r - 将 mutate_at 与 mutate_if 一起使用

Question

我正在我的包中创建一个通用函数。目标是找到百分比列，然后使用 parse_number如果他们是 character列。我一直无法使用 mutate_at 找到解决方案和 ifelse .我在下面粘贴了一个reprex。

 library(tidyverse)


df <- tibble::tribble(
  ~name, ~pass_percent, ~attendance_percent, ~grade,
  "Jon",         "90%",                0.85,    "B",
  "Jim",        "100%",                   1,    "A"
  )

percent_names <- df %>% select(ends_with("percent"))%>% names()


# Error due to attendance_percent already being in numeric value

if (percent_names %>% length() > 0) {
    df <-
      df %>%
      dplyr::mutate_at(percent_names, readr::parse_number)
  }
#> Error in parse_vector(x, col_number(), na = na, locale = locale, trim_ws = trim_ws): is.character(x) is not TRUE

Answer 1

您的 attendance_percent变量是数字，不是字符和 parse_number只需要字符变量，见 here .所以一个解决方案是:

edited_parse_number <- function(x, ...) {
  if (mode(x) == 'numeric') {
    x
  } else {
    parse_number(x, ...)
  }
}


df %>%
  dplyr::mutate_at(vars(percent_names), edited_parse_number)

#  name  pass_percent attendance_percent grade
#  <chr>        <dbl>              <dbl> <chr>
#1 Jon             90               0.85 B    
#2 Jim            100               1    A   

或者

如果您不想使用该额外函数，请在开始时提取字符变量:

percent_names <- df %>% 
  select(ends_with("percent")) %>% 
  select_if(is.character) %>% 
  names()
percent_names
# [1] "pass_percent"


df %>%
  dplyr::mutate_at(vars(percent_names), parse_number)
#   name  pass_percent attendance_percent grade
#   <chr>        <dbl>              <dbl> <chr>
# 1 Jon             90               0.85 B    
# 2 Jim            100               1    A