函数 NSE 内的 group_by dplyr

我使用时遇到问题dplyr and group_by在管道函数调用中。

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函数，如下所示：

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="__"))))
      }

并从中吸取建议post https://stackoverflow.com/questions/24606282/passing-data-frame-to-mutate-within-function, used $在函数内的 mutate 操作期间引用 data.frame 的变量时的表示法。

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))

我猜测group_by需要调整以采用命名变量，否则我必须使用.dot参考列的符号，尽管我只是把它扔到风中......

您需要学习如何使用 1) SE 版本dplyr动词，例如group_by_ and mutate_2）神秘的lazyeval::interp。请阅读vignette("nse")小心。

那么我们可以得出：

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

注意我们需要替换top_n with a filter_声明，因为没有top_n_存在。看看源头top_n很明显如何filter_应构造语句。

或者如果你想变得更奇特，你可以编写一个 NSE 版本order_bars:

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，您只需完整编写 SE 版本，然后将 NSE 版本链接到 SE 版本lazyeval。我将把它作为练习留给读者。