有没有办法在 tidyverse 生态系统中使用循环？答案

【问题标题】：Is there a way to use a loop inside the tidyverse ecosystem?有没有办法在 tidyverse 生态系统中使用循环？
【发布时间】：2021-06-28 04:45:09
【问题描述】：

我尝试循环遍历 mutate 和 fct_recode 函数来重新编码这个问题，但没有成功。有没有办法以 tidyverse 或任何其他方式解决它？抱歉冗长-我想尽可能清楚。

set.seed(2021)

df <- tibble(

  a1 = factor(ifelse(sign(rnorm(30))==-1, 0, 1), labels = c("", "Yes")),
  a2 = factor(ifelse(sign(rnorm(30))==-1, 0, 1), labels = c("", "Yes")),
  b1 = factor(ifelse(sign(rnorm(30))==-1, 0, 1), labels = c("", "Yes")),
  b2 = factor(ifelse(sign(rnorm(30))==-1, 0, 1), labels = c("", "Yes")),
  d1 = gl(2, 15, labels = c("Males", "Females")),
  d2 = factor(ifelse(sign(rnorm(30))==-1, 0, 1), labels = c("Python", "R"))
)

df %>% count(a1)

# Replace "" with No

myvars <- c("a1", "a2", "b1", "b2")


df %>% mutate(a1 = factor(if_else(a1 == "", "No",
                                  as.character(a1))),
              a2 = factor(if_else(a1 == "", "No",
                                as.character(a1))),
              b1= factor(if_else(b1== "", "No",
                                  as.character(b1))))

#---- Attempt one (forloop)

# Desire for all the variables

for (i in myvars){

  df <- df %>%
    mutate(i = factor(if_else(i == "", "No",
                              as.character(i))))
}

df %>% count(a1) # No replacement

【问题讨论】：

标签： r for-loop dplyr tidyverse

【解决方案1】：

使用data.table

library(data.table)
setDT(df)[, (myvars) := lapply(.SD, function(x) factor(fifelse(x == "", "No",
             as.character(x))), .SDcols = myvars]

在我们需要的 OP 的 for 循环中

for (i in myvars){

df <- df %>%
    mutate(!! i := factor(if_else(!! rlang::sym(i) == "", "No",
                          as.character(!! rlang::sym(i)))))
  } 

df %>% 
    count(a1)
# A tibble: 2 x 2
#  a1        n
#* <fct> <int>
#1 No       12
#2 Yes      18

或者我们可以使用fct_recode

library(forcats)
df %>%
    mutate(across(all_of(myvars), ~ fct_recode(., No = '')))

【讨论】：

【解决方案2】：

tidyverse 有自己的“循环”功能。

要将函数应用于多列，您可以使用across。

library(dplyr)

df <- df %>% mutate(across(all_of(myvars), 
               ~factor(if_else(. == "", "No", as.character(.)))))

df %>% count(a1)

#  a1        n
#  <fct> <int>
#1 No       12
#2 Yes      18

【讨论】：