12

我正在尝试创建一个用于创建棒棒糖图的函数ggplot2。我想将所有参数传递...给aes()within geom_point()。但是，我想排除size参数传递到aes()内部geom_segment()（如果您查看a()下面的输出，原因很明显）。因此，我捕获...usingrlang::enquos()而不是按原样传递它。a()在我将 to 传递dots给this 的函数aes()中ggplot()没有问题。但是在功能中b()我得到了错误Can't use '!!!' at top level.

我被困在这一点上，希望能提供任何解决此问题的意见。

library(ggplot2)
data("mtcars")

d <- dplyr::count(mtcars, cyl, am)

a <- function(data, x, y, ...) {
  x <- rlang::enquo(x)
  y <- rlang::enquo(y)
  dots <- rlang::enquos(...)

  ggplot(data, aes(!!x, !!y, !!!dots)) +
    geom_segment(aes(y = 0, xend = !!x, yend = !!y)) +
    geom_point()
}

b <- function(data, x, y, ...) {
  x <- rlang::enquo(x)
  y <- rlang::enquo(y)

  dots <- rlang::enquos(...)
  segment_args <- dots[names(dots) != "size"]

  ggplot(data, aes(!!x, !!y)) +
    geom_segment(aes(y = 0, xend = !!x, yend = !!y, !!!segment_args)) +
    geom_point(aes(!!!dots))
}

a(d, cyl, n, color = factor(am), size = am)



b(d, cyl, n, color = factor(am), size = am)
#> Error: Can't use `!!!` at top level.

这是我的sessionInfo()：

R version 3.5.2 (2018-12-20)
Platform: x86_64-apple-darwin16.7.0 (64-bit)
Running under: macOS Sierra 10.12.1

Matrix products: default
BLAS: /System/Library/Frameworks/Accelerate.framework/Versions/A/Frameworks/vecLib.framework/Versions/A/libBLAS.dylib
LAPACK: /usr/local/Cellar/openblas/0.3.5/lib/libopenblasp-r0.3.5.dylib

locale:
[1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods  
[7] base     

other attached packages:
[1] ggplot2_3.2.1

loaded via a namespace (and not attached):
 [1] Rcpp_1.0.3       digest_0.6.18    withr_2.1.2     
 [4] assertthat_0.2.0 crayon_1.3.4     dplyr_0.8.3     
 [7] grid_3.5.2       R6_2.3.0         gtable_0.2.0    
[10] magrittr_1.5     scales_1.0.0     pillar_1.4.2    
[13] rlang_0.4.2      lazyeval_0.2.1   rstudioapi_0.10 
[16] labeling_0.3     tools_3.5.2      glue_1.3.0      
[19] purrr_0.3.3      munsell_0.5.0    compiler_3.5.2  
[22] pkgconfig_2.0.2  colorspace_1.4-0 tidyselect_0.2.5
[25] tibble_2.1.3
4

4 回答 4

10

显然这是一个已知问题，aes()您可以在此处验证。一种解决方法是：

b <- function(data, x, y, ...) {
  x <- rlang::enquo(x)
  y <- rlang::enquo(y)

  dots <- rlang::enquos(...)
  segment_args <- dots[names(dots) != "size"]

  ggplot(data, aes(!!x, !!y)) +
    geom_segment(aes(, y = 0, xend = !!x, yend = !!y, !!!segment_args)) +
    geom_point(aes(, , !!!dots))
}

注意 中的单逗号geom_segment()和 中的双逗号geom_point()。

于 2020-01-03T16:18:16.283 回答
4

如果您按照 rlang 的说明进行操作，您将获得更多详细信息：

> rlang::last_error()
<error>
message: Can't use `!!!` at top level.
class:   `rlang_error`
backtrace:
 1. global::b(d, cyl, n, color = factor(am), size = am)
 4. ggplot2::aes(y = 0, xend = !!x, yend = !!y, !!!segment_args)
 5. rlang::enquos(x = x, y = y, ..., .ignore_empty = "all")
 6. rlang:::endots(...)
 7. rlang:::map(...)
 8. base::lapply(.x, .f, ...)
 9. rlang:::FUN(X[[i]], ...)
Call `rlang::last_trace()` to see the full backtrace

