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将列表中的每个数据帧绑定到另一个列表中的各个数据帧[R]

rbind loops list r

C. Guff · 技术社区 · 1 年前

我有两个由数据帧组成的列表(在示例中,按列表列出3个数据帧)。我想绑定这两个列表的数据帧行。

我想结婚 rbind() 具有 Map() , map() 或 lapply() :

Map(rbind,List1,List2)

然而,当我应用该代码时,我只将第一个列表的一个数据帧绑定到第二个列表中的另一个。在最终输出中只给出三个最终数据帧。而我希望第一个列表的每个数据帧被第二个列表的每一个数据帧绑定。给定9个数据帧的输出,而不是我实际拥有的3个。

如果我有 清单1 由组成 df-a , df-b , df-c ,和 清单2 由组成 df-1 , df-2 , df-3

我想在列表中有数据帧的最终组合:

a-1,
a-2,
a-3,
b-1,
b-2,
b-3,
c-1,
c-2,
c-3

相反,使用上面的代码,我只有:

a-1,
b-2,
c-3

我该如何解决此问题?

在下面,您将找到一个列表示例:

列表1

List1 <- structure(list(structure(list(Values = c(3.95464079265625, 4.28848139983694, 
4.59263295848293, 4.87348316186162, 5.13543268303855, 5.38166290292844, 
5.61456044441555, 5.8359690473681, 6.04734746369445, 6.24987292533157
), array = c(10, 12, 14, 16, 18, 20, 22, 24, 26, 28), ID = c("1", 
"1", "1", "1", "1", "1", "1", "1", "1", "1")), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -10L)), structure(list(
    Values = c(6.82179666846063, 7.09317719563801, 7.33103172027803, 
    7.54350812736648, 7.73603145714462, 7.91240924925746, 8.07542410682578, 
    8.22717585659922, 8.36929092723351, 8.50305660066192), array = c(10, 
    12, 14, 16, 18, 20, 22, 24, 26, 28), ID = c("2", "2", "2", 
    "2", "2", "2", "2", "2", "2", "2")), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -10L)), structure(list(
    Values = c(4.30943511800407, 4.62131690026944, 4.90255443339214, 
    5.15997082435108, 5.39822432778288, 5.62065617021581, 5.8297555323974, 
    6.02743344284351, 6.21519337899337, 6.39424241723516), array = c(10, 
    12, 14, 16, 18, 20, 22, 24, 26, 28), ID = c("3", "3", "3", 
    "3", "3", "3", "3", "3", "3", "3")), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -10L))), ptype = structure(list(
    Values = numeric(0), array = numeric(0), ID = character(0)), class = c("tbl_df", 
"tbl", "data.frame"), row.names = integer(0)), class = c("vctrs_list_of", 
"vctrs_vctr", "list"))

列表2

List2 <- structure(list(structure(list(Values = c(8.17331050066691, 8.87216395029884, 
9.50945470046081, 10.098386534235, 10.6480633233427, 11.1650711748644, 
11.6543539639589, 12.1197331206333, 12.5642336602292, 12.9902979406118
), array = c(10, 12, 14, 16, 18, 20, 22, 24, 26, 28), ID = c("1", 
"1", "1", "1", "1", "1", "1", "1", "1", "1")), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -10L)), structure(list(
    Values = c(7.98726323968068, 8.65441558454706, 9.26177728829764, 
    9.82224562340532, 10.3446999917905, 10.8355558754145, 11.2996236117639, 
    11.7406176335271, 12.1614755496734, 12.5645671491611), array = c(10, 
    12, 14, 16, 18, 20, 22, 24, 26, 28), ID = c("2", "2", "2", 
    "2", "2", "2", "2", "2", "2", "2")), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -10L)), structure(list(
    Values = c(7.34557601230256, 8.06135956556851, 8.72069616000395, 
    9.33527354603708, 9.9132220632988, 10.4604689570926, 10.9814938935561, 
    11.4797806661262, 11.9581023062164, 12.4187090723963), array = c(10, 
    12, 14, 16, 18, 20, 22, 24, 26, 28), ID = c("3", "3", "3", 
    "3", "3", "3", "3", "3", "3", "3")), class = c("tbl_df", 
"tbl", "data.frame"), row.names = c(NA, -10L))), ptype = structure(list(
    Values = numeric(0), array = numeric(0), ID = character(0)), class = c("tbl_df", 
"tbl", "data.frame"), row.names = integer(0)), class = c("vctrs_list_of", 
"vctrs_vctr", "list"))

2 回复 | 直到 1 年前

Andre Wildberg 1 年前

看看这是否适合你

Map(\(x) lapply(List2, \(y) rbind(x, y)), List1)
[[1]]
[[1]][[1]]
# A tibble: 20 Ã 3
   Values array ID   
    <dbl> <dbl> <chr>
 1   3.95    10 1    
 2   4.29    12 1    
 3   4.59    14 1    
 4   4.87    16 1    
 5   5.14    18 1    
 6   5.38    20 1    
 7   5.61    22 1    
 8   5.84    24 1    
 9   6.05    26 1    
10   6.25    28 1    
11   8.17    10 1    
12   8.87    12 1    
13   9.51    14 1    
14  10.1     16 1    
15  10.6     18 1    
16  11.2     20 1    
17  11.7     22 1    
18  12.1     24 1    
19  12.6     26 1    
20  13.0     28 1

...

它实际上只是一个替身 lapply

lapply(List1, \(x) lapply(List2, \(y) rbind(x, y)))

margusl 1 年前

一种选择是首先使用创建组合 expand.grid() / tidyr::expand_grid() 然后 rbind() 通过 pmap() .

直接使用框架列表:

expand.grid(List1, List2) |> 
  purrr::pmap(rbind) |>
  str()
#> List of 9
#>  $ :Classes 'tbl_df', 'tbl' and 'data.frame':    20 obs. of  3 variables:
#>   ..$ Values: num [1:20] 3.95 4.29 4.59 4.87 5.14 ...
#>   ..$ array : num [1:20] 10 12 14 16 18 20 22 24 26 28 ...
#>   ..$ ID    : chr [1:20] "1" "1" "1" "1" ...

#>   /../

#>  $ :Classes 'tbl_df', 'tbl' and 'data.frame':    20 obs. of  3 variables:
#>   ..$ Values: num [1:20] 4.31 4.62 4.9 5.16 5.4 ...
#>   ..$ array : num [1:20] 10 12 14 16 18 20 22 24 26 28 ...
#>   ..$ ID    : chr [1:20] "3" "3" "3" "3" ...

或者通过索引:

expand.grid(idx_1 = seq_along(List1), idx_2 = seq_along(List2)) |> 
  purrr::pmap(\(idx_1, idx_2) rbind(List1[[idx_1]], List2[[idx_2]])) |>
  str()
#> List of 9
#>  $ :Classes 'tbl_df', 'tbl' and 'data.frame':    20 obs. of  3 variables:
#>   ..$ Values: num [1:20] 3.95 4.29 4.59 4.87 5.14 ...
#>   ..$ array : num [1:20] 10 12 14 16 18 20 22 24 26 28 ...
#>   ..$ ID    : chr [1:20] "1" "1" "1" "1" ...

#>   /../

#>  $ :Classes 'tbl_df', 'tbl' and 'data.frame':    20 obs. of  3 variables:
#>   ..$ Values: num [1:20] 4.31 4.62 4.9 5.16 5.4 ...
#>   ..$ array : num [1:20] 10 12 14 16 18 20 22 24 26 28 ...
#>   ..$ ID    : chr [1:20] "3" "3" "3" "3" ...

^{创建于2024-06-09

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