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Add empty columns to a dataframe with specified names from a vector (6 answers)

Closed 21 mins ago.

我的问题与现有问题的不同之处在于,我想创建新的栏目,编号为<>条码>:<>直线>而不是,取决于现有的栏目。

一些 du数据:

library(dplyr)
dat <- tibble(
    a = 1:5,
    b = LETTERS[1:5]
)

我知道,我可以设立一栏,如一栏。

dat <- dat %>%
    mutate(foo = NA, bar = NA, bar2 = NA)

并且,我可以 修改<>>>>,更方便地使用<条码>。

new_vars <- c("foo", "bar", "bar2") dat <- dat %>% mutate(across(all_of(new_vars), ~ replace(., is.na(.), 0)))

但是,在不以类似方式提及现有各栏的情况下,我如何创建新的栏目? E.g. 增加填满的新栏目 NA:

tibble( a = 1:5, b = LETTERS[1:5] ) %>% # mutate(across(all_of(new_vars), ~ function(.x) NA)) # Error mutate(across(all_of(new_vars), NA)) # Error

开放给任何<条码>逆向>替代物。

Answer 1

类似于,这一答案是。

new_vars <- c("foo", "bar", "bar2")

tibble(
  a = 1:5,
  b = LETTERS[1:5]
) %>% 
  mutate(!!!setNames(rep(NA, length(new_vars)), new_vars))
# or (thanks @joran)
# tibble::add_column(!!!setNames(rep(NA, length(new_vars)), new_vars))

产出

     a b     foo   bar   bar2 
  <int> <chr> <lgl> <lgl> <lgl>
1     1 A     NA    NA    NA   
2     2 B     NA    NA    NA   
3     3 C     NA    NA    NA   
4     4 D     NA    NA    NA   
5     5 E     NA    NA    NA

Answer 2

我使用了与下一位研究员一样的verse,但为了避免做事情,我们所花的时间太长,正变得sil不清。

这里。管道友好。

library(dplyr)
dat <- tibble(
  a = 1:5,
  b = LETTERS[1:5]
)

new_vars <- c("foo", "bar", "bar2")

# ?
# dat[new_vars] <- NA

add_vars <- function(df,vars,val){
  df[vars] <- val
  df
}

dat |>
  add_vars(df = _,vars = new_vars,val = NA)

You could even use an anonymous function (but only with the magrittr pipe):

dat %>%
  ((x) {x[new_vars] <- NA; x})

这还使用<条码>功能(x)syntax。

Answer 3

Using dplyr::bind_cols() and pipes:

library(dplyr)

tibble(a = 1:5,
       b = LETTERS[1:5]) %>% 
bind_cols(., setNames(lapply(new_vars, function(x) x = NA), new_vars))

结果:

# A tibble: 5 × 5
      a b     foo   bar   bar2 
  <int> <chr> <lgl> <lgl> <lgl>
1     1 A     NA    NA    NA   
2     2 B     NA    NA    NA   
3     3 C     NA    NA    NA   
4     4 D     NA    NA    NA   
5     5 E     NA    NA    NA

Although I think the second answer to this question, on which this is based, is just as good.

如果你真的想要<条码>,则Ritchie在评论中回答。

Answer 4

Maybe this is the style you re looking for:

library(dplyr)

dat <- tibble(
    a = 1:5,
    b = LETTERS[1:5]
)

new_vars <- c("foo", "bar", "bar2")

dat %>% 
    purrr::reduce(new_vars, ~mutate(.x, {{.y}} := 0), .init = .)

Instead of using across() we use purrr::reduce() which will loop over the new_vars. We apply the mutate function to the output of the previous iteration. We want to start with .init = dat, but we pipe it in.

You could even use reduce2 if you wanted to have different values for each of the new_vars.

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