Flextable 背景无法找到列或识别 {{ }} 或 。在我的职责范围内

Question

我正在构建一个使用 flextable 创建表格的函数。我想指定颜色渐变，但有些值不适用。我希望这些 NA 值（更改为“-”）是透明的。该数据集包含某人对特定年份的评级（示例中为 rating_2023），但我希望代码能够灵活地适应不同的年份，而不是硬编码特定年份。

由于某种原因，当我想检测某个值是否为 NA 时，flextable 无法理解

.

、

{{}}

或

enquo()

。

具体来说，这是我的示例数据：

library(dplyr)
library(flextable)
library(scales)

sample_data <- tibble(emp_id = c(1, 1, 1, 1,
                                 2, 2, 2, 2, 
                                 3, 3, 3, 3,
                                 4, 4, 4, 4),
                      survey_period = c("q4", "q3", "q2", "q1",
                                        "q4", "q3", "q2", "q1",
                                        "q4", "q3", "q2", "q1",
                                        "q4", "q3", "q2", "q1"),
                      rating_2023 = c(100, 90, 40, NA,
                                 -10, -20, 0, 10,
                                 80, -10, NA, NA,
                                 90, 50, -20, NA))

当我在

i = ~is.na(rating_2023),

中的 rating_2023 中进行硬编码时，我的函数就可以工作了

make_table <- function(data, time_period){
  
  gen_rating <- paste0("rating_", time_period)
  
  gradient <- col_numeric(
    palette = c("red", "transparent", "green"),
    domain = c(-100, 0, 100)
  )
  
  data %>%
    flextable() %>%
    colformat_num(na_str = " - ") %>%
    bg(
      bg = gradient,
      j = 3,
      part = "body"
    ) %>%
    bg(
      bg = "transparent",
      i = ~is.na(rating_2023),
      j = gen_rating
    )
  
}

make_table(sample_data, time_period = "2023")

但是当我尝试通过管道输入对象时，它会产生

Error in eval(as.call(f[[2]]), envir = data) : object 'gen_rating' not found

。当我尝试使用

.

时，我也得到了类似的答案：

make_table <- function(data, time_period){
  
  gen_rating <- paste0("rating_", time_period)
  
  gradient <- col_numeric(
    palette = c("red", "transparent", "green"),
    domain = c(-100, 0, 100)
  )
  
  data %>%
    flextable() %>%
    colformat_num(na_str = " - ") %>%
    bg(
      bg = gradient,
      j = 3,
      part = "body"
    ) %>%
    bg(
      bg = "transparent",
      i = ~is.na({{gen_rating}}),
      j = gen_rating
    )
  
}

make_table(sample_data, time_period = "2023")

为什么这不起作用？有没有解决方案，或者我今年被迫进行硬编码？

Answer 1

使用

as.formula()

或

reformulate()

- 还有一个带有“rlang”的解决方案，但我不记得如何编码：

library(dplyr)
library(flextable)
library(scales)

sample_data <- tibble(
  emp_id = c(
    1, 1, 1, 1,
    2, 2, 2, 2,
    3, 3, 3, 3,
    4, 4, 4, 4
  ),
  survey_period = c(
    "q4", "q3", "q2", "q1",
    "q4", "q3", "q2", "q1",
    "q4", "q3", "q2", "q1",
    "q4", "q3", "q2", "q1"
  ),
  rating_2023 = c(
    100, 90, 40, NA,
    -10, -20, 0, 10,
    80, -10, NA, NA,
    90, 50, -20, NA
  )
)

make_table <- function(data, time_period) {
  gen_rating <- paste0("rating_", time_period)

  gradient <- col_numeric(
    palette = c("red", "transparent", "green"),
    domain = c(-100, 0, 100)
  )

  data %>%
    flextable() %>%
    colformat_num(na_str = " - ") %>%
    bg(
      bg = gradient,
      j = 3,
      part = "body"
    ) %>%
    bg(
      bg = "transparent",
      i = reformulate(sprintf("is.na(`%s`)", gen_rating)),
      j = gen_rating
    )
}

make_table(sample_data, time_period = "2023")