Question

问题

我有两篇针对不同模型的参数的小标题，我希望找到一种方法来生成这些模型的前80个输出的列表的列表列。这里的示例是100个模型，但是将来，我将研究10,000多个模型，这就是为什么我希望将它们的输出与它们连续分组的原因。

我尝试过的事情

library(tidyverse)
new.tibble <- tibble(a = rep(1:5, 20))
new.tibble <- new.tibble %>% add_column(b = 1:100)

my.vector <- c(1:80)

# What I want
new.tibble <- new.tibble %>% mutate(c = lapply(my.vector, function(x) {a ^ (x / b)}))

我已经用different apply functions尝试了此操作（适用，适用，适用等），但它似乎不起作用。当我运行最后一行时，得到以下内容：

> new.tibble <- new.tibble %>% mutate(c = lapply(my.vector, function(x) {a ^ (x / b)}))
Error: Column `c` must be length 100 (the number of rows) or one, not 80

这使我相信，该mutate总共仅生成80个输出，而不是每行80个输出，并且按照我的意愿将这80个输出存储在该行的列表中。我尝试让我的小标题按行排列，看看是否有帮助：

> row.tibble <- rowwise(new.tibble)
> row.tibble <- row.tibble %>% mutate(c = lapply(my.vector, function(x){a ^ (x / b)}))
Error: Column `c` must be length 1 (the group size), not 80

它没有。我知道建立一个while循环并不难，只需生成不同的输出作为自己的单独列表即可，但是有10,000多个列表，每个列表对应于行中的模型，我想列表列将是组织输出的最佳方法。我也尝试过使用as.list强制返回列表的输出，但这并没有达到我的预期：

> row.tibble <- row.tibble %>% mutate(c = as.list(lapply(my.vector, FUN = function(x){a ^ (x / b)})))
Error: Column `c` must be length 1 (the group size), not 80
> new.tibble <- new.tibble %>% mutate(c = as.list(lapply(my.vector, FUN = function(x){a ^ (x / b)})))
Error: Column `c` must be length 100 (the number of rows) or one, not 80

我尝试避开不适，并尝试直接获得所需的输出，但这没有用：

> new.tibble %>% mutate(c = as.list(a ^ (my.vector / b)))
# A tibble: 100 x 3
       a     b c        
   <int> <int> <list>   
 1     1     1 <dbl [1]>
 2     2     2 <dbl [1]>
 3     3     3 <dbl [1]>
 4     4     4 <dbl [1]>
 5     5     5 <dbl [1]>
 6     1     6 <dbl [1]>
 7     2     7 <dbl [1]>
 8     3     8 <dbl [1]>
 9     4     9 <dbl [1]>
10     5    10 <dbl [1]>
# ... with 90 more rows
Warning message:
In my.vector/b :
  longer object length is not a multiple of shorter object length

> row.tibble %>% mutate(c = as.list(a ^ (my.vector / b)))
Error: Column `c` must be length 1 (the group size), not 80

其他信息

> sessionInfo()
R version 4.0.0 (2020-04-24)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 10 x64 (build 18363)

Matrix products: default

locale:
[1] LC_COLLATE=English_United States.1252  LC_CTYPE=English_United States.1252    LC_MONETARY=English_United States.1252 LC_NUMERIC=C                          
[5] LC_TIME=English_United States.1252    

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
[1] forcats_0.5.0   stringr_1.4.0   dplyr_0.8.5     purrr_0.3.4     readr_1.3.1     tidyr_1.0.3     tibble_3.0.1    ggplot2_3.3.0   tidyverse_1.3.0

loaded via a namespace (and not attached):
 [1] Rcpp_1.0.4.6     cellranger_1.1.0 pillar_1.4.4     compiler_4.0.0   dbplyr_1.4.3     tools_4.0.0      jsonlite_1.6.1   lubridate_1.7.8  lifecycle_0.2.0 
[10] nlme_3.1-147     gtable_0.3.0     lattice_0.20-41  pkgconfig_2.0.3  rlang_0.4.6      reprex_0.3.0     cli_2.0.2        DBI_1.1.0        rstudioapi_0.11 
[19] haven_2.2.0      withr_2.2.0      xml2_1.3.2       httr_1.4.1       fs_1.4.1         generics_0.0.2   vctrs_0.2.4      hms_0.5.3        grid_4.0.0      
[28] tidyselect_1.0.0 glue_1.4.0       R6_2.4.1         fansi_0.4.1      readxl_1.3.1     modelr_0.1.7     magrittr_1.5     backports_1.1.6  scales_1.1.1    
[37] ellipsis_0.3.0   rvest_0.3.5      assertthat_0.2.1 colorspace_1.4-1 utf8_1.1.4       stringi_1.4.6    munsell_0.5.0    broom_0.5.6      crayon_1.3.4

Answer 1

您是否正在寻找这种结果？

new.tibble <- new.tibble %>% 
  mutate(c = map2(.x = a, .y = b, .f = ~.x^(my.vector/.y) ))

输出：

head(new.tibble)
# A tibble: 6 x 3
      a     b c         
  <int> <int> <list>    
1     1     1 <dbl [80]>
2     2     2 <dbl [80]>
3     3     3 <dbl [80]>
4     4     4 <dbl [80]>
5     5     5 <dbl [80]>
6     1     6 <dbl [80]>

有没有一种（整洁的）方式来创建自定义列表列？

1 个答案: