问题描述
我在管道函数调用中使用 dplyr 和 group_by 有问题。
I am having problems using dplyr and group_by within a piped function call.
可重复的示例
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="__"))))
我已经实现了以下功能，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="__"))))
}
，使用 $ 符号。
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 需要进行调整才能采取命名变量，或者我必须使用 .dot 符号来引用列，尽管我只是把它扔在风中...
I suspect group_by needs to be tweaked to take either named variables, or I have to use .dot notation to reference columns, although I'm just throwing this out into the wind...
推荐答案
您将需要学习如何使用1）SE版本的 dplyr 动词，如 group_by _ 和 mutate _ 和2）神秘的 lazyeval :: interp 。请仔细阅读 vignette（nse）。
You'll need to learn how to use 1) the SE versions of dplyr verbs such as group_by_ and mutate_ and 2) the enigmatic lazyeval::interp. Please read vignette("nse") carefully.
然后我们可以来：
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，需要用 filter _ 语句替换 top_n ，因为没有 top_n _ 存在。看看 top_n 的来源显而易见，应该如何构建 filter _ 语句。
Note that we need to replace top_n with a filter_ statement, since no top_n_ exists. Looking at the source of top_n makes it obvious how the filter_ statement should be constructed.
或者如果你想得到喜好，你可以写一个NSE版本的 order_bars ：
Or if you want to get fancy, you can write a NSE version of 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，您只会完全编写SE版本，并将NSE版本链接到SE版本，并使用 lazyeval 。我会把它作为练习留给读者。
Ideally, you would write only the SE version fully, and link the NSE version to the SE version with lazyeval. I'll leave that as an exercise to the reader.
这篇关于group_by dplyr在函数NSE中的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！