Question

我有一个嵌套的对象列表，我想首先重命名一些变量并对其绑定对象，但只选择一些变量。在下面的示例中，我想在第二个对象中将列A重命名为a，在第三个对象中将w重命名为x，然后排绑定所有三个对象，仅选择列a和x。

数据：

df <- list(structure(list(a = 1:3, 
x = c(-1.99, -1.11, -0.34), 
y = c("C", "B", "A")), .Names = c("a", "x",                                                                                                  "y"), class = c("tbl_df", "tbl", "data.frame"), row.names = c(NA,                                                                                                                                                              -3L)), structure(list(a = 1:3, x = c(-0.44, -1.07,                                                                                                                                                                                                  -0.23)), .Names = c("A", "x"), class = c("tbl_df",                                                                                                                                                                                                                                           "tbl", "data.frame"), row.names = c(NA, -3L)), 
structure(list(a = 1:3, x = c(-0.62, -0.60, -0.06), 
y = c(3L, 2L, 1L)), .Names = c("a", "w", "y"), class = c("tbl_df",                                                                                                                                                                                                                                                                                                             "tbl", "data.frame"), row.names = c(NA, -3L)))

列表结构：

   > lapply(df, names)
[[1]]
[1] "a" "x" "y"

[[2]]
[1] "A" "x"

[[3]]
[1] "a" "w" "y"

然后，行绑定：

library(plyr)
df2 <- ldply(df, data.frame)

Answer 1

您可以通过以下方式实现这一目标：

library(plyr)
df = lapply(df, function(x) {plyr::rename(x,c("A"="a","w"="x"),warn_missing = F)})
df2 <- ldply(lapply(df, function(x) {x[,c("a","x")]}), data.frame)

输出：

  a     x
1 1 -1.99
2 2 -1.11
3 3 -0.34
4 1 -0.44
5 2 -1.07
6 3 -0.23
7 1 -0.62
8 2 -0.60
9 3 -0.06

希望这有帮助。

Answer 2

使用purrr（map），dplyr（rename，select，bind_rows，%>%）和magrittr（ %<>%，%>%））：

library(purrr)
library(dplyr)
library(magrittr)
df[[2]] %<>% rename(.,a = A)
df[[3]] %<>% rename(.,x = w)
df %>% map_df(. %>% select("a","x"))
# # A tibble: 9 x 2
#       a     x
#   <int> <dbl>
# 1     1 -1.99
# 2     2 -1.11
# 3     3 -0.34
# 4     1 -0.44
# 5     2 -1.07
# 6     3 -0.23
# 7     1 -0.62
# 8     2 -0.60
# 9     3 -0.06

或在基地R：

names(df[[2]])[names(df[[2]]) == "A"] <- "a"
names(df[[3]])[names(df[[3]]) == "w"] <- "x"
do.call(rbind,lapply(df,"[",c("a","x")))

Answer 3

另一个想法可能是创建一个带有替换值的命名向量v，遍历列表，如果匹配则重命名并选择所需的列。

v <- c("a" = "A", "x" = "w")

map_df(df, .f = ~ rename_if(
  .x, 
  .p = names(.x) %in% v, 
  .f = funs(stringi::stri_replace_all_fixed(., v, names(v), vectorize_all = FALSE))) %>% 
    select(names(v))
)

给出了：

## A tibble: 9 x 2
#      a     x
#  <int> <dbl>
#1     1 -1.99
#2     2 -1.11
#3     3 -0.34
#4     1 -0.44
#5     2 -1.07
#6     3 -0.23
#7     1 -0.62
#8     2 -0.60
#9     3 -0.06

重命名嵌套列表中的列和行绑定

3 个答案: