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用dplyr中的list元素创建一个dataframe

dplyr r

Laura · 技术社区 · 6 年前

这是我的数据帧:

    df<-list(structure(list(Col1 = structure(1:6, .Label = c("A", "B", 
"C", "D", "E", "F"), class = "factor"), Col2 = structure(c(1L, 
2L, 3L, 2L, 4L, 5L), .Label = c("B", "C", "D", "F", "G"), class = "factor")), class = "data.frame", row.names = c(NA, 
-6L)), structure(list(Col1 = structure(c(1L, 4L, 5L, 6L, 2L, 
3L), .Label = c("A", "E", "H", "M", "N", "P"), class = "factor"), 
    Col2 = structure(c(1L, 2L, 3L, 2L, 4L, 5L), .Label = c("B", 
    "C", "D", "F", "G"), class = "factor")), class = "data.frame", row.names = c(NA, 
-6L)), structure(list(Col1 = structure(c(1L, 4L, 6L, 5L, 2L, 
3L), .Label = c("A", "W", "H", "M", "T", "U"), class = "factor"), 
    Col2 = structure(c(1L, 2L, 3L, 2L, 4L, 5L), .Label = c("B", 
    "C", "D", "S", "G"), class = "factor")), class = "data.frame", row.names = c(NA, 
-6L)))

我想提取col1= df[[1]][1] 作为数据帧。然后这个列表的第二个位置的col1我要合并到 测向[[1]][1] ,那么我将有一个包含2列的数据帧。在这之后,我想把列表第三个位置的第1列合并到有两列的dataframe,然后我将有一个有3列的dataframe。

换句话说,我的数据帧应该有3列,即列表中每个条目的所有第一列。

dplyr包可以帮我做这个吗?

有什么帮助吗?

5 回复 | 直到 6 年前

Rui Barradas 6 年前

lapply 提取名为 "Col1 一气呵成。然后设置结果的名称。

col1 <- as.data.frame(lapply(df, '[[', "Col1"))
names(col1) <- letters[seq_along(col1)]

col1
#  a b c
#1 A A A
#2 B M M
#3 C N U
#4 D P T
#5 E E W
#6 F H H

选择您可能会发现更好的任何其他列名。

dplyr 可能是这样的

df %>% 
  unlist(recursive = FALSE) %>%
  as.data.frame %>%
  select(., starts_with("Col1"))
#  Col1 Col1.1 Col1.2
#1    A      A      A
#2    B      M      M
#3    C      N      U
#4    D      P      T
#5    E      E      W
#6    F      H      H

acylam 6 年前

与 map_dfc 从 purrr

library(purrr)

map_dfc(df, `[`, 1)

输出:

  Col1 Col11 Col12
1    A     A     A
2    B     M     M
3    C     N     U
4    D     P     T
5    E     E     W
6    F     H     H

Calum You 6 年前

替代使用 map_dfc 利用 purrr 的简明元素提取语法,允许按名称或位置指定元素的元素。例如,第一种方法相当于

map_dfc(df, `[[`, 1)

这与 [ 因为这些列不会被命名为 Col1 V 取而代之的是名字,因为名字像 Col11 Col12 可能会让人困惑。

df <- list(structure(list(Col1 = structure(1:6, .Label = c("A", "B", "C", "D", "E", "F"), class = "factor"), Col2 = structure(c(1L, 2L, 3L, 2L, 4L, 5L), .Label = c("B", "C", "D", "F", "G"), class = "factor")), class = "data.frame", row.names = c(NA, -6L)), structure(list(Col1 = structure(c(1L, 4L, 5L, 6L, 2L, 3L), .Label = c("A", "E", "H", "M", "N", "P"), class = "factor"), Col2 = structure(c(1L, 2L, 3L, 2L, 4L, 5L), .Label = c("B", "C", "D", "F", "G"), class = "factor")), class = "data.frame", row.names = c(NA, -6L)), structure(list(Col1 = structure(c(1L, 4L, 6L, 5L, 2L, 3L), .Label = c("A", "W", "H", "M", "T", "U"), class = "factor"), Col2 = structure(c(1L, 2L, 3L, 2L, 4L, 5L), .Label = c("B", "C", "D", "S", "G"), class = "factor")), class = "data.frame", row.names = c(NA, -6L)))

library(purrr)
map_dfc(df, 1)
#> # A tibble: 6 x 3
#>   V1    V2    V3   
#>   <fct> <fct> <fct>
#> 1 A     A     A    
#> 2 B     M     M    
#> 3 C     N     U    
#> 4 D     P     T    
#> 5 E     E     W    
#> 6 F     H     H
map_dfc(df, "Col1")
#> # A tibble: 6 x 3
#>   V1    V2    V3   
#>   <fct> <fct> <fct>
#> 1 A     A     A    
#> 2 B     M     M    
#> 3 C     N     U    
#> 4 D     P     T    
#> 5 E     E     W    
#> 6 F     H     H

创建日期:2018-09-19 reprex package (第0.2.0版)。

Sal-laS 6 年前

res<-1:nrow(df[[1]][1])

for(i in 1:length(df)){
  print ( as.vector(df[[i]][1]))
  res<-cbind(res,as.data.frame(df[[i]][1]))
}
res$res<-NULL

所以,输出是:

  Col1 Col1 Col1
1    A    A    A
2    B    M    M
3    C    N    U
4    D    P    T
5    E    E    W
6    F    H    H

Chriss Paul 6 年前

使用 dplyr

library(dplyr)
df %>% 
  sapply('[[',1) %>%
  as.data.frame
#returns
  V1 V2 V3
1  A  A  A
2  B  M  M
3  C  N  U
4  D  P  T
5  E  E  W
6  F  H  H