使用字符串作为列名时如何使用 dplyr::arrange(desc())? [英] How to use dplyr::arrange(desc()) when using a string as column name?

问题描述

如何使用 dplyr::arrange(dplyr::desc()) 并传入一个字符串作为列名?

How can I use dplyr::arrange(dplyr::desc()) and pass in a string as the column name?

这是一个示例数据集:

df <- data.frame(a = 1:3, b = 3:1)

有效的例子:

df %>% dplyr::arrange(b)
df %>% dplyr::arrange_("b")
df %>% dplyr::arrange(dplyr::desc(b))

但我似乎无法同时使用 arrange 和 desc 的字符串，这是我尝试过的两个版本，但都不起作用:

But I can't seem to use a string with both arrange and desc, these are the two version I tried that don't work:

df %>% dplyr::arrange(dplyr::desc("b"))
df %>% dplyr::arrange_(dplyr::desc("b"))

谢谢！

推荐答案

tl;dr : df %>%排列(desc(!!sym("b")))

首先，dplyr 动词的标准评估已被弃用，因此:

First of all the standard evaluations of dplyr verbs are deprecated, so instead of:

library(dplyr)
x <- "b"
df %>% arrange_(x)

现在建议输入:

library(dplyr)
library(rlang)
df %>% arrange(!!sym(x))

请参阅 ?arrange_ ，它链接到名为 Deprecated SE 版本的主要动词的帮助主题. 并提供了一些详细信息.

See ?arrange_ , it links to a help topic named Deprecated SE versions of main verbs. and offers some details.

从那里到降序，很容易适应新的公式:

From there to sort descending it is straightforward to adapt the new formulation :

df %>% arrange(desc(!!sym(x)))

如果您的数据未分组，这些也能正常工作:

These work as well if your data is not grouped:

df %>% arrange(desc(.[[x]]))
df %>% arrange(desc(.data[[x]]))

<小时>

仅供参考，要使其与 arrange_ 一起使用，我们可以执行以下操作，但最好使用上述方法！

---

FYI, to make it work with arrange_ we could have done the following, better use the approach above however!

df %>% arrange_(paste0("desc(",x,")"))

如果我们有 OP 示例中的数字变量，则可以简化:

Which can be simplified if we have numeric variables like in OP's example:

df %>% arrange_(paste0("-",x))

或者使用lazyeval::interp

df %>% arrange_(interp(~desc(y),y=as.name(x)))

或者如@shyam-saladi 建议的那样:

Or as @shyam-saladi proposes:

desc_ <- function(x) lazyeval::interp(~desc(var), var = as.name(x))
# or just
# desc_ <- function(x) paste0("desc(",x,")")
df %>% arrange_(desc_(x))

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