[英]list of named lists to data.frame
I have a list of named lists of the following form from a JSON object: 我有一个JSON对象的以下表单的命名列表列表:
my_list = list(list(a = 10, b = "blah"),
list(a = 15, b = "stuff"))
Each element of the outer list is a named list and I want to convert it to a data.frame of the following form with the column names intact: 外部列表的每个元素都是一个命名列表,我想将它转换为以下形式的data.frame,列名完整:
a b
10 "blah"
15 "stuff"
On the surface, I can achieve this by doing to_df = data.frame(do.call(rbind, my_list))
. 从表面上看,我可以通过执行
to_df = data.frame(do.call(rbind, my_list))
来实现这一点。
However, if I were to try to extract an individual column using to_df$a
or to_df[,1]
I would get a list instead of a vector as normally expected from a data.frame: 但是,如果我尝试使用
to_df$a
或to_df[,1]
来提取单个列to_df[,1]
我会得到一个列表而不是像data.frame通常所期望的那样的向量:
> to_df[,1]
[[1]]
[1] 10
[[2]]
[1] 15
Instead of: 代替:
> to_df[,1]
[1] 10 15
An old post on the R mailing list suggested the following solution: to_df = as.data.frame(t(sapply(my_list, rbind)))
. R邮件列表上的旧帖子提出了以下解决方案:
to_df = as.data.frame(t(sapply(my_list, rbind)))
。 But not only does this not transfer over the column names, it still has the same issue of returning a list instead of a vector when looking at individual columns using to_df[,1]
. 但是这不仅不会转移到列名,它仍然具有使用
to_df[,1]
查看单个列时返回列表而不是向量的相同问题。
What's the best way to achieve this? 实现这一目标的最佳方法是什么? Is there a
dplyr
way? 有一种
dplyr
方式吗?
EDIT: Thanks for all the solutions, it appears the trick is to lapply
and transform each element of the list to a data.frame
and then bind them together using dplyr or do.call
. 编辑:感谢所有的解决方案,似乎诀窍是将列表的每个元素
lapply
并转换为data.frame
,然后使用dplyr或do.call
将它们绑定在一起。 Alternatively, data.table
does most of the work with a single call to rbindlist
. 另外,
data.table
也最具有单一通话的工作,以rbindlist
。
I prefer rbindlist
from the data.table package. 我更喜欢
rbindlist
从data.table包。 It's simple, fast, and returns a data frame/table. 它简单,快速,并返回数据框/表。
data.table::rbindlist(my_list)
# a b
# 1: 10 blah
# 2: 15 stuff
Another advantage of rbindlist()
is that it will automatically fill in missing values with NA
. rbindlist()
另一个优点是它会自动用NA
填充缺失值。
To remove the data.table
class, you can just wrap in as.data.frame()
要删除
data.table
类,只需包装as.data.frame()
as.data.frame(data.table::rbindlist(my_list))
It looks like you can do this with bind_rows
from the development version of dplyr , dplyr_0.4.2.9002, as of two days ago. 看起来您可以使用
bind_rows
的开发版本dplyr_0.4.2.9002中的bind_rows执行此操作,截至两天前。
library(dplyr)
bind_rows(my_list)
Source: local data frame [2 x 2]
a b
1 10 blah
2 15 stuff
在基地R你可能会这样做
df<-do.call(rbind,lapply(my_list,data.frame))
Fast pure base
R way to do it if the columns are of different types and you want to preserve the types 如果列具有不同类型并且您希望保留类型,则使用快速纯碱
base
R方式
# sample data
set.seed(46823239)
list_of_lists <-
replicate(
100, list(a = rnorm(100), b = sample.int(100, 100, replace = TRUE),
c = factor(sample(letters, 100, replace = TRUE))),
simplify = FALSE)
str( # show first two lists
list_of_lists[1:2])
#R> List of 2
#R> $ :List of 3
#R> ..$ a: num [1:100] -0.0439 -0.4487 -0.5682 -0.8062 1.5074 ...
#R> ..$ b: int [1:100] 59 91 63 87 61 72 92 77 62 41 ...
#R> ..$ c: Factor w/ 26 levels "a","b","c","d",..: 4 16 5 14 25 17 25 4 4 20 ...
#R> $ :List of 3
#R> ..$ a: num [1:100] 0.356 1.239 -0.926 -0.673 -1.168 ...
#R> ..$ b: int [1:100] 62 21 90 20 41 99 57 6 83 22 ...
#R> ..$ c: Factor w/ 26 levels "a","b","c","d",..: 15 16 17 6 3 13 21 16 3 11 ...
# define functions to stack
f1 <- function(x){
. <- function(...){
args <- list(...)
if(is.factor(args[[1]]))
# see https://stackoverflow.com/a/3449403/5861244
return(factor(do.call(c, lapply(args, as.character))))
do.call(c, args)
}
out <- NULL
for(i in 1:length(x[[1]]))
out <- c(out, list(do.call(., lapply(x, "[[", i))))
out <- data.frame(out)
names(out) <- names(x[[1]])
out
}
f2 <- function(x)
# simple alternative from http://r.789695.n4.nabble.com/Convert-list-of-lists-lt-gt-data-frame-td860048.html
do.call(rbind, lapply(x, data.frame))
# show output
all.equal( # yields the same
f1(list_of_lists), f2(list_of_lists))
#R> [1] TRUE
all.equal(
f1(list_of_lists), data.table::rbindlist(list_of_lists),
check.attributes = FALSE)
#R> [1] TRUE
out <- f1(list_of_lists)
head(out, 5)
#R> a b c
#R> 1 -0.04391595 59 d
#R> 2 -0.44866652 91 p
#R> 3 -0.56815817 63 e
#R> 4 -0.80622044 87 n
#R> 5 1.50736514 61 y
sapply(out, class)
#R> a b c
#R> "numeric" "integer" "factor"
# benchmark
microbenchmark::microbenchmark(
f1(list_of_lists), f2(list_of_lists), data.table::rbindlist(list_of_lists))
#R> Unit: microseconds
#R> expr min lq mean median uq max neval
#R> f1(list_of_lists) 1259.850 1426.3685 1633.127 1531.0590 1643.257 7086.211 100
#R> f2(list_of_lists) 31348.099 34293.8720 61224.476 37003.7930 92775.162 153318.869 100
#R> data.table::rbindlist(list_of_lists) 652.246 786.7645 1040.994 872.6905 1022.221 4063.994 100
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.