CODEWITHSUNDEEP
rdataframetibble
pd441
pd441

Reputation: 2763

Create empty tibble/data frame with column names coming from a vector

I would like to create an empty data frame where the column names are coming from a character vector.

for example, if this was my vector:

 vec <- letters[1:3]

I would like create an empty data frame like the following:

 df <- tibble('a' = character(), 'b' = character(), 'c' = character())

however, I would like to iterate through the items in the vector to populate the dataframe names and not have to manually specify each one. In reality my vector has 40+ names.

I've tried the following by they don't work:

 df <- tibble(vec[1:3])
 df <- tibble(vec)
 df <- tibble(for (i in 1:3){
   vec[i]
 })

Any help on this would be greatly appreciated!

Upvotes: 10

Views: 9814

Answers (5)

Marek
Marek

Reputation: 50704

You could use read_csv from readr package, which could read from text vector.

First you need to paste your vector and then read it:

readr::read_csv(I(paste(vec, collapse=",")), col_types = "c")

In pipeline:

vec %>%
    paste(collapse = ",") %>% # makes "a,b,c"
    I() %>% # needed for read_ to treat as string not a file
    readr::read_csv(col_types = "c") # columns types as characters

Upvotes: 0

s_baldur
s_baldur

Reputation: 33