在 dplyr::funs 的命名参数中,我可以引用其他参数的名称吗?

In a named argument to dplyr::funs, can I reference the names of other arguments?

考虑以下几点:

library(tidyverse)

df <- tibble(x = rnorm(100), y = rnorm(100, 10, 2), z = x * y)

df %>% 
mutate_all(funs(avg = mean(.), dev = sd(.), scaled = (. - mean(.)) / sd(.)))

有没有办法通过引用 avgdev 列来避免调用 meansd 两次。我的想法是

df %>% 
mutate_all(funs(avg = mean(.), dev = sd(.), scaled = (. - avg) / dev))

显然这行不通,因为没有列 avgdev,但是 x_avgx_devy_avgy_dev,等等

funs 中有什么好方法可以使用 rlang 工具以编程方式创建这些列引用,这样我就可以将之前命名参数创建的列引用到 funs(当.x时,我会参考x_meanx_dev来计算x_scaled,等等)?

我认为如果将数据转换为长格式会更容易

library(tidyverse)

set.seed(111)
df <- tibble(x = rnorm(100), y = rnorm(100, 10, 2), z = x * y)

df %>% 
  gather(key, value) %>% 
  group_by(key) %>% 
  mutate(avg    = mean(value),
         sd     = sd(value),
         scaled = (value - avg) / sd)
#> # A tibble: 300 x 5
#> # Groups:   key [3]
#>    key    value     avg    sd scaled
#>    <chr>  <dbl>   <dbl> <dbl>  <dbl>
#>  1 x      0.235 -0.0128  1.07  0.232
#>  2 x     -0.331 -0.0128  1.07 -0.297
#>  3 x     -0.312 -0.0128  1.07 -0.279
#>  4 x     -2.30  -0.0128  1.07 -2.14 
#>  5 x     -0.171 -0.0128  1.07 -0.148
#>  6 x      0.140 -0.0128  1.07  0.143
#>  7 x     -1.50  -0.0128  1.07 -1.39 
#>  8 x     -1.01  -0.0128  1.07 -0.931
#>  9 x     -0.948 -0.0128  1.07 -0.874
#> 10 x     -0.494 -0.0128  1.07 -0.449
#> # ... with 290 more rows

reprex package (v0.2.1.9000)

创建于 2018-11-04

这看起来有点费解,但它确实有效:

scaled <- function(col_name, x, y) {
  col_name <- deparse(substitute(col_name))
  avg <- eval.parent(as.symbol(paste0(col_name, x)))
  dev <- eval.parent(as.symbol(paste0(col_name, y)))
  (eval.parent(as.symbol(col_name)) - avg) / dev
}

df %>%
  mutate_all(funs(avg = mean(.), dev = sd(.), scaled = scaled(., "_avg", "_dev"))) 

这可能对你有用:

avg <- quo(mean(.))
dev <- quo(sd(.))
res <- df %>% 
  mutate_all(funs(avg = !!avg, dev = !!dev, scaled = (. - !!avg) / !!dev))

确认它有效:

res0 <- df %>% 
  mutate_all(funs(avg = mean(.), dev = sd(.), scaled = (. - mean(.)) / sd(.)))
identical(res, res0)
# [1] TRUE