将 NULL 分配给 R 中的列表元素? [英] Assigning NULL to a list element in R?

问题描述

我发现这种行为很奇怪，希望更多有经验的用户分享他们的想法和解决方法.在 R 中运行下面的代码示例:

I found this behaviour odd and wanted more experienced users to share their thoughts and workarounds. On running the code sample below in R:

sampleList <- list()
d<- data.frame(x1 = letters[1:10], x2 = 1:10, stringsAsFactors = FALSE)
for(i in 1:nrow(d)) {
        sampleList[[i]] <- d$x1[i]
}

print(sampleList[[1]])
#[1] "a"
print(sampleList[[2]])
#[1] "b"
print(sampleList[[3]])
#[1] "c"
print(length(sampleList))
#[1] 10

sampleList[[2]] <- NULL
print(length(sampleList))
#[1] 9
print(sampleList[[2]])
#[1] "c"
print(sampleList[[3]])
#[1] "d"

列表元素向上移动.也许这正如预期的那样，但我正在尝试实现一个函数，在该函数中我合并列表的两个元素并删除一个.我基本上想丢失该列表索引或将其设为 NULL.

The list elements get shifted up. Maybe this is as expected, but I am trying to implement a function where I merge two elements of a list and drop one. I basically want to lose that list index or have it as NULL.

有什么办法可以为它分配 NULL 而看不到上述行为吗?

Is there any way I can assign NULL to it and not see the above behaviour?

感谢您的建议.

推荐答案

好问题.

查看R-FAQ:

在 R 中，如果 x 是一个列表，则 x[i] <- NULL 和 x[[i]] <- NULL 从 x 中删除指定的元素.其中第一个与 S 不兼容，在 S 中它是空操作.(请注意，您可以使用 x[i] <- list(NULL) 将元素设置为 NULL.)

In R, if x is a list, then x[i] <- NULL and x[[i]] <- NULL remove the specified elements from x. The first of these is incompatible with S, where it is a no-op. (Note that you can set elements to NULL using x[i] <- list(NULL).)

考虑以下示例:

> t <- list(1,2,3,4)
> t[[3]] <- NULL          # removing 3'd element (with following shifting)
> t[2] <- list(NULL)      # setting 2'd element to NULL.
> t
[[1]]
[2] 1

[[2]]
NULL

[[3]]
[3] 4

更新:

正如 the R Inferno 的作者所评论的，在处理时可能会有更微妙的情况与 NULL.考虑非常通用的代码结构:

As the author of the R Inferno commented, there can be more subtle situations when dealing with NULL. Consider pretty general structure of code:

# x is some list(), now we want to process it.
> for (i in 1:n) x[[i]] <- some_function(...)

现在请注意，如果 some_function() 返回 NULL，你可能不会得到你想要的:有些元素会消失.你应该使用 lapply 函数.看看这个玩具示例:

Now be aware, that if some_function() returns NULL, you maybe will not get what you want: some elements will just disappear. you should rather use lapply function. Take a look at this toy example:

> initial <- list(1,2,3,4)
> processed_by_for <- list(0,0,0,0)
> processed_by_lapply <- list(0,0,0,0)
> toy_function <- function(x) {if (x%%2==0) return(x) else return(NULL)}
> for (i in 1:4) processed_by_for[[i]] <- toy_function(initial[[i]])
> processed_by_lapply <- lapply(initial, toy_function)
> processed_by_for
  [[1]]
  [1] 0

  [[2]]
  [1] 2

  [[3]]
  NULL

  [[4]]
  [1] 4

> processed_by_lapply
  [[1]]
  NULL

  [[2]]
  [1] 2

  [[3]]
  NULL

  [[4]]
  [1] 4

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