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
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>
其他回答
一个简短的(但可能不是最快的)方法是使用基底r，因为数据帧只是一个长度相等的向量的列表。因此，你的输入列表和一个30 x 132 data.frame之间的转换将是:
df <- data.frame(l)
从这里我们可以将其转置为132 x 30的矩阵，并将其转换回数据帧:
new_df <- data.frame(t(df))
一句话:
new_df <- data.frame(t(data.frame(l)))
行名看起来很讨厌，但是您总是可以用
行名称（new_df） <- 1：nrow（new_df）
下面这个简单的命令对我有用:
myDf <- as.data.frame(myList)
参考(Quora的答案)
> myList <- list(a = c(1, 2, 3), b = c(4, 5, 6))
> myList
$a
[1] 1 2 3
$b
[1] 4 5 6
> myDf <- as.data.frame(myList)
a b
1 1 4
2 2 5
3 3 6
> class(myDf)
[1] "data.frame"
但如果不清楚如何将列表转换为数据帧，则会失败:
> myList <- list(a = c(1, 2, 3), b = c(4, 5, 6, 7))
> myDf <- as.data.frame(myList)
函数错误(…)， row.names = NULL，检查。rows = FALSE, check.names = TRUE，: 参数暗示不同的行数:3,4
注意:答案是朝着问题的标题，可能会跳过问题的一些细节
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>
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。
从不同的角度;
install.packages("smotefamily")
library(smotefamily)
library(dplyr)
data_example = sample_generator(5000,ratio = 0.80)
genData = BLSMOTE(data_example[,-3],data_example[,3])
#There are many lists in genData. If we want to convert one of them to dataframe.
sentetic=as.data.frame.array(genData$syn_data)
# as.data.frame.array seems to be working.
