传递参数给dplyr函数

本文介绍了传递参数给dplyr函数的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我想使用 dplyr 参数化以下计算，它可以找出哪些值 Sepal.Length 与更多关联超过一个值 Sepal.Width ：

I want to parameterise the following computation using dplyr that finds which values of Sepal.Length are associated with more than one value of Sepal.Width:

library(dplyr) iris %>% group_by(Sepal.Length) %>% summarise(n.uniq=n_distinct(Sepal.Width)) %>% filter(n.uniq > 1)

通常我会写这样的东西：

Normally I would write something like this:

not.uniq.per.group <- function(data, group.var, uniq.var) { iris %>% group_by(group.var) %>% summarise(n.uniq=n_distinct(uniq.var)) %>% filter(n.uniq > 1) }

但是，这种方法抛出错误，因为 dplyr 使用非标准评估。这个函数应该如何写？

However, this approach throws errors because dplyr uses non-standard evaluation. How should this function be written?

推荐答案

你需要使用标准的评估版本的 dplyr 函数（只是附加'_'到函数名，即。 group_by _ & 总结_ ）并将字符串传递给你的函数，然后你需要变成符号。要参数summaryise的参数，您需要使用 lazyeval 包中定义的 interp（）。具体地：

You need to use the standard evaluation versions of the dplyr functions (just append '_' to the function names, ie. group_by_ & summarise_) and pass strings to your function, which you then need to turn into symbols. To parameterise the argument of summarise_, you will need to use interp(), which is defined in the lazyeval package. Concretely:

library(dplyr) library(lazyeval) not.uniq.per.group <- function(df, grp.var, uniq.var) { df %>% group_by_(grp.var) %>% summarise_( n_uniq=interp(~n_distinct(v), v=as.name(uniq.var)) ) %>% filter(n_uniq > 1) } not.uniq.per.group(iris, "Sepal.Length", "Sepal.Width")

dplyr 小插曲为非更多细节的标准评估。

See the dplyr vignette for non standard evaluation for more details.