将列表转换为数据框架的最有效方法是什么?

Very often I want to convert a list wherein each index has identical element types to a data frame. For example, I may have a list:

我经常想要转换一个列表，其中每个索引都有相同的元素类型到一个数据帧。例如，我可能有一个列表:

> my.list
[[1]]
[[1]]$global_stdev_ppb
[1] 24267673

[[1]]$range
[1] 0.03114799

[[1]]$tok
[1] "hello"

[[1]]$global_freq_ppb
[1] 211592.6


[[2]]
[[2]]$global_stdev_ppb
[1] 11561448

[[2]]$range
[1] 0.08870838

[[2]]$tok
[1] "world"

[[2]]$global_freq_ppb
[1] 1002043

I want to convert this list to a data frame where each index element is a column. The natural (to me) thing to go is to is use do.call:

我想将这个列表转换为一个数据帧，其中每个索引元素都是一个列。(对我来说)最自然的事情就是用doto。

> my.matrix<-do.call("rbind", my.list)
> my.matrix
     global_stdev_ppb range      tok     global_freq_ppb
[1,] 24267673         0.03114799 "hello" 211592.6       
[2,] 11561448         0.08870838 "world" 1002043

Straightforward enough, but when I attempt to cast this matrix as a data frame, the columns remain list elements, rather than vectors:

很简单，但是当我尝试将这个矩阵作为一个数据框架时，列仍然是列表元素，而不是向量:

> my.df<-as.data.frame(my.matrix, stringsAsFactors=FALSE)
> my.df[,1]
[[1]]
[1] 24267673

[[2]]
[1] 11561448

Currently, to get the data frame cast properly I am iterating over each column using unlist and as.vector, then recasting the data frame as such:

目前，为了正确地获得数据框架，我使用unlist和as迭代每个列。向量，然后重铸数据框如下:

new.list<-lapply(1:ncol(my.matrix), function(x) as.vector(unlist(my.matrix[,x])))
my.df<-as.data.frame(do.call(cbind, new.list), stringsAsFactors=FALSE)

This, however, seem very inefficient. Is there are better way to do this?

然而，这似乎非常低效。有更好的办法吗?

6 个解决方案

#1

I think you want:

我认为你想要的:

> do.call(rbind, lapply(my.list, data.frame, stringsAsFactors=FALSE))
  global_stdev_ppb      range   tok global_freq_ppb
1         24267673 0.03114799 hello        211592.6
2         11561448 0.08870838 world       1002043.0
> str(do.call(rbind, lapply(my.list, data.frame, stringsAsFactors=FALSE)))
'data.frame':   2 obs. of  4 variables:
 $ global_stdev_ppb: num  24267673 11561448
 $ range           : num  0.0311 0.0887
 $ tok             : chr  "hello" "world"
 $ global_freq_ppb : num  211593 1002043

#2

Another option is:

另一个选项是:

data.frame(t(sapply(mylist, `[`)))

but this simple manipulation results in a data frame of lists:

但是这个简单的操作导致了列表的数据框架:

> str(data.frame(t(sapply(mylist, `[`))))
'data.frame':   2 obs. of  3 variables:
 $ a:List of 2
  ..$ : num 1
  ..$ : num 2
 $ b:List of 2
  ..$ : num 2
  ..$ : num 3
 $ c:List of 2
  ..$ : chr "a"
  ..$ : chr "b"

An alternative to this, along the same lines but now the result same as the other solutions, is:

另一种方法，沿着相同的直线但现在的结果和其他解一样，是:

data.frame(lapply(data.frame(t(sapply(mylist, `[`))), unlist))

[Edit: included timings of @Martin Morgan's two solutions, which have the edge over the other solution that return a data frame of vectors.] Some representative timings on a very simple problem:

[编辑:包括@Martin Morgan的两个解决方案的时间安排，它们比返回向量数据帧的另一个解决方案具有优势。[英语背诵文选关于一个非常简单的问题的一些有代表性的时间:

mylist <- list(list(a = 1, b = 2, c = "a"), list(a = 2, b = 3, c = "b"))

