如何使用dplyr引用变量而不是列 [英] How to refer to variable instead of column with dplyr

问题描述

使用dplyr：filter时，我经常计算一个保存可行选择的局部变量：

When using dplyr:filter, I often compute a local variable that holds the viable choices:

df <- as_tibble(data.frame(id=c("a","b"), val=1:6))
ids <- c("b","c")
filter(df, id %in% ids)
# giving id %in% c("b","c")

但是，如果偶然地数据集具有相同名称的列，则无法实现预期的目的：

However, if the dataset by chance has a column with the same name, this fails to achieve the intended purpose:

df$ids <- "a"
filter(df, id %in% ids)
# giving id %in% "a"

我应该如何显式引用ids变量而不是ids列？

How should I explicitly refer to the ids variable instead of the ids column?

推荐答案

用取消引号！告诉 filter 在调用环境中查找而不是在数据框：

Unquote with !! to tell filter to look in the calling environment instead of the data frame:

library(tidyverse)

df <- data_frame(id = rep(c("a","b"), 3), val = 1:6)
ids <- c("b", "c")

df %>% filter(id %in% ids)
#> # A tibble: 3 x 2
#>      id   val
#>   <chr> <int>
#> 1     b     2
#> 2     b     4
#> 3     b     6

df <- df %>% mutate(ids = "a")

df %>% filter(id %in% ids)
#> # A tibble: 3 x 3
#>      id   val   ids
#>   <chr> <int> <chr>
#> 1     a     1     a
#> 2     a     3     a
#> 3     a     5     a

df %>% filter(id %in% !!ids)
#> # A tibble: 3 x 3
#>      id   val   ids
#>   <chr> <int> <chr>
#> 1     b     2     a
#> 2     b     4     a
#> 3     b     6     a

当然，避免此类问题的更好方法是不要使用相同的名称全球环境中的向量。

Of course, the better way to avoid such issues is to not put identically-named vectors in your global environment.

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