Question

这是（using `rlang` for conditional labelling in `ggplot` using `ggrepel`）的后续问题，它解决了我在自定义函数中遇到的问题，该函数使用表达式在标记数据点时过滤掉数据。但是答案提出了另一个我不知道如何解决的问题。

这是自定义函数，它使用rlang评估用户输入的表达式以在将标签附加到数据点时过滤掉数据。当不在列表列中使用时，此功能可以正常工作。例如-

# loading needed libraries
library(tidyverse)
library(ggplot2)
library(ggrepel)

# custom function
label_adder <- function(data, x, y, label.var, exp = NULL) {
  param_list <- as.list(match.call())

  if ("exp" %in% names(param_list)) {
    my_exp <- rlang::enquo(exp)
  }
  else {
    a <- "dplyr::row_number(x = .) > 0"
    my_exp <- rlang::quo(!!rlang::sym(a))
  }

  plot <-
    ggplot(mapping = aes(
      x = !!rlang::enquo(x),
      y = !!rlang::enquo(y)
    )) +
    geom_point(data = data) +
    geom_smooth(data = data, method = "lm") +
    geom_label_repel(
      data = data %>% filter(!!my_exp),
      mapping = aes(label = !!rlang::enquo(label.var))
    )
  return(plot)
}

# using the function
label_adder(
  data = datasets::iris,
  x = Sepal.Length,
  y = Sepal.Width,
  label.var = Species,
  exp = Sepal.Length > 7
)

但是当我对purrr::map使用相同的函数时，它将失败。

# creating a list column
df.listcol <- datasets::iris %>%
  dplyr::mutate(.data = ., Species2 = Species) %>% # just creates a copy of this variable
  dplyr::group_by(.data = ., Species) %>%
  tidyr::nest(data = .)

# running function on dataframe with list columns
df.listcol %>% # creates a nested dataframe with list column called `data`
  dplyr::mutate( # creating a new list column of ggstatsplot outputs
    .data = .,
    plot = data %>%
      purrr::map(
        .x = .,
        .f = ~label_adder(
          data = .,
          x = Sepal.Length,
          y = Sepal.Width
        )
      )
  )
#> Error in mutate_impl(.data, dots): Evaluation error: Evaluation error: object 'dplyr::row_number(x = .) > 0' not found..

但是，如果我通过指定label.var和exp来使用该函数，则效果很好。

# running function on dataframe with list columns
df.listcol %>% # creates a nested dataframe with list column called `data`
  dplyr::mutate( # creating a new list column of ggstatsplot outputs
    .data = .,
    plot = data %>%
      purrr::map(
        .x = .,
        .f = ~label_adder(
          data = .,
          x = Sepal.Length,
          y = Sepal.Width,
          label.var = Species,
          exp = Sepal.Length > 7
        )
      )
  )
#> # A tibble: 3 x 3
#>   Species    data              plot    
#>   <fct>      <list>            <list>  
#> 1 setosa     <tibble [50 x 5]> <S3: gg>
#> 2 versicolor <tibble [50 x 5]> <S3: gg>
#> 3 virginica  <tibble [50 x 5]> <S3: gg>

所以我的问题是，为什么在未指定label.var和exp的情况下该函数失败，以及如何解决此问题？

由reprex package（v0.2.0.9000）创建于2018-08-31。

Answer 1

因此，在这里我们编辑函数以使过滤更加灵活。

# loading needed libraries
library(tidyverse)
library(ggplot2)
library(ggrepel)

# custom function
label_adder <- function(data, x, y, label.var, exp = NULL) {
  param_list <- as.list(match.call())
  label_data <- data %>% {if ("exp" %in% names(param_list)) filter(., !!enquo(exp)) else .}

  plot <-
    ggplot(mapping = aes(
      x = !!rlang::enquo(x),
      y = !!rlang::enquo(y)
    )) +
    geom_point(data = data) +
    geom_smooth(data = data, method = "lm") +
    geom_label_repel(
      data = label_data,
      mapping = aes(label = !!rlang::enquo(label.var))
    )
  return(plot)
}

df.listcol <- datasets::iris %>%
  dplyr::mutate(.data = ., Species2 = Species) %>% # just creates a copy of this variable
  dplyr::group_by(.data = ., Species) %>%
  tidyr::nest(data = .)



test <- df.listcol %>% mutate(plot = map(data, ~label_adder(., x = Sepal.Length, label.var = Species2, y = Sepal.Width)))
test
#> # A tibble: 3 x 3
#>   Species    data              plot    
#>   <fct>      <list>            <list>  
#> 1 setosa     <tibble [50 × 5]> <S3: gg>
#> 2 versicolor <tibble [50 × 5]> <S3: gg>
#> 3 virginica  <tibble [50 × 5]> <S3: gg>
test$plot[[2]]

test2 <- df.listcol %>% mutate(plot = map(data, ~label_adder(., x = Sepal.Length, label.var = Species2, y = Sepal.Width, exp = Sepal.Length > 6.5)))
test2
#> # A tibble: 3 x 3
#>   Species    data              plot    
#>   <fct>      <list>            <list>  
#> 1 setosa     <tibble [50 × 5]> <S3: gg>
#> 2 versicolor <tibble [50 × 5]> <S3: gg>
#> 3 virginica  <tibble [50 × 5]> <S3: gg>
test2$plot[[2]]

由reprex package（v0.2.0）于2018-08-31创建。

正确评估列表列内的`rlang`表达式

1 个答案: