在R中加入dplyr时嵌套重复变量

Question

我参加的是有重复的,我列的数据帧(tibbles)不要想加入.下面的示例是我通常会做的(加入i,但不是a或b):

library(dplyr)

df1 <- tibble(i = letters[1:3], a = 1:3,   b = 4:6)
df2 <- tibble(i = letters[1:3], a = 11:13, b = 14:16)

d <- full_join(df1, df2, by ="i")
d
#> # A tibble: 3 × 5
#>       i   a.x   b.x   a.y   b.y
#>   <chr> <int> <int> <int> <int>
#> 1     a     1     4    11    14
#> 2     b     2     5    12    15
#> 3     c     3     6    13    16

我希望这些重复的变量作为嵌套列表返回,例如下面创建的输出:

tibble(
  i = letters[1:3],
  a = list(c(1, 11), c(2, 12), c(3, 13)),
  b = list(c(4, 14), c(5, 15), c(6, 16))
)
#> # A tibble: 3 × 3
#>       i         a         b
#>   <chr>    <list>    <list>
#> 1     a <dbl [2]> <dbl [2]>
#> 2     b <dbl [2]> <dbl [2]>
#> 3     c <dbl [2]> <dbl [2]>

有没有一种简单的方法来做这样的事情？

除此之外,我一直在使用各种stringr和tidyr方法(不成功).这是一个抛出错误的示例:

library(stringr)
library(tidyr)

# Find any variables with .x or .y
dup_var <- d %>% select(matches("\\.[xy]")) %>% names()

# Condense to the stems (original names) of these variables
dup_var_stems <- dup_var %>% str_replace("(\\.[x|y])+", "") %>% unique()

# For each stem, try to nest relevant data into a single variable
for (stem in dup_var_stems) {
  d <- d %>% nest_(key_col = stem, nest_cols = names(d)[str_detect(names(d), paste0(stem, "[$|\\.]"))])
}

UPDATE

在得到@Sotos和@conor的答案之后,我会提到解决方案需要在许多数据帧上推广到多个连接和重复列.下面是两个列(i和j)在五个数据帧上进行连接的示例.这会创建五个重复的列版本,a并且b还有很多独特的列c:g.一个问题是,没有后缀复制了这么多的数据帧导致复制版本.x,.x.x等.如此简单的正则表达式匹配.x|.y将错过列的无后缀版本.

library(dplyr)
library(purrr)


id_cols <- tibble(i = c("x", "x", "y", "y"),
                  j = c(1, 2, 1, 2))

df1 <- id_cols %>% cbind(tibble(a = 1:4, b = 5:8, c = 21:24))
df2 <- id_cols %>% cbind(tibble(a = 2:5, b = 6:9, d = 31:34))
df3 <- id_cols %>% cbind(tibble(a = 2:5, b = 6:9, e = 31:34))
df4 <- id_cols %>% cbind(tibble(a = 2:5, b = 6:9, f = 31:34))
df5 <- id_cols %>% cbind(tibble(a = 2:5, b = 6:9, g = 31:34))
datalist <- list(df1, df2, df3, df4, df5)

d <- reduce(datalist, full_join, by = c("i", "j"))
d
#>   i j a.x b.x  c a.y b.y  d a.x.x b.x.x  e a.y.y b.y.y  f a b  g
#> 1 x 1   1   5 21   2   6 31     2     6 31     2     6 31 2 6 31
#> 2 x 2   2   6 22   3   7 32     3     7 32     3     7 32 3 7 32
#> 3 y 1   3   7 23   4   8 33     4     8 33     4     8 33 4 8 33
#> 4 y 2   4   8 24   5   9 34     5     9 34     5     9 34 5 9 34

Answer 1

这是一次尝试,

library(dplyr)
library(tidyr)

melt(d, id.vars = 'i') %>% 
   group_by(a = sub('\\..*', '', variable), i) %>% 
   summarise(new = list(value)) %>% 
   spread(a, new)

# A tibble: 3 × 3
#      i         a         b
#* <chr>    <list>    <list>
#1     a <int [2]> <int [2]>
#2     b <int [2]> <int [2]>
#3     c <int [2]> <int [2]>

#With structure
Classes ‘tbl_df’, ‘tbl’ and 'data.frame':   3 obs. of  3 variables:
 $ i: chr  "a" "b" "c"
 $ a:List of 3
  ..$ : int  1 11
  ..$ : int  2 12
  ..$ : int  3 13
 $ b:List of 3
  ..$ : int  4 14
  ..$ : int  5 15
  ..$ : int  6 16

#Or via reshape2 package

library(dplyr)
library(reshape2)

d1 <- melt(d, id.vars = 'i') %>% 
         group_by(a = sub('\\..*', '', variable), i) %>% 
         summarise(new = list(value))

d2 <- dcast(d1, i ~ a, value.var = 'new')
#d2
#  i     a     b
#1 a 1, 11 4, 14
#2 b 2, 12 5, 15
#3 c 3, 13 6, 16

#with structure:
str(d2)
'data.frame':   3 obs. of  3 variables:
 $ i: chr  "a" "b" "c"
 $ a:List of 3
  ..$ : int  1 11
  ..$ : int  2 12
  ..$ : int  3 13
 $ b:List of 3
  ..$ : int  4 14
  ..$ : int  5 15
  ..$ : int  6 16

编辑

跟随你的想法,

library(dplyr)
library(reshape2)
library(purrr)
library(tidyr)

df <- melt(d, id.vars = c(names(d)[!grepl('a|b', names(d))]))

dots <- names(df)[!grepl('value', names(df))] %>% map(as.symbol)

df %>% mutate(variable = sub('\\..*', '', variable)) %>%
     group_by_(.dots = dots) %>%
     summarise(new = list(value)) %>%
     spread(variable, new) %>%
     ungroup()
# A tibble: 4 × 9
#      i     j     c     d     e     f     g         a         b
#* <chr> <dbl> <int> <int> <int> <int> <int>    <list>    <list>
#1     x     1    21    31    31    31    31 <int [5]> <int [5]>
#2     x     2    22    32    32    32    32 <int [5]> <int [5]>
#3     y     1    23    33    33    33    33 <int [5]> <int [5]>
#4     y     2    24    34    34    34    34 <int [5]> <int [5]>