如何多次 pivot_longer 相同的列？
How to pivot_longer same columns multiple times?
数据
我尝试使用谷歌搜索并在 SO 中搜索此问题的答案，但答案通常只是说明如何将列分成两列，而不是按特定标准对列进行分组。首先，这是我的数据的输入 hwk:
structure(list(V1 = structure(c(4, 4, 2, 2, 2, 2, 2, 2, 4, 4,
2, 3, 2, 3, 4, 2, 2, 2, 3, 3, 2, 3, 1, 3, 3, 3, 3, 4, 1, 2, 4,
1, 2, 3, 2, 3, 1, 1, 2, 2, 4, 3, 2, 1, 2, 3, 3, 4, 3, 3, 2, 3,
1, 4, 3, 2, 3, 4, 1, 3, 3, 3, 2, 2, 1, 2, 3, 4, 4, 2, 4, 3, 2,
3, 3, 3, 3, 2, 4, 3, 3, 3, 2, 2, 3, 4, 2, 4, 4, 2, 2, 3, 3), format.spss = "F8.0"),
V2 = structure(c(4, 4, 3, 4, 3, 4, 3, 2, 4, 1, 3, 3, 3, 4,
3, 3, 2, 3, 4, 3, 1, 4, 2, 3, 4, 2, 4, 3, 3, 2, 3, 2, 3,
3, 4, 3, 3, 3, 3, 3, 3, 2, 4, 2, 2, 2, 4, 3, 4, 4, 2, 4,
2, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 4, 3, 3, 4, 4, 4, 4, 4,
3, 4, 3, 3, 3, 4, 2, 4, 3, 4, 3, 3, 2, 3, 3, 4, 3, 4, 3,
4, 4, 3), format.spss = "F8.0"), V3 = structure(c(4, 4, 4,
4, 4, 4, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4), format.spss = "F8.0"),
V4 = structure(c(4, 4, 3, 4, 3, 4, 2, 1, 3, 2, 3, 1, 4, 4,
2, 3, 2, 2, 2, 4, 1, 2, 2, 2, 3, 2, 3, 2, 2, 1, 3, 1, 1,
2, 4, 1, 1, 2, 3, 2, 2, 1, 1, 1, 3, 2, 4, 3, 3, 3, 3, 3,
3, 4, 3, 1, 4, 3, 4, 3, 2, 3, 2, 1, 4, 1, 4, 1, 2, 4, 4,
4, 3, 3, 3, 2, 2, 1, 4, 3, 2, 3, 2, 1, 3, 4, 1, 2, 4, 3,
4, 2, 2), format.spss = "F8.0"), V5 = structure(c(3, 3, 3,
4, 3, 4, 3, 1, 1, 1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 3, 2,
2, 2, 2, 4, 2, 3, 2, 3, 4, 1, 4, 2, 3, 3, 2, 2, 3, 2, 2,
3, 3, 2, 3, 3, 3, 2, 2, 2, 3, 2, 3, 3, 2, 2, 3, 3, 2, 3,
2, 2, 3, 3, 3, 2, 3, 3, 3, 4, 3, 2, 3, 3, 3, 3, 3, 3, 4,
3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 4, 3, 3), format.spss = "F8.0"),
V6 = structure(c(4, 4, 3, 4, 3, 4, 4, 1, 3, 3, 3, 3, 2, 3,
4, 2, 4, 3, 3, 3, 3, 4, 4, 3, 3, 3, 4, 4, 4, 3, 4, 4, 3,
3, 3, 4, 2, 2, 3, 3, 3, 4, 2, 4, 3, 4, 4, 4, 3, 4, 2, 4,
3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 3, 1, 4, 4, 4, 4, 4, 4,
4, 3, 4, 4, 4, 4, 2, 4, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 3,
4, 4, 4), format.spss = "F8.0"), V7 = structure(c(4, 4, 2,
4, 2, 4, 4, 3, 3, 3, 2, 2, 4, 4, 3, 3, 1, 4, 3, 3, 1, 2,
4, 3, 4, 2, 4, 4, 3, 3, 2, 2, 3, 2, 4, 3, 3, 3, 3, 3, 3,
1, 4, 3, 2, 2, 4, 3, 4, 4, 2, 4, 2, 3, 4, 3, 3, 3, 4, 3,
4, 4, 3, 4, 4, 3, 4, 4, 4, 4, 3, 4, 4, 4, 3, 3, 4, 3, 4,
