函数 NSE 中的 group_by dplyr答案

【问题标题】：group_by dplyr within a function NSE函数 NSE 中的 group_by dplyr
【发布时间】：2016-12-28 11:58:03
【问题描述】：

我在管道函数调用中使用 dplyr 和 group_by 时遇到问题。

可重现的示例：

使用以下数据：

ex_data<- structure(list(word1 = c("no", "not", "not", "no", "not", "not", 
"not", "not", "no", "not", "no", "not", "not", "not", "no", "not", 
"no", "no", "not", "not", "not", "no", "not", "without", "never", 
"no", "not", "no", "no", "not", "not", "not", "no", "no", "no", 
"not", "not", "without", "never", "no", "not", "not", "not", 
"not", "not", "never", "no", "no", "not", "not"), word2 = c("doubt", 
"like", "help", "no", "want", "wish", "allow", "care", "harm", 
"sorry", "great", "leave", "pretend", "worth", "pleasure", "love", 
"danger", "want", "afraid", "doubt", "fail", "good", "forget", 
"feeling", "forget", "matter", "avoid", "chance", "hope", "forgotten", 
"miss", "perfectly", "bad", "better", "opportunity", "admit", 
"fair", "delay", "failed", "wish", "dislike", "distress", "refuse", 
"regret", "trust", "want", "evil", "greater", "better", "blame"
), score = c(-1L, 2L, 2L, -1L, 1L, 1L, 1L, 2L, -2L, -1L, 3L, 
-1L, -1L, 2L, 3L, 3L, -2L, 1L, -2L, -1L, -2L, 3L, -1L, 1L, -1L, 
1L, -1L, 2L, 2L, -1L, -2L, 3L, -3L, 2L, 2L, -1L, 2L, -1L, -2L, 
1L, -2L, -2L, -2L, -2L, 1L, 1L, -3L, 3L, 2L, -2L), n = c(102L, 
99L, 82L, 60L, 45L, 39L, 36L, 23L, 22L, 21L, 19L, 18L, 18L, 17L, 
16L, 16L, 15L, 15L, 15L, 14L, 14L, 13L, 13L, 13L, 12L, 12L, 12L, 
11L, 11L, 10L, 10L, 10L, 9L, 9L, 9L, 9L, 9L, 9L, 8L, 8L, 8L, 
8L, 8L, 8L, 8L, 7L, 7L, 7L, 7L, 7L), contribution = c(-102L, 
198L, 164L, -60L, 45L, 39L, 36L, 46L, -44L, -21L, 57L, -18L, 
-18L, 34L, 48L, 48L, -30L, 15L, -30L, -14L, -28L, 39L, -13L, 
13L, -12L, 12L, -12L, 22L, 22L, -10L, -20L, 30L, -27L, 18L, 18L, 
-9L, 18L, -9L, -16L, 8L, -16L, -16L, -16L, -16L, 8L, 7L, -21L, 
21L, 14L, -14L)), .Names = c("word1", "word2", "score", "n", 
"contribution"), row.names = c(NA, -50L), class = c("tbl_df", 
"tbl", "data.frame"))

常规的典型管道操作按预期工作：

outside_result<- ex_data %>% 
  mutate(word2=reorder(word2,contribution)) %>% 
  group_by(word1) %>% 
  top_n(10,abs(contribution)) %>% 
  group_by(word1,word2) %>% 
  arrange(desc(contribution)) %>% 
  ungroup() %>% 
  mutate(word2 = factor(paste(word2,word1, sep = "__"),
                              levels=rev(paste(word2,word1,sep="__"))))

我已将上述内容实现为如下所示的函数：

order_bars <- function(df,facetPanel,barCategory,value){
        df %>% mutate(barCategory=reorder(barCategory,value)) %>% 
          group_by(facetPanel) %>% 
          top_n(10,abs(value)) %>% 
          group_by(facetPanel,barCategory) %>% 
          arrange(desc(value)) %>% 
          ungroup() %>% 
          mutate(barCategory = factor(paste(barCategory,facetPanel, sep = "__"),
                                     levels=rev(paste(barCategory,facetPanel,sep="__"))))
      }

并从这个post 中获得建议，在函数内的变异操作期间引用data.frame 的变量时使用$ 表示法。

inside_result<-order_bars(ex_data,ex_data$word1,ex_data$word2,ex_data$contribution)

R 抛出以下错误：

Error: unknown variable to group by : facetPanel
Called from: resolve_vars(new_groups, tbl_vars(.data))

我怀疑 group_by 需要调整以获取命名变量，或者我必须使用 .dot 表示法来引用列，尽管我只是把它扔到风中......

