# 2     1 exit      1998-02-01 NA           10        4        2
# 3     1 dob       1997-01-01 NA           NA       NA       NA

为了比较，原始数据集看起来像

df
#   id  test_date test_result  exit_date   dob_date exit_result_1 exit_result_2 exit_result_3
# 1  1 1998-01-01         yes 1998-02-01 1997-01-01            10             4             2