使用 dplyr 时，是否可以替代 plyr 的“重估”功能？

Question

我喜欢 revalue 函数是 plyr 用于替换字符串。它简单易记。

但是，我已将新代码迁移到 dplyr，它似乎没有 revalue 函数。在 dplyr 中，以前用 revalue 做的事情被接受的成语是什么？

Answer 1

从 dplyr 版本 dplyr_0.5.0 开始有一个 recode 函数可用，它看起来与 plyr 中的 revalue 非常相似。

根据recode 文档示例部分构建的示例：

set.seed(16)
x = sample(c("a", "b", "c"), 10, replace = TRUE)
x
 [1] "a" "b" "a" "b" "b" "a" "c" "c" "c" "a"

recode(x, a = "Apple", b = "Bear", c = "Car")

   [1] "Car"   "Apple" "Bear"  "Apple" "Car"   "Apple" "Apple" "Car"   "Car"   "Apple"

如果你只定义了一些你想重新编码的值，默认情况下，其余的都用NA填充。

recode(x, a = "Apple", c = "Car")
 [1] "Car"   "Apple" NA      "Apple" "Car"   "Apple" "Apple" "Car"   "Car"   "Apple"

可以使用.default 参数更改此行为。

recode(x, a = "Apple", c = "Car", .default = x)
 [1] "Car"   "Apple" "b"     "Apple" "Car"   "Apple" "Apple" "Car"   "Car"   "Apple"

如果您想用其他东西替换缺失值，还有一个 .missing 参数。

Answer 2

我们可以使用来自base R 的chartr 来做到这一点

chartr("ac", "AC", x)

数据

x <- c("a", "b", "c")

Answer 3

我想评论@aosmith 的答案，但缺乏声誉。现在看来dplyr的recode函数的默认设置是不影响未指定的级别。

x = sample(c("a", "b", "c"), 10, replace = TRUE)
x
[1] "c" "c" "b" "b" "a" "b" "c" "c" "c" "b"

recode(x , a = "apple", b = "banana" )

[1] "c"      "c"      "banana" "banana" "apple"  "banana" "c"      "c"      "c"      "banana"

要将所有未指定的级别更改为NA，应包含参数.default = NA_character_。

recode(x, a = "apple", b = "banana", .default = NA_character_)

[1] "apple"  "banana" "apple"  "banana" "banana" "apple"  NA       NA       NA       "apple" 

Answer 4

我觉得很方便的一个替代方法是 data.tables 的 mapvalues 函数 例如

df[, variable := mapvalues(variable, old = old_names_string_vector, new = new_names_string_vector)]

Answer 5

R 基础解决方案

您可以为此使用base 中的ifelse()。函数参数是ifelse(test, yes, no)。举个例子：

(x <- sample(c("a", "b", "c"), 5, replace = TRUE))
[1] "c" "a" "b" "a" "a"

ifelse(x == "a", "Apple", x)
[1] "c"     "Apple" "b"     "Apple" "Apple"

如果您想重新编码多个值，您可以像这样以嵌套方式使用该函数：

ifelse(x == "a", "Apple", ifelse(x == "b", "Banana", x))
[1] "c"      "Apple"  "Banana" "Apple"  "Apple"

自带功能

有许多必须重新编码的值会使ifelse() 的编码变得混乱。因此，这里有一个自己的函数：

my_revalue <- function(x, ...){
  reval <- list(...)

  from <- names(reval)
  to <- unlist(reval)

  out <- eval(parse(text= paste0("{", paste0(paste0("x[x ==", "'", from,"'", "]", "<-", "'", to, "'"), collapse= ";"), ";x", "}")))

  return(out)
}

现在我们可以非常快速地更改多个值：

my_revalue(vec= x, "a" = "Apple", "b" = "Banana", "c" = "Cranberry")
[1] "Cranberry" "Apple"     "Banana"      "Apple"     "Apple"