list 将列表转换为数据框的最有效方法是什么？

What is the most efficient way to cast a list as a data frame?

提问by DrewConway

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:

我经常想将其中每个索引具有相同元素类型的列表转换为数据框。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:

我想将此列表转换为数据框，其中每个索引元素都是一列。自然（对我而言）的事情是使用do.call：

> 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，但是当我尝试将此矩阵转换为数据框时，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 unlistand as.vector, then recasting the data frame as such:

目前，为了正确转换数据框，我使用unlistand遍历每一列as.vector，然后重新转换数据框：

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。有没有更好的方法来做到这一点？

回答by Joshua Ulrich

I think you want:

我想你想要：

> 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

回答by Gavin Simpson

Another option is:

另一种选择是：

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会产生一个列表数据框：

> 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:

与此相同的替代方案，但现在结果与其他解决方案相同，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，它们比返回向量数据帧的另一个解决方案更具优势。] 关于一个非常简单的问题的一些代表性时间：

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

回答by JD Long

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:

我不能告诉你这在内存或速度方面是“最有效的”，但在编码方面非常有效：

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()

带有 data.frame() 的 lapply() 步骤将每个列表项转换为单行数据框，然后与 rbind() 配合使用

回答by Kevin Ushey

Although this question has long since been answered, it's worth pointing out the data.tablepackage has rbindlistwhich accomplishes this task veryquickly:

尽管这个问题早就得到了回答，但值得指出的是可以非常快速地完成此任务的data.table软件包：rbindlist

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

回答by Martin Morgan

This

这个

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

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

是一个函数，它返回一个提取 x 的第 i 个元素的函数。所以

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

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

获取一个命名的（感谢 Map！）可以制成数据框的向量列表

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

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

对于它的速度更快，通常对使用unlist(lapply(...), use.names=FALSE)作为

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

A more general variant is

更一般的变体是

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?

列表列表结构什么时候出现？也许有一个更早的步骤可以用更矢量化的东西代替迭代？

回答by Yi Li

The dplyr package's bind_rowsis efficient.

dplyr 包bind_rows是高效的。

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

回答by sbha

Not sure where they rank as far as efficiency, but depending on the structure of your lists there are some tidyverseoptions. A bonus is that they work nicely with unequal length lists:

不确定它们在效率方面的排名，但根据您的列表结构，有一些tidyverse选项。一个好处是它们可以很好地处理不等长的列表：

l <- list(a = list(var.1 = 1, var.2 = 2, var.3 = 3)
        , b = list(var.1 = 4, var.2 = 5)
        , c = list(var.1 = 7, var.3 = 9)
        , d = list(var.1 = 10, var.2 = 11, var.3 = NA))

df <- dplyr::bind_rows(l)
df <- purrr::map_df(l, dplyr::bind_rows)
df <- purrr::map_df(l, ~.x)

# all create the same data frame:
# A tibble: 4 x 3
  var.1 var.2 var.3
  <dbl> <dbl> <dbl>
1     1     2     3
2     4     5    NA
3     7    NA     9
4    10    11    NA

And you can also mix vectors and data frames:

您还可以混合向量和数据框：

library(dplyr)
bind_rows(
  list(a = 1, b = 2),
  data_frame(a = 3:4, b = 5:6),
  c(a = 7)
)

# A tibble: 4 x 2
      a     b
  <dbl> <dbl>
1     1     2
2     3     5
3     4     6
4     7    NA