【问题讨论】：

标签： r function dplyr

【解决方案1】：

您需要学习如何使用 1) dplyr 动词的 SE 版本，例如 group_by_ 和 mutate_，以及 2) 神秘的 lazyeval::interp。请仔细阅读vignette("nse")。

那么我们可以来：

order_bars <- function(df, facetPanel, barCategory, value){
  require(lazyeval)
  df %>% 
    mutate_(barCategory = interp(~reorder(x, y), x = as.name(barCategory), 
                                 y = as.name(value))) %>% 
    group_by_(facetPanel) %>% 
    filter_(interp(~min_rank(desc(abs(x))) <= 10, x = as.name(value))) %>% 
    group_by_(facetPanel, barCategory) %>% 
    arrange_(interp(~desc(x), x = as.name(value))) %>% 
    ungroup() %>% 
    mutate_(barCategory = interp(
      ~factor(paste(x, y, sep = "__"), levels = rev(paste(x, y, sep = "__"))),
      x = as.name(barCategory), y = as.name(facetPanel)))
}

order_bars(ex_data, 'word1', 'word2', 'contribution')

# A tibble: 25 × 6
   word1    word2 score     n contribution  barCategory
   <chr>    <chr> <int> <int>        <int>       <fctr>
1    not     like     2    99          198    like__not
2    not     help     2    82          164    help__not
3     no    great     3    19           57    great__no
4     no pleasure     3    16           48 pleasure__no
5    not     love     3    16           48    love__not
6    not     care     2    23           46    care__not
7    not     want     1    45           45    want__not
8    not     wish     1    39           39    wish__not
9     no     good     3    13           39     good__no
10   not    allow     1    36           36   allow__not

请注意，我们需要将top_n 替换为filter_ 语句，因为不存在top_n_。查看top_n 的源代码就可以清楚地知道应该如何构造filter_ 语句。

或者如果你想变得花哨，你可以写一个order_bars的NSE版本：

order_bars <- function(df,facetPanel,barCategory,value){
  facetPanel <- substitute(facetPanel)
  barCategory <- substitute(barCategory)
  value <- substitute(value)

  require(lazyeval)
  df %>% 
    mutate_(barCategory = interp(~reorder(x, y), x = barCategory, y = value)) %>% 
    group_by_(facetPanel) %>% 
    filter_(interp(~min_rank(desc(abs(x))) <= 10, x = value)) %>% 
    group_by_(facetPanel, barCategory) %>% 
    arrange_(interp(~desc(x), x = value)) %>% 
    ungroup() %>% 
    mutate_(barCategory = interp(
      ~factor(paste(x, y, sep = "__"), levels = rev(paste(x, y, sep = "__"))),
      x = barCategory, y = facetPanel))
}

order_bars(ex_data, word1, word2, contribution)

理想情况下，您应该只编写完整的 SE 版本，并使用 lazyeval 将 NSE 版本链接到 SE 版本。我将把它作为练习留给读者。

【讨论】：

这是我第一次遇到 SE，这是一个很好的学习机会……感谢您的指导！
awwww 不错的答案 axeman 但伙计，SE 评估的东西太可怕了
此行为已更改。 “dplyr 现在使用整洁的评估语义。NSE 动词仍然捕获它们的参数，但您现在可以取消引用这些参数的一部分。这提供了 NSE 动词的完全可编程性。因此，下划线版本现在是多余的。”见programming with dplyr

【解决方案2】：

使用 rlang_0.4.0 和 dplyr_0.8.2，我们可以使用 tidy-evaluation 运算符 ({{...}}) 或 curly-curly，它将引用和取消引用抽象为单个插值步骤。

library(rlang)
library(dplyr)
order_barsN <- function(df, facetPanel, barCategory, value) {
    df %>% 
        mutate(barCategory = reorder({{barCategory}}, {{value}}))%>%
        group_by({{facetPanel}}) %>%
        filter(min_rank(desc(abs({{value}}))) <= 10) %>%
        group_by({{facetPanel}}, {{barCategory}}) %>%
        arrange(desc({{value}})) %>%
        ungroup %>%
        mutate(barCategory = factor(str_c({{barCategory}}, {{facetPanel}}, sep="__"),
                levels = rev(str_c({{barCategory}}, {{facetPanel}}, sep="__"))))

        }


out2 <- order_barsN(ex_data, word1, word2, contribution)

-检查之前的答案

out1 <- order_bars(ex_data, word1, word2, contribution)
identical(out1, out2)
#[1] TRUE

【讨论】：