3, 3, 3, 3, 2, 2, 4, 4, 4, 4, 2, 4, 4, 3), format.spss = "F8.0"),
V8 = structure(c(4, 4, 2, 1, 2, 1, 1, 1, 3, 3, 2, 3, 2, 3,
4, 2, 2, 2, 3, 3, 2, 3, 1, 3, 3, 3, 3, 4, 1, 2, 4, 1, 2,
3, 2, 3, 1, 1, 2, 2, 3, 1, 1, 1, 2, 3, 3, 4, 3, 3, 2, 3,
1, 3, 4, 2, 3, 4, 1, 3, 3, 3, 2, 2, 1, 2, 3, 4, 4, 2, 4,
3, 4, 4, 4, 4, 3, 2, 4, 3, 3, 3, 2, 2, 3, 4, 2, 4, 4, 2,
1, 3, 4), format.spss = "F8.0"), V9 = structure(c(4, 4, 4,
4, 4, 4, 4, 4, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 2, 3, 4, 4,
4, 4, 3, 4, 4, 4, 4, 4, 4, 4, 3, 3, 3, 4, 3, 2, 4, 3, 4,
4, 4, 4, 4, 4, 3, 3, 3, 3, 4, 4, 4, 4, 4, 3, 4, 3, 2, 4,
3, 3, 4, 4, 4, 3, 4, 4, 4, 4, 4, 3, 4, 3, 4, 3, 4, 4, 4,
4, 3, 4, 4, 4, 4, 4, 3, 2, 4, 4, 4, 4, 4), format.spss = "F8.0"),
V10 = structure(c(4, 4, 2, 4, 2, 4, 3, 2, 3, 3, 3, 2, 4,
4, 2, 2, 1, 3, 4, 4, 1, 4, 2, 3, 3, 2, 4, 3, 2, 3, 3, 1,
3, 2, 4, 3, 2, 3, 3, 3, 3, 1, 2, 4, 2, 3, 4, 4, 3, 3, 2,
4, 2, 4, 3, 3, 4, 3, 4, 3, 4, 4, 4, 1, 4, 3, 3, 4, 3, 4,
4, 3, 3, 3, 3, 3, 4, 1, 4, 3, 3, 3, 3, 2, 3, 4, 4, 2, 4,
2, 4, 4, 3), format.spss = "F8.0"), V11 = structure(c(3,
3, 1, 4, 1, 4, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 2, 2, 2, 1,
2, 3, 1, 2, 3, 3, 2, 1, 2, 2, 2, 3, 2, 2, 3, 2, 1, 2, 2,
1, 1, 4, 3, 1, 3, 2, 3, 1, 2, 1, 2, 1, 2, 2, 1, 2, 2, 3,
2, 2, 2, 2, 2, 2, 1, 1, 1, 3, 3, 4, 2, 1, 2, 2, 3, 3, 3,
3, 4, 3, 2, 3, 3, 2, 2, 2, 2, 1, 3, 1, 4, 1, 3), format.spss = "F8.0"),
V12 = structure(c(4, 4, 3, 2, 3, 2, 3, 1, 3, 3, 3, 3, 2,
3, 3, 2, 4, 3, 3, 4, 4, 3, 3, 4, 4, 3, 3, 3, 4, 3, 4, 4,
3, 3, 3, 4, 2, 2, 3, 3, 3, 4, 2, 4, 3, 4, 4, 4, 3, 4, 2,
4, 3, 3, 3, 3, 4, 3, 3, 2, 2, 1, 1, 3, 1, 4, 4, 4, 4, 4,
4, 4, 3, 3, 2, 2, 2, 2, 4, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4,
3, 2, 3, 4), format.spss = "F8.0")), class = c("tbl_df",
"tbl", "data.frame"), row.names = c(NA, -93L))
目前的解决方案
所以我希望第一个变量是焦虑量表的问题，每个变量的分数：
hwk2 <- hwk %>%
pivot_longer(cols = 1:3,
names_to = "Anxiety Test",
values_to = "Anxiety Score")
hwk2
这给了我这个：
# A tibble: 279 x 11
V4 V5 V6 V7 V8 V9 V10 V11 V12 `Anxiety Test` `Anxiety Score`
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <chr> <dbl>
1 4 3 4 4 4 4 4 3 4 V1 4
2 4 3 4 4 4 4 4 3 4 V2 4
3 4 3 4 4 4 4 4 3 4 V3 4
4 4 3 4 4 4 4 4 3 4 V1 4
5 4 3 4 4 4 4 4 3 4 V2 4
6 4 3 4 4 4 4 4 3 4 V3 4
7 3 3 3 2 2 4 2 1 3 V1 2
8 3 3 3 2 2 4 2 1 3 V2 3
9 3 3 3 2 2 4 2 1 3 V3 4
10 4 4 4 4 1 4 4 4 2 V1 2
# ... with 269 more rows
但我希望剩下的是抑郁症，所以我再次转向：
hwk3 <- hwk2 %>%
pivot_longer(cols = 1:9,
names_to = "Depression Test",
values_to = "Depression Score")
hwk3
现在给了我这个：
`Anxiety Test` `Anxiety Score` `Depression Test` `Depression Score`
<chr> <dbl> <chr> <dbl>
1 V1 4 V4 4
2 V1 4 V5 3
3 V1 4 V6 4
4 V1 4 V7 4
5 V1 4 V8 4
6 V1 4 V9 4
7 V1 4 V10 4
8 V1 4 V11 3
9 V1 4 V12 4
10 V2 4 V4 4
# ... with 2,501 more rows
我想要的
好吧，现在我想将指定测试为焦虑测试或抑郁测试的单列和分数的单列分组。所以基本上它看起来像这样：
Test Test_Type Score
V1 Anxiety 4
V4 Depression 3
如果有一个不那么迂回的方式来做到这一点，那就太好了。同样重要的是一个解决方案，它不会改变分数的实际值，而只是将它们分组到我上面提到的类别中。
您在寻找：
library(tidyverse)
hwk %>%
rename_with(.cols = everything(),
.fn = ~str_c(., c(rep('_Anxiety', 3), rep('_Depression', 9)))) %>%
pivot_longer(cols = everything(),
names_to = c('Test', 'Test_Type'),
names_pattern = '(.*)_(.*)',
values_to = 'Score')
# A tibble: 1,116 x 3
Test Test_Type Score
<chr> <chr> <dbl>
1 V1 Anxiety 4
2 V2 Anxiety 4
3 V3 Anxiety 4
4 V4 Depression 4
5 V5 Depression 3
6 V6 Depression 4
7 V7 Depression 4
8 V8 Depression 4
9 V9 Depression 4
10 V10 Depression 4
# ... with 1,106 more rows
数据
我尝试使用谷歌搜索并在 SO 中搜索此问题的答案，但答案通常只是说明如何将列分成两列，而不是按特定标准对列进行分组。首先，这是我的数据的输入 hwk:
structure(list(V1 = structure(c(4, 4, 2, 2, 2, 2, 2, 2, 4, 4,
2, 3, 2, 3, 4, 2, 2, 2, 3, 3, 2, 3, 1, 3, 3, 3, 3, 4, 1, 2, 4,
1, 2, 3, 2, 3, 1, 1, 2, 2, 4, 3, 2, 1, 2, 3, 3, 4, 3, 3, 2, 3,
1, 4, 3, 2, 3, 4, 1, 3, 3, 3, 2, 2, 1, 2, 3, 4, 4, 2, 4, 3, 2,
3, 3, 3, 3, 2, 4, 3, 3, 3, 2, 2, 3, 4, 2, 4, 4, 2, 2, 3, 3), format.spss = "F8.0"),
V2 = structure(c(4, 4, 3, 4, 3, 4, 3, 2, 4, 1, 3, 3, 3, 4,
3, 3, 2, 3, 4, 3, 1, 4, 2, 3, 4, 2, 4, 3, 3, 2, 3, 2, 3,
3, 4, 3, 3, 3, 3, 3, 3, 2, 4, 2, 2, 2, 4, 3, 4, 4, 2, 4,
2, 3, 3, 3, 3, 3, 4, 3, 3, 3, 3, 4, 3, 3, 4, 4, 4, 4, 4,
3, 4, 3, 3, 3, 4, 2, 4, 3, 4, 3, 3, 2, 3, 3, 4, 3, 4, 3,
4, 4, 3), format.spss = "F8.0"), V3 = structure(c(4, 4, 4,
4, 4, 4, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4,
4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4), format.spss = "F8.0"),
V4 = structure(c(4, 4, 3, 4, 3, 4, 2, 1, 3, 2, 3, 1, 4, 4,
2, 3, 2, 2, 2, 4, 1, 2, 2, 2, 3, 2, 3, 2, 2, 1, 3, 1, 1,
2, 4, 1, 1, 2, 3, 2, 2, 1, 1, 1, 3, 2, 4, 3, 3, 3, 3, 3,
3, 4, 3, 1, 4, 3, 4, 3, 2, 3, 2, 1, 4, 1, 4, 1, 2, 4, 4,
