R - 如何将此嵌套的 for 循环转换为可以改变列表的 lapply 函数

Question

我的数据看起来像这样

aList <- list(a1 = c("apple", "banana", "orange", "strawberry", "cherry"),
              a2 = c("banana", "cherry", "apple"),
              a3 = c("apple", "strawberry", "pineapple"),
              a4 = c("raspberry", "strawberry", "apple"),
              a5 = c("pineapple", "lemon", "orange", "banana", "apple"),
              a6 = c("lemon", "apple", "blueberry"),
              a7 = c("watermelon", "apple", "banana", "mango"),
              a8 = c("mango", "cherry", "apple", "lemon"),
              a9 = c("orange", "banana", "strawberry"),
              a10 = c("mango", "strawberry"))

我想把它变成垂直格式，就像你运行这段代码时发生的那样：

vertical_data <- list()
for (x in names(aList)) {
  for (y in aList[[x]]) {
    if (is.null(vertical_data[[y]])) {
      vertical_data[[y]] <- x
    } else {
      vertical_data[[y]] <- c(x, vertical_data[[y]])
    }
  }
}
vertical_data

我希望每个条目都告诉我特定水果的产地。

使用双 for 循环很容易做到这一点。但是当我用嵌套的 lapply 函数做同样的事情时，它看起来根本没有修改列表（即vertical_data）。这是为什么？我想使用 apply 函数执行此操作的原因是它更快。我的实际数据集将包含数千个项目和“水果”。使用 for 循环会花费很长时间。

非常感谢您的帮助。

谢谢

Answer 1

我们可以在unlisted 数据上使用split

split(rep(names(aList), lengths(aList)), unlist(aList))

---

或者另一种选择是将stack 转换为两列“data.frame”，然后执行split

with(stack(aList), split(as.character(ind), values))
#$apple
#[1] "a1" "a2" "a3" "a4" "a5" "a6" "a7" "a8"

#$banana
#[1] "a1" "a2" "a5" "a7" "a9"

#$blueberry
#[1] "a6"

#$cherry
#[1] "a1" "a2" "a8"

#$lemon
#[1] "a5" "a6" "a8"

#$mango
#[1] "a7"  "a8"  "a10"

#$orange
#[1] "a1" "a5" "a9"

#$pineapple
#[1] "a3" "a5"

#$raspberry
#[1] "a4"

#$strawberry
#[1] "a1"  "a3"  "a4"  "a9"  "a10"

#$watermelon
#[1] "a7"

---

或者正如@rawr 提到的那样

unstack(stack(aList)[2:1])

---

关于lapply 和for 循环内的分配，它是基于环境的。在for循环中，赋值修改了全局env中的对象，但是在lapply中，它是一个自包含的env或者必须做<<-（不建议）或者指定env为全局env

vertical_data <- list()
lapply(names(aList), function(x) lapply(aList[[x]], 
      function(y) if (is.null(vertical_data[[y]])) {
         vertical_data[[y]] <<- x
         } else {vertical_data[[y]] <<- c(x, vertical_data[[y]])
         }))

Answer 2

我们可以使用enframe将名称列表转换为数据框，然后根据value拆分name。

tibble::enframe(aList) %>% tidyr::unnest(value) %>% {split(.$name, .$value)}

#$apple
#[1] "a1" "a2" "a3" "a4" "a5" "a6" "a7" "a8"

#$banana
#[1] "a1" "a2" "a5" "a7" "a9"

#$blueberry
#[1] "a6"

#$cherry
#[1] "a1" "a2" "a8"

#$lemon
#[1] "a5" "a6" "a8"

#$mango
#[1] "a7"  "a8"  "a10"

#$orange
#[1] "a1" "a5" "a9"

#$pineapple
#[1] "a3" "a5"

#$raspberry
#[1] "a4"

#$strawberry
#[1] "a1"  "a3"  "a4"  "a9"  "a10"

#$watermelon
#[1] "a7"