正在更新 fct_relevel 解决方案，因为 across 接管了 mutate_at

Question

的解决方案不再有效，因为 mutate_at 已被弃用。我很难使用 mutate(across(...), fct_relevel(., ....) 拼凑更新的解决方案，因为我收到一个错误

"Problem with `mutate()` input `..2`.
i `..2 = ~...`.
x `..2` must be a vector, not a `formula` object."

但是没有公式

假设我有

f1 <- factor(c("a", "b", "c", "d"))
g2 <- factor(c("a", "b", "c", "d"))
t3 <- factor(c("a", "b", "c", "d"))
f5 <- factor(c("m", "l", "f", "p"))
y8 <- factor(c("a", "b", "c", "d"))
df <- tibble(f1, g2, t3, f5, y8)

并且我想要 fct_relevel f1、g2 和 y8（非顺序变量）以便您将因子排序为“c”、“d”、“b”、“a”对于所有这三个变量。

Answer 1

大概是这样的：

library(dplyr) # 1.0.7
library(forcats) # 0.5.1

f1 <- factor(c("a", "b", "c", "d"))
g2 <- factor(c("a", "b", "c", "d"))
t3 <- factor(c("a", "b", "c", "d"))
f5 <- factor(c("m", "l", "f", "p"))
y8 <- factor(c("a", "b", "c", "d"))
df <- tibble(f1, g2, t3, f5, y8)


df_1 <- df %>%
  mutate(across(c(f1, g2, y8), forcats::fct_relevel, "c", "d", "b", "a"))

levels(df$f1)
#> [1] "c" "d" "b" "a"
levels(df$g2)
#> [1] "c" "d" "b" "a"
levels(df$t3)
#> [1] "a" "b" "c" "d"
levels(df$f5)
#> [1] "f" "l" "m" "p"
levels(df$y8)
#> [1] "c" "d" "b" "a"

^{由 reprex package (v2.0.0)}

于 2021-07-25 创建

下面我再添加两个等同于上面的语法。

# option 2
df_2 <- df %>%
  mutate(across(c(f1, g2, y8), ~forcats::fct_relevel(., "c", "d", "b", "a")))

# option 3
df_3 <- df %>%
  mutate(across(c(f1, g2, y8), ~forcats::fct_relevel(.x, "c", "d", "b", "a")))

identical(df_1, df_2) # TRUE 
identical(df_1, df_3) # TRUE

上述fct_relevel()的操作方式是将每一列（即f1、g2、y8）视为传递给[=16的向量=]，而不是 . 或 .x。它“知道”做这样的“堵塞”，因为我们在函数之前使用 ~。

正在更新 fct_relevel 解决方案，因为 across 接管了 mutate_at

Updating fct_relevel solution since across took over mutate_at

r

pipe

dplyr

forcats

across