如何在 cols = 以特定前缀开头的任何列时使用 tidyr pivot_longer

Question

每周，我都会得到一个原始数据集，我需要从中生成报告。我想编写一个每周都可以运行的 R 脚本。不幸的是，根据收集的标本，每周原始数据的列组略有不同。这是一周一周的情况示例。

library(readr)
w1 <- read_csv("species, males, females - fed, females - unfed
a,2,0,3
b,5,7,2
c,8,4,9")
w2 <- read_csv("species, males, females - mixed
a,2,0
b,5,7
c,8,4")

> w1
# A tibble: 3 x 4
  species males `females - fed` `females - unfed`
  <chr>   <dbl>           <dbl>             <dbl>
1 a           2               0                 3
2 b           5               7                 2
3 c           8               4                 9
> w2
# A tibble: 3 x 3
  species males `females - mixed`
  <chr>   <dbl>             <dbl>
1 a           2                 0
2 b           5                 7
3 c           8                 4

这是我通常使用的方式 pivot_longer:

library(tidyr)
w1 %>% pivot_longer(cols = c(males, `females - fed`, `females - unfed`),
             names_to = c("sex","feeding_status"),
             names_sep = " - ",
             values_to = "quantity")

# A tibble: 9 x 4
  species sex     feeding_status quantity
  <chr>   <chr>   <chr>             <dbl>
1 a       males   NA                    2
2 a       females fed                   0
3 a       females unfed                 3
4 b       males   NA                    5
5 b       females fed                   7
6 b       females unfed                 2
7 c       males   NA                    8
8 c       females fed                   4
9 c       females unfed                 9

我如何为 pivot_longer 编写适用于 w1、w2 和 w3 的代码？（编辑以包括 w3 -- 见下文）

我试过 (select(., starts_with("females")) 但无法找出正确的语法。pivot_longer 文档提到 names_pattern()和extract()，看起来很有前途，但我不知道如何使用它们。谢谢！

编辑：为了回应 akrun 的回答，我意识到我必须提供稍微复杂一些的样本数据。该代码还需要处理偶尔出现在名为 "unknown sex" 的数据集中的列，如下所示：

w3 <- read_csv("species, males, females - mixed, unknown sex
a,2,0,4
b,5,7,0
c,8,4,23")

> w3
# A tibble: 3 x 4
  species males `females - mixed` `unknown sex`
  <chr>   <dbl>             <dbl>         <dbl>
1 a           2                 0             4
2 b           5                 7             0
3 c           8                 4            23

akrun 下面建议的用于解决 w1 和 w2 的代码导致 w3 的 "unknown sex" 列中的值加倍：

w3 %>%
  pivot_longer(cols = c(males, starts_with('females')),
               names_to = c("sex", "feeding_status"), names_sep=" - ")

# A tibble: 6 x 5
  species `unknown sex` sex     feeding_status value
  <chr>           <dbl> <chr>   <chr>          <dbl>
1 a                   4 males   NA                 2
2 a                   4 females mixed              0
3 b                   0 males   NA                 5
4 b                   0 females mixed              7
5 c                  23 males   NA                 8
6 c                  23 females mixed              4

Answer 1

常用的列是'species'，所以我们可以用-

library(dplyr)
library(tidyr)
 w1 %>% 
    pivot_longer(cols = -species, 
       names_to = c("sex","feeding_status"),
          names_sep = " - ",
          values_to = "quantity")

或select_helpers

之一

w1 %>%
    pivot_longer(cols = c(males, starts_with('females')),
      names_to = c("sex", "feeding_status"), names_sep=" - ")
# A tibble: 9 x 4
#  species sex     feeding_status value
#  <chr>   <chr>   <chr>          <dbl>
#1 a       males   <NA>               2
#2 a       females fed                0
#3 a       females unfed              3
#4 b       males   <NA>               5
#5 b       females fed                7
#6 b       females unfed              2
#7 c       males   <NA>               8
#8 c       females fed                4
#9 c       females unfed              9

如果我们想包括多个案例，那么matches在另一个选项

w3 %>% 
   pivot_longer(cols = c(males, matches('^(females|unknown)')), 
   names_to = c("sex", "feeding_status"), names_sep=" - ")
# A tibble: 9 x 4
#  species sex         feeding_status value
#  <chr>   <chr>       <chr>          <dbl>
#1 a       males       <NA>               2
#2 a       females     mixed              0
#3 a       unknown sex <NA>               4
#4 b       males       <NA>               5
#5 b       females     mixed              7
#6 b       unknown sex <NA>               0
#7 c       males       <NA>               8
#8 c       females     mixed              4
#9 c       unknown sex <NA>              23

如何在 cols = 以特定前缀开头的任何列时使用 tidyr pivot_longer

How to use tidyr pivot_longer when cols = any column that starts with a certain prefix

r

select

dplyr

names

startswith