然后

> rlang::last_trace()
    █
 1. └─global::b(d, cyl, n, color = factor(am), size = am)
 2.   ├─ggplot2::geom_segment(aes(y = 0, xend = !!x, yend = !!y, !!!segment_args))
 3.   │ └─ggplot2::layer(...)
 4.   └─ggplot2::aes(y = 0, xend = !!x, yend = !!y, !!!segment_args)
 5.     └─rlang::enquos(x = x, y = y, ..., .ignore_empty = "all")
 6.       └─rlang:::endots(...)
 7.         └─rlang:::map(...)
 8.           └─base::lapply(.x, .f, ...)
 9.             └─rlang:::FUN(X[[i]], ...)

所以看来问题出在!!!segment_args

编辑1：只是玩，但由于segment_args当前是一个值，我尝试了以下，错误确实消失了：

b <- function(data, x, y, ...) {
  x <- rlang::enquo(x)
  y <- rlang::enquo(y)

  dots <- rlang::enquos(...)
  print(dots)
  segment_args <- dots[[setdiff(names(dots), "size")]]
  print(names(dots))

  print(segment_args)

  ggplot(data, aes(!!x, !!y)) +
    geom_segment(aes(y = 0, xend = !!x, yend = !!y, !!segment_args)) +
    geom_point(aes(!!!dots))
}

这只能确认问题出在使用 !!! 因为上面现在给出了一个错误aes(!!!dots)，这取决于在示例中segment_args中只有一个元素这一事实，但它可能会为进一步调查提供帮助

于 2020-01-03T15:01:31.813 回答
2

我认为您不再需要引用/取消引用。相反，您可以使用双括号{{ x }}并将点保留为点...

以下工作并且更容易理解：

b <- function(data, x, y, ...) {
  ggplot(data, aes( {{x}} , {{y}} )) +
    geom_segment(aes(y = 0, xend = {{x}}, yend = {{y}}, ...)) +
    geom_point(aes(...))
}
于 2020-01-03T12:38:25.333 回答
1

编辑 2：

您可以覆盖 的size值geom_segment，这样您就不必在之前操作带引号的点：

b <- function(data, x, y, ...) {
  x <- enquo(x)
  y <- enquo(y)
  dots <- enquos(...)

  ggplot(data, aes(!!x, !!y, !!!dots)) +
    geom_segment(aes(y = 0, xend = !!x, yend = !!y), size = 1) +
    geom_point(aes())
}

b(d, cyl, n)
b(d, cyl, n, color = factor(am))
b(d, cyl, n, color = factor(am), size = am)

编辑：鉴于我对提供明确论点的评论，我尝试了这个，它似乎有效

b <- function(data, x, y, color, size) {
  x <- enquo(x)
  y <- enquo(y)
  color <- enquo(color)
  size <- enquo(size)

  ggplot(data, aes(!!x, !!y, color = !!color)) +
    geom_segment(aes(y = 0, xend = !!x, yend = !!y)) +
    geom_point(aes(size=!!size))
}

鉴于您的示例，我建议使用以下解决方法，其中在函数中创建所需的变量而不是...从geom_xxx.

library(dplyr)
library(rlang)
library(ggplot2)

data("mtcars")
d <- dplyr::count(mtcars, cyl, am)

b <- function(data, x, y, aspect) {
  x <- enquo(x)
  y <- enquo(y)
  aspect <- enquo(aspect)

  data <- data %>% mutate(
    color = factor(!!aspect),
    size = !!aspect
  )

  ggplot(data, aes(!!x, !!y, color = color)) +
    geom_segment(aes(y = 0, xend = !!x, yend = !!y)) +
    geom_point(aes(size=size))
}

b(d, cyl, n, am)
于 2020-01-03T15:05:56.317 回答