> ## @Joshua Ulrich's solution:
> system.time(replicate(1000, do.call(rbind, lapply(mylist, data.frame,
+                                     stringsAsFactors=FALSE))))
   user  system elapsed 
  1.740   0.001   1.750

> ## @JD Long's solution:
> system.time(replicate(1000, do.call(rbind, lapply(mylist, data.frame))))
   user  system elapsed 
  2.308   0.002   2.339

> ## my sapply solution No.1:
> system.time(replicate(1000, data.frame(t(sapply(mylist, `[`)))))
   user  system elapsed 
  0.296   0.000   0.301

> ## my sapply solution No.2:
> system.time(replicate(1000, data.frame(lapply(data.frame(t(sapply(mylist, `[`))), 
+                                               unlist))))
   user  system elapsed 
  1.067   0.001   1.091

> ## @Martin Morgan's Map() sapply() solution:
> f = function(x) function(i) sapply(x, `[[`, i)
> system.time(replicate(1000, as.data.frame(Map(f(mylist), names(mylist[[1]])))))
   user  system elapsed 
  0.775   0.000   0.778

> ## @Martin Morgan's Map() lapply() unlist() solution:
> f = function(x) function(i) unlist(lapply(x, `[[`, i), use.names=FALSE)
> system.time(replicate(1000, as.data.frame(Map(f(mylist), names(mylist[[1]])))))
   user  system elapsed 
  0.653   0.000   0.658

#3

I can't tell you this is the "most efficient" in terms of memory or speed, but it's pretty efficient in terms of coding:

我不能说这是内存或速度方面的“最有效”，但在编码方面它是相当高效的:

my.df <- do.call("rbind", lapply(my.list, data.frame))

the lapply() step with data.frame() turns each list item into a single row data frame which then acts nice with rbind()

使用data.frame()的lapply()步骤将每个列表项转换为单个行数据帧，然后使用rbind()进行良好的操作

#4

Although this question has long since been answered, it's worth pointing out the data.table package has rbindlist which accomplishes this task very quickly:

虽然这个问题很久以前就有人回答过了，但值得指出的是数据。table package有rbindlist，可以很快完成这个任务:

library(microbenchmark)
library(data.table)
l <- replicate(1E4, list(a=runif(1), b=runif(1), c=runif(1)), simplify=FALSE)

microbenchmark( times=5,
  R=as.data.frame(Map(f(l), names(l[[1]]))),
  dt=data.frame(rbindlist(l))
)

gives me

给我

Unit: milliseconds
 expr       min        lq    median        uq       max neval
    R 31.060119 31.403943 32.278537 32.370004 33.932700     5
   dt  2.271059  2.273157  2.600976  2.635001  2.729421     5

#5

This

这

f = function(x) function(i) sapply(x, `[[`, i)

is a function that returns a function that extracts the i'th element of x. So

是一个返回一个提取x的第i个元素的函数

Map(f(mylist), names(mylist[[1]]))

gets a named (thanks Map!) list of vectors that can be made into a data frame

获取可构成数据帧的已命名(感谢映射!)向量列表

as.data.frame(Map(f(mylist), names(mylist[[1]])))

For speed it's usually faster to use unlist(lapply(...), use.names=FALSE) as

对于速度，使用unlist通常会更快(lapply(…)，use.names=FALSE)。

f = function(x) function(i) unlist(lapply(x, `[[`, i), use.names=FALSE)

A more general variant is

一个更普遍的变体是

f = function(X, FUN) function(...) sapply(X, FUN, ...)

When do the list-of-lists structures come up? Maybe there's an earlier step where an iteration could be replaced by something more vectorized?

列表结构什么时候出现?也许有一个早期的步骤，迭代可以被更矢量化的东西代替?

#6

The dplyr package's bind_rows is efficient.

dplyr包的bind_rows是有效的。

one <- mtcars[1:4, ]
two <- mtcars[11:14, ]
system.time(dplyr::bind_rows(one, two))
   user  system elapsed 
  0.001   0.000   0.001