rapply到R中的嵌套数据帧列表

时间:2021-12-24 19:32:45

i have a nested list whose fundamental element is data frames, and i want to traverse this list recursively to do some computation of each data frame, finally to get a nested list of results in the same structure as the input. I know "rapply" is exactly for such kind of task, but i met a problem that, rapply actually goes even deeper than i want, i.e. it decomposes every data frame and applies to each column instead (because a data frame itself is a list in R).

我有一个嵌套列表,其基本元素是数据帧,我想递归遍历此列表以对每个数据帧进行一些计算,最后得到与输入结构相同的嵌套结果列表。我知道“rapply”正是为了这样的任务,但是我遇到了一个问题,rapply实际上比我想要的更深,即它分解每个数据帧并应用于每一列(因为数据帧本身就是一个列表)在R)。

One workaround i can think about is to convert each data frame to matrix, but it will force to uniform the data types, so i don't like it really. I want to know if there is any way to control the recursive depth of rapply. Any idea? Thanks.

我能想到的一个解决方法是将每个数据帧转换为矩阵,但它会强制统一数据类型,所以我真的不喜欢它。我想知道是否有任何方法可以控制rapply的递归深度。任何想法?谢谢。

1 个解决方案

#1


6  

1. wrap in proto

用原型包裹

When creating your list structure try wrapping the data frames in proto objects:

创建列表结构时,尝试在proto对象中包装数据框:

library(proto)
L <- list(a = proto(DF = BOD), b = proto(DF = BOD))
rapply(L, f = function(.) colSums(.$DF), how = "replace")

giving:

赠送:

$a
  Time demand 
    22     89 

$b
  Time demand 
    22     89 

Wrap the result of your function in a proto object too if you want to further rapply it;

如果你想进一步提升它,你也可以将你的函数的结果包装在一个proto对象中;

f <- function(.) proto(result = colSums(.$DF))
out <- rapply(L, f = f, how = "replace")
str(out)

giving:

赠送:

List of 2
 $ a:proto object 
 .. $ result: Named num [1:2] 22 89 
 ..  ..- attr(*, "names")= chr [1:2] "Time" "demand" 
 $ b:proto object 
 .. $ result: Named num [1:2] 22 89 
 ..  ..- attr(*, "names")= chr [1:2] "Time" "demand" 

2. write your own rapply alternative

2.写自己的替代品

recurse <- function (L, f) {
    if (inherits(L, "data.frame")) f(L)
    else lapply(L, recurse, f)
}

L <- list(a = BOD, b = BOD)
recurse(L, colSums)

This gives:

这给出了:

$a
  Time demand 
    22     89 

$b
  Time demand 
    22     89 

ADDED: second approach

增加:第二种方法

#1


6  

1. wrap in proto

用原型包裹

When creating your list structure try wrapping the data frames in proto objects:

创建列表结构时,尝试在proto对象中包装数据框:

library(proto)
L <- list(a = proto(DF = BOD), b = proto(DF = BOD))
rapply(L, f = function(.) colSums(.$DF), how = "replace")

giving:

赠送:

$a
  Time demand 
    22     89 

$b
  Time demand 
    22     89 

Wrap the result of your function in a proto object too if you want to further rapply it;

如果你想进一步提升它,你也可以将你的函数的结果包装在一个proto对象中;

f <- function(.) proto(result = colSums(.$DF))
out <- rapply(L, f = f, how = "replace")
str(out)

giving:

赠送:

List of 2
 $ a:proto object 
 .. $ result: Named num [1:2] 22 89 
 ..  ..- attr(*, "names")= chr [1:2] "Time" "demand" 
 $ b:proto object 
 .. $ result: Named num [1:2] 22 89 
 ..  ..- attr(*, "names")= chr [1:2] "Time" "demand" 

2. write your own rapply alternative

2.写自己的替代品

recurse <- function (L, f) {
    if (inherits(L, "data.frame")) f(L)
    else lapply(L, recurse, f)
}

L <- list(a = BOD, b = BOD)
recurse(L, colSums)

This gives:

这给出了:

$a
  Time demand 
    22     89 

$b
  Time demand 
    22     89 

ADDED: second approach

增加:第二种方法