我有一个嵌套的数据列表。它的长度是132，每一项是一个长度为20的列表。是否有一种快速的方法可以将这种结构转换为具有132行和20列数据的数据帧?
下面是一些示例数据:
l <- replicate(
132,
as.list(sample(letters, 20)),
simplify = FALSE
)
当前回答
这是最后对我有用的方法:
do.call（“rbind”， lapply（S1， as.data.frame））
其他回答
尝试折叠::unlist2d ('unlist to data.frame'的简写):
l <- replicate(
132,
list(sample(letters, 20)),
simplify = FALSE
)
library(collapse)
head(unlist2d(l))
.id.1 .id.2 V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 V17 V18 V19 V20
1 1 1 e x b d s p a c k z q m u l h n r t o y
2 2 1 r t i k m b h n s e p f o c x l g v a j
3 3 1 t r v z a u c o w f m b d g p q y e n k
4 4 1 x i e p f d q k h b j s z a t v y l m n
5 5 1 d z k y a p b h c v f m u l n q e i w j
6 6 1 l f s u o v p z q e r c h n a t m k y x
head(unlist2d(l, idcols = FALSE))
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11 V12 V13 V14 V15 V16 V17 V18 V19 V20
1 e x b d s p a c k z q m u l h n r t o y
2 r t i k m b h n s e p f o c x l g v a j
3 t r v z a u c o w f m b d g p q y e n k
4 x i e p f d q k h b j s z a t v y l m n
5 d z k y a p b h c v f m u l n q e i w j
6 l f s u o v p z q e r c h n a t m k y x
该方法使用一个tidyverse包(purrr)。
列表:
x <- as.list(mtcars)
将其转换为数据帧(更具体地说是tibble):
library(purrr)
map_df(x, ~.x)
编辑时间:2021年5月30日
这实际上可以通过dplyr中的bind_rows()函数实现。
x <- as.list(mtcars)
dplyr::bind_rows(x)
A tibble: 32 x 11
mpg cyl disp hp drat wt qsec vs am gear carb
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 21 6 160 110 3.9 2.62 16.5 0 1 4 4
2 21 6 160 110 3.9 2.88 17.0 0 1 4 4
3 22.8 4 108 93 3.85 2.32 18.6 1 1 4 1
4 21.4 6 258 110 3.08 3.22 19.4 1 0 3 1
5 18.7 8 360 175 3.15 3.44 17.0 0 0 3 2
6 18.1 6 225 105 2.76 3.46 20.2 1 0 3 1
7 14.3 8 360 245 3.21 3.57 15.8 0 0 3 4
8 24.4 4 147. 62 3.69 3.19 20 1 0 4 2
9 22.8 4 141. 95 3.92 3.15 22.9 1 0 4 2
10 19.2 6 168. 123 3.92 3.44 18.3 1 0 4 4
# ... with 22 more rows
Reshape2产生与上面的plyr示例相同的输出:
library(reshape2)
l <- list(a = list(var.1 = 1, var.2 = 2, var.3 = 3)
, b = list(var.1 = 4, var.2 = 5, var.3 = 6)
, c = list(var.1 = 7, var.2 = 8, var.3 = 9)
, d = list(var.1 = 10, var.2 = 11, var.3 = 12)
)
l <- melt(l)
dcast(l, L1 ~ L2)
收益率:
L1 var.1 var.2 var.3
1 a 1 2 3
2 b 4 5 6
3 c 7 8 9
4 d 10 11 12
如果你几乎没有像素，你可以在一行w/ recast()中完成这一切。
对于像从嵌套JSON中获得的3级或更多级别的深度嵌套列表的一般情况:
{
"2015": {
"spain": {"population": 43, "GNP": 9},
"sweden": {"population": 7, "GNP": 6}},
"2016": {
"spain": {"population": 45, "GNP": 10},
"sweden": {"population": 9, "GNP": 8}}
}
考虑一下melt()将嵌套列表转换为高格式的方法:
myjson <- jsonlite:fromJSON(file("test.json"))
tall <- reshape2::melt(myjson)[, c("L1", "L2", "L3", "value")]
L1 L2 L3 value
1 2015 spain population 43
2 2015 spain GNP 9
3 2015 sweden population 7
4 2015 sweden GNP 6
5 2016 spain population 45
6 2016 spain GNP 10
7 2016 sweden population 9
8 2016 sweden GNP 8
接着是dcast()，然后再次扩大到一个整洁的数据集，其中每个变量组成一个a列，每个观察值组成一行:
wide <- reshape2::dcast(tall, L1+L2~L3)
# left side of the formula defines the rows/observations and the
# right side defines the variables/measurements
L1 L2 GNP population
1 2015 spain 9 43
2 2015 sweden 6 7
3 2016 spain 10 45
4 2016 sweden 8 9
tibble包有一个函数enframe()，它通过将嵌套的列表对象强制转换为嵌套的tibble(“整齐的”数据帧)对象来解决这个问题。下面是R for Data Science的一个简单例子:
x <- list(
a = 1:5,
b = 3:4,
c = 5:6
)
df <- enframe(x)
df
#> # A tibble: 3 × 2
#> name value
#> <chr> <list>
#> 1 a <int [5]>
#> 2 b <int [2]>
#> 3 c <int [2]>
Since you have several nests in your list, l, you can use the unlist(recursive = FALSE) to remove unnecessary nesting to get just a single hierarchical list and then pass to enframe(). I use tidyr::unnest() to unnest the output into a single level "tidy" data frame, which has your two columns (one for the group name and one for the observations with the groups value). If you want columns that make wide, you can add a column using add_column() that just repeats the order of the values 132 times. Then just spread() the values.
library(tidyverse)
l <- replicate(
132,
list(sample(letters, 20)),
simplify = FALSE
)
l_tib <- l %>%
unlist(recursive = FALSE) %>%
enframe() %>%
unnest()
l_tib
#> # A tibble: 2,640 x 2
#> name value
#> <int> <chr>
#> 1 1 d
#> 2 1 z
#> 3 1 l
#> 4 1 b
#> 5 1 i
#> 6 1 j
#> 7 1 g
#> 8 1 w
#> 9 1 r
#> 10 1 p
#> # ... with 2,630 more rows
l_tib_spread <- l_tib %>%
add_column(index = rep(1:20, 132)) %>%
spread(key = index, value = value)
l_tib_spread
#> # A tibble: 132 x 21
#> name `1` `2` `3` `4` `5` `6` `7` `8` `9` `10` `11`
#> * <int> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr> <chr>
#> 1 1 d z l b i j g w r p y
#> 2 2 w s h r i k d u a f j
#> 3 3 r v q s m u j p f a i
#> 4 4 o y x n p i f m h l t
#> 5 5 p w v d k a l r j q n
#> 6 6 i k w o c n m b v e q
#> 7 7 c d m i u o e z v g p
#> 8 8 f s e o p n k x c z h
#> 9 9 d g o h x i c y t f j
#> 10 10 y r f k d o b u i x s
#> # ... with 122 more rows, and 9 more variables: `12` <chr>, `13` <chr>,
#> # `14` <chr>, `15` <chr>, `16` <chr>, `17` <chr>, `18` <chr>,
#> # `19` <chr>, `20` <chr>
