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使用dplyr将R中的元组列表除以R中的元组列表

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假设我有一个嵌套的tibble，格式如下：

# A tibble: 3 x 3
AccountNumber Tibble1          Tibble2         
         <int> <list>           <list>          
1             1 <tibble [1 x 3]> <tibble [1 x 3]>
2             2 <tibble [1 x 3]> <tibble [1 x 3]>
3             3 <tibble [1 x 3]> <tibble [1 x 3]>

这可以通过以下代码生成：

library(tidyverse)

tibble1 <- tibble(AccountNumber = 1:3, A_1 = 1, B_1 = 2, C_1 = 3) %>%
  group_by(AccountNumber) %>%
  nest(.key = "Tibble1")

tibble2 <- tibble(AccountNumber = 1:3, A_2 = 4, B_2 = 5, C_2 = 6) %>%
  group_by(AccountNumber) %>%
  nest(.key = "Tibble2")

tibble_joined <- left_join(tibble1, tibble2, by = "AccountNumber")
  • How would I create a third list of tibbles by dividing Tibble1 by Tibble 2?

基本上我想要以下格式：

# A tibble: 3 x 3
AccountNumber Tibble1          Tibble2            Tibble3(Tibble2 / Tibble1)
         <int> <list>           <list>            <list>
1             1 <tibble [1 x 3]> <tibble [1 x 3]> <tibble [1 x 3]>
2             2 <tibble [1 x 3]> <tibble [1 x 3]> <tibble [1 x 3]>
3             3 <tibble [1 x 3]> <tibble [1 x 3]> <tibble [1 x 3]>

...其中Tibble3只是Tibble 2与Tibble 1之比：

  • 每列为

  • 每个帐号

到目前为止，我的尝试是：

tibble_joined %>%
  group_by(AccountNumber) %>%
  mutate(Tibble3 = tibble(tibble2 / tibble1))

和

tibble_joined %>%
  group_by(AccountNumber) %>%
  summarise(Tibble3 = tibble2 / tibble1)

这两个都给出了这个错误：

Error in mutate_impl(.data, dots) : 
  Evaluation error: non-numeric argument to binary operator.

我试图找到这个问题的优雅解决方案，但我找不到任何东西 .

================================================== =======================

我完全清楚我的问题可以通过以下方式解决：

tibble_Main %>%
  group_by(AccountNumber) %>%
  unnest() %>%
  mutate(A_Ratio = A_2 / A_1,
         B_Ratio = B_2 / B_1,
         C_Ratio = C_2 / C_2)

...生成以下内容：

# A tibble: 3 x 10
# Groups: AccountNumber [3]
  AccountNumber   A_1   B_1   C_1   A_2   B_2   C_2 A_Ratio B_Ratio C_Ratio
          <int> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>   <dbl>   <dbl>   <dbl>
1             1  1.00  2.00  3.00  4.00  5.00  6.00    4.00    2.50    1.00
2             2  1.00  2.00  3.00  4.00  5.00  6.00    4.00    2.50    1.00
3             3  1.00  2.00  3.00  4.00  5.00  6.00    4.00    2.50    1.00

...但这看起来很麻烦，并且会因许多专栏而烦恼 .

r dplyr tibble

1 回答

  • 1

    我们可以使用 purrr 从 purrr 将 tibble 除以另一个 tibble

    library(purrr)
res <- tibble_joined %>% 
          mutate(Tibble3 = map2(Tibble1, Tibble2,  ~ as_tibble( .y/.x) %>%
                             rename_all(funs(sub('_.*', "_ratio", .)))))
res
# A tibble: 3 x 4
#   AccountNumber Tibble1          Tibble2          Tibble3         
#          <int> <list>           <list>           <list>          
#1             1 <tibble [1 x 3]> <tibble [1 x 3]> <tibble [1 x 3]>
#2             2 <tibble [1 x 3]> <tibble [1 x 3]> <tibble [1 x 3]>
#3             3 <tibble [1 x 3]> <tibble [1 x 3]> <tibble [1 x 3]>

res$Tibble3
#[[1]]
# A tibble: 1 x 3
#  A_ratio B_ratio C_ratio
#    <dbl>   <dbl>   <dbl>
#1    4.00    2.50    2.00

#[[2]]
# A tibble: 1 x 3
#  A_ratio B_ratio C_ratio
#    <dbl>   <dbl>   <dbl>
#1    4.00    2.50    2.00

#[[3]]
# A tibble: 1 x 3
#  A_ratio B_ratio C_ratio
#    <dbl>   <dbl>   <dbl>
#1    4.00    2.50    2.00

    注意： purrr 是 tidyverse 包的一部分

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