4, 3, 3, 3, 2, 2, 1, 4, 3, 2, 3, 2, 1, 3, 4, 1, 2, 4, 3,
4, 2, 2), format.spss = "F8.0"), V5 = structure(c(3, 3, 3,
4, 3, 4, 3, 1, 1, 1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 3, 2,
2, 2, 2, 4, 2, 3, 2, 3, 4, 1, 4, 2, 3, 3, 2, 2, 3, 2, 2,
3, 3, 2, 3, 3, 3, 2, 2, 2, 3, 2, 3, 3, 2, 2, 3, 3, 2, 3,
2, 2, 3, 3, 3, 2, 3, 3, 3, 4, 3, 2, 3, 3, 3, 3, 3, 3, 4,
3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 4, 3, 3), format.spss = "F8.0"),
V6 = structure(c(4, 4, 3, 4, 3, 4, 4, 1, 3, 3, 3, 3, 2, 3,
4, 2, 4, 3, 3, 3, 3, 4, 4, 3, 3, 3, 4, 4, 4, 3, 4, 4, 3,
3, 3, 4, 2, 2, 3, 3, 3, 4, 2, 4, 3, 4, 4, 4, 3, 4, 2, 4,
3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 3, 1, 4, 4, 4, 4, 4, 4,
4, 3, 4, 4, 4, 4, 2, 4, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 3,
4, 4, 4), format.spss = "F8.0"), V7 = structure(c(4, 4, 2,
4, 2, 4, 4, 3, 3, 3, 2, 2, 4, 4, 3, 3, 1, 4, 3, 3, 1, 2,
4, 3, 4, 2, 4, 4, 3, 3, 2, 2, 3, 2, 4, 3, 3, 3, 3, 3, 3,
1, 4, 3, 2, 2, 4, 3, 4, 4, 2, 4, 2, 3, 4, 3, 3, 3, 4, 3,
4, 4, 3, 4, 4, 3, 4, 4, 4, 4, 3, 4, 4, 4, 3, 3, 4, 3, 4,
3, 3, 3, 3, 2, 2, 4, 4, 4, 4, 2, 4, 4, 3), format.spss = "F8.0"),
V8 = structure(c(4, 4, 2, 1, 2, 1, 1, 1, 3, 3, 2, 3, 2, 3,
4, 2, 2, 2, 3, 3, 2, 3, 1, 3, 3, 3, 3, 4, 1, 2, 4, 1, 2,
3, 2, 3, 1, 1, 2, 2, 3, 1, 1, 1, 2, 3, 3, 4, 3, 3, 2, 3,
1, 3, 4, 2, 3, 4, 1, 3, 3, 3, 2, 2, 1, 2, 3, 4, 4, 2, 4,
3, 4, 4, 4, 4, 3, 2, 4, 3, 3, 3, 2, 2, 3, 4, 2, 4, 4, 2,
1, 3, 4), format.spss = "F8.0"), V9 = structure(c(4, 4, 4,
4, 4, 4, 4, 4, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 2, 3, 4, 4,
4, 4, 3, 4, 4, 4, 4, 4, 4, 4, 3, 3, 3, 4, 3, 2, 4, 3, 4,
4, 4, 4, 4, 4, 3, 3, 3, 3, 4, 4, 4, 4, 4, 3, 4, 3, 2, 4,
3, 3, 4, 4, 4, 3, 4, 4, 4, 4, 4, 3, 4, 3, 4, 3, 4, 4, 4,
4, 3, 4, 4, 4, 4, 4, 3, 2, 4, 4, 4, 4, 4), format.spss = "F8.0"),
V10 = structure(c(4, 4, 2, 4, 2, 4, 3, 2, 3, 3, 3, 2, 4,
4, 2, 2, 1, 3, 4, 4, 1, 4, 2, 3, 3, 2, 4, 3, 2, 3, 3, 1,
3, 2, 4, 3, 2, 3, 3, 3, 3, 1, 2, 4, 2, 3, 4, 4, 3, 3, 2,
4, 2, 4, 3, 3, 4, 3, 4, 3, 4, 4, 4, 1, 4, 3, 3, 4, 3, 4,
4, 3, 3, 3, 3, 3, 4, 1, 4, 3, 3, 3, 3, 2, 3, 4, 4, 2, 4,
2, 4, 4, 3), format.spss = "F8.0"), V11 = structure(c(3,
3, 1, 4, 1, 4, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 2, 2, 2, 1,
2, 3, 1, 2, 3, 3, 2, 1, 2, 2, 2, 3, 2, 2, 3, 2, 1, 2, 2,
1, 1, 4, 3, 1, 3, 2, 3, 1, 2, 1, 2, 1, 2, 2, 1, 2, 2, 3,
2, 2, 2, 2, 2, 2, 1, 1, 1, 3, 3, 4, 2, 1, 2, 2, 3, 3, 3,
3, 4, 3, 2, 3, 3, 2, 2, 2, 2, 1, 3, 1, 4, 1, 3), format.spss = "F8.0"),
V12 = structure(c(4, 4, 3, 2, 3, 2, 3, 1, 3, 3, 3, 3, 2,
3, 3, 2, 4, 3, 3, 4, 4, 3, 3, 4, 4, 3, 3, 3, 4, 3, 4, 4,
3, 3, 3, 4, 2, 2, 3, 3, 3, 4, 2, 4, 3, 4, 4, 4, 3, 4, 2,
4, 3, 3, 3, 3, 4, 3, 3, 2, 2, 1, 1, 3, 1, 4, 4, 4, 4, 4,
4, 4, 3, 3, 2, 2, 2, 2, 4, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4,
3, 2, 3, 4), format.spss = "F8.0")), class = c("tbl_df",
"tbl", "data.frame"), row.names = c(NA, -93L))
目前的解决方案
所以我希望第一个变量是焦虑量表的问题，每个变量的分数：
hwk2 <- hwk %>%
pivot_longer(cols = 1:3,
names_to = "Anxiety Test",
values_to = "Anxiety Score")
hwk2
这给了我这个：
# A tibble: 279 x 11
V4 V5 V6 V7 V8 V9 V10 V11 V12 `Anxiety Test` `Anxiety Score`
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <chr> <dbl>
1 4 3 4 4 4 4 4 3 4 V1 4
2 4 3 4 4 4 4 4 3 4 V2 4
3 4 3 4 4 4 4 4 3 4 V3 4
4 4 3 4 4 4 4 4 3 4 V1 4
5 4 3 4 4 4 4 4 3 4 V2 4
6 4 3 4 4 4 4 4 3 4 V3 4
7 3 3 3 2 2 4 2 1 3 V1 2
8 3 3 3 2 2 4 2 1 3 V2 3
9 3 3 3 2 2 4 2 1 3 V3 4
10 4 4 4 4 1 4 4 4 2 V1 2
# ... with 269 more rows
但我希望剩下的是抑郁症，所以我再次转向：
hwk3 <- hwk2 %>%
pivot_longer(cols = 1:9,
names_to = "Depression Test",
values_to = "Depression Score")
hwk3
现在给了我这个：
`Anxiety Test` `Anxiety Score` `Depression Test` `Depression Score`
<chr> <dbl> <chr> <dbl>
1 V1 4 V4 4
2 V1 4 V5 3
3 V1 4 V6 4
4 V1 4 V7 4
5 V1 4 V8 4
6 V1 4 V9 4
7 V1 4 V10 4
8 V1 4 V11 3
9 V1 4 V12 4
10 V2 4 V4 4
# ... with 2,501 more rows
我想要的
好吧，现在我想将指定测试为焦虑测试或抑郁测试的单列和分数的单列分组。所以基本上它看起来像这样：
Test Test_Type Score
V1 Anxiety 4
V4 Depression 3
如果有一个不那么迂回的方式来做到这一点，那就太好了。同样重要的是一个解决方案，它不会改变分数的实际值，而只是将它们分组到我上面提到的类别中。
您在寻找：
library(tidyverse)
hwk %>%
rename_with(.cols = everything(),
.fn = ~str_c(., c(rep('_Anxiety', 3), rep('_Depression', 9)))) %>%
pivot_longer(cols = everything(),
names_to = c('Test', 'Test_Type'),
names_pattern = '(.*)_(.*)',
values_to = 'Score')
# A tibble: 1,116 x 3
Test Test_Type Score
<chr> <chr> <dbl>
1 V1 Anxiety 4
2 V2 Anxiety 4
3 V3 Anxiety 4
4 V4 Depression 4
5 V5 Depression 3
6 V6 Depression 4
7 V7 Depression 4
8 V8 Depression 4
9 V9 Depression 4
10 V10 Depression 4
# ... with 1,106 more rows