Question

我建立了一个似乎可以正常工作的函数，但我不明白为什么。

我最初的问题是获取一个包含总体计数的data.frame并将其扩展以重新创建原始总体。如果您事先知道列名，这很容易。

      library(tidyverse)

      set.seed(121)

      test_counts <- tibble(Population = letters[1:4], Length = c(1,1,2,1), 
         Number = sample(1:100, 4))

      expand_counts_v0 <- function(Length, Population, Number) { 
            tibble(Population = Population, 
                   Length = rep(Length, times = Number))

      }


      test_counts %>% pmap_dfr(expand_counts_v0) %>%   # apply it
                 group_by(Population, Length) %>%    # test it
                   summarise(Number = n()) %>%  
                   ungroup %>%
                  { all.equal(., test_counts)}
      # [1] TRUE

但是，我想将其概括为一个不需要在data.frame的列名上知道的函数，并且我对NSE感兴趣，所以我写道：

test_counts1 <- tibble(Population = letters[1:4], 
                 Length = c(1,1,2,1), 
                 Number = sample(1:100, 4),
                 Height = c(100, 50, 45, 90),
                 Width = c(700, 50, 60, 90)
               )


expand_counts_v1 <- function(df, count = NULL) { 
     countq <- enexpr(count)
     names <- df %>% select(-!!countq) %>% names 
     namesq <- names %>% map(as.name)

     cols <- map(namesq, ~ expr(rep(!!., times = !!countq))
          ) %>% set_names(namesq)

      make_tbl <- function(...) {
                         expr(tibble(!!!cols)) %>% eval(envir = df)
      }

      df %>% pmap_dfr(make_tbl)
}

但是，当我测试此功能时，它似乎将行重复了4次：

   test_counts %>% expand_counts_v1(count = Number) %>% 
                   group_by(Population, Length) %>%
                   summarise(Number = n()) %>%
                   ungroup %>%
                   { sum(.$Number)/sum(test_counts$Number)}
   # [1] 4

这使我猜出了一个解决方案，即

   expand_counts_v2 <- function(df, count = NULL) { 
             countq <- enexpr(count)
             names <- df %>% select(-!!countq) %>% names 
             namesq <- names %>% map(as.name)

             cols <- map(namesq, ~ expr(rep(!!., times = !!countq))
              ) %>% set_names(namesq)

              make_tbl <- function(...) {
                          expr(tibble(!!!cols)) %>% eval(envir = df)
       }

      df %>% make_tbl
   }

这似乎可行：

 test_counts %>% expand_counts_v2(count = Number) %>% 
                 group_by(Population, Length) %>%
                 summarise(Number = n()) %>%
                 ungroup %>%
                { all.equal(., test_counts)}
 # [1] TRUE 

  test_counts1 %>% expand_counts_v2(count = Number) %>% 
                      group_by(Population, Length, Height, Width) %>%
                      summarise(Number = n()) %>%
                      ungroup %>%
                    { all.equal(., test_counts1)}
   # [1] TRUE

但是我不明白为什么。即使我不再使用pmap，它如何评估每一行？该功能需要应用到每一行才能正常工作，因此必须采用某种方式，但我看不出它是如何实现的。

编辑

Artem对发生的事情进行了正确的解释后，我意识到我可以做到这一点

expand_counts_v2 <- function(df, count = NULL) { 
      countq <- enexpr(count)
      names <- df %>% select(-!!countq) %>% names 
      namesq <- names %>% map(as.name)

      cols <- map(namesq, ~ expr(rep(!!., times = !!countq))
                  ) %>% set_names(namesq)

    expr(tibble(!!!cols)) %>% eval_tidy(data = df)
}

摆脱了不必要的mk_tbl函数。但是，正如Artem所说，这只是真正有效，因为rep是矢量化的。因此，它正在工作，但无法通过重写_v0函数并对其进行映射来完成，这是我试图复制的过程。最终，我发现了rlang :: new_function并写道：

expand_counts_v3 <- function(df, count = NULL) { 
      countq <- enexpr(count)
      names <- df %>% select(-!!countq) %>% names 
      namesq <- names %>% map(as.name)

      cols <- map(namesq, ~ expr(rep(!!., times = !!countq))
                  ) %>% set_names(namesq)

      all_names <- df %>% names %>% map(as.name) 
    args <- rep(0, times = length(all_names)) %>% as.list %>% set_names(all_names)

    correct_function <- new_function(args,     # this makes the function as in _v0
                                     expr(tibble(!!!cols))  )
    pmap_dfr(df, correct_function)     # applies it as in _v0
}

更长，也许更丑，但是按照我最初想要的方式工作。

Answer 1

问题出在eval( envir = df )中，该问题将整个数据帧暴露给make_tbl()。请注意，您永远不会在...中使用make_tbl()参数。相反，该函数有效地计算了

with( df, tibble(Population = rep(Population, times = Number), 
                 Length = rep(Length, times=Number)) )

无论您提供什么参数。当您通过pmap_dfr()调用该函数时，它实际上会计算上述四次（每行一次）并按行连接结果，从而导致您观察到的条目重复。当您删除pmap_dfr()时，该函数将被调用一次，但是由于rep本身是矢量化的（请尝试执行rep( test_counts$Population, test_counts$Number )以了解我的意思），make_tbl()会在一口气。

试图了解eval（expr，envir = df）的工作原理

1 个答案: