长幅和宽幅练习

Question

我有以下问题：

给出以下示例数据集：

 structure(list(idpat = c("1", "2", "2", "1", "1"), idA = c("100a", 
"100b", "100c", "100b", "100a"), redA = c(1, 0, 1, 0, 1), idB = c("200a", 
"200b", "200a", "200b", "200a"), redB = c(0, 1, 1, 1, 0)), class = "data.frame", row.names = c(NA, 
-5L))

  idpat  idA redA  idB redB
1     1 100a    1 200a    0
2     2 100b    0 200b    1
3     2 100c    1 200a    1
4     1 100b    0 200b    1
5     1 100a    1 200a    0

我想获得：

 structure(list(idpat = c(1, 1, 2, 2, 2, 2, 1, 1, 1, 1), id = c("100a", 
"200a", "100b", "200b", "100c", "200a", "100b", "200b", "100a", 
"200a"), red = c(1, 0, 0, 1, 1, 1, 0, 1, 1, 0)), class = "data.frame", row.names = c(NA, 
-10L))

   idpat   id red
1      1 100a   1
2      1 200a   0
3      2 100b   0
4      2 200b   1
5      2 100c   1
6      2 200a   1
7      1 100b   0
8      1 200b   1
9      1 100a   1
10     1 200a   0

我正在尝试以下操作，但无法得到我所需要的。这个pivot_longer对我来说有点棘手：

df1|>pivot_longer(cols =c("idpat","idA","idB"),names_to = c("id") ,values_to = "vals")

Answer 1

为了完整性，采用

data.table

方法

library(data.table)
melt(setDT(mydata), measure.vars = patterns(id = "^id[AB]", red = "^red[AB]"))
#    idpat variable   id red
# 1:     1        1 100a   1
# 2:     2        1 100b   0
# 3:     2        1 100c   1
# 4:     1        1 100b   0
# 5:     1        1 100a   1
# 6:     1        2 200a   0
# 7:     2        2 200b   1
# 8:     2        2 200a   1
# 9:     1        2 200b   1
#10:     1        2 200a   0
     

Answer 2

您可以使用

names_pattern

参数和针对多个值列的特殊

".value"

来实现您想要的结果，如下所示：

df1 |>
  tidyr::pivot_longer(
    cols = -idpat,
    names_to = c(".value", "what"),
    names_pattern = "^(.*?)(.)$"
  )
#> # A tibble: 10 × 4
#>    idpat what  id      red
#>    <chr> <chr> <chr> <dbl>
#>  1 1     A     100a      1
#>  2 1     B     200a      0
#>  3 2     A     100b      0
#>  4 2     B     200b      1
#>  5 2     A     100c      1
#>  6 2     B     200a      1
#>  7 1     A     100b      0
#>  8 1     B     200b      1
#>  9 1     A     100a      1
#> 10 1     B     200a      0

Answer 3

这是一个基本解决方案。首先创建列表

L

，它定义哪些输入列与哪些输出列关联。我们使用输入列号，但也可以使用相应的输入列名称。

L <- list(idpat = c(1, 1), id = c(2, 4), red = c(3, 5))
data.frame(lapply(L, function(x) c(t(df1[x]))))

##     idpat   id red
## 1       1 100a   1
## 2       1 200a   0
## 3       2 100b   0
## 4       2 200b   1
## 5       2 100c   1
## 6       2 200a   1
## 7       1 100b   0
## 8       1 200b   1
## 9       1 100a   1
## 10      1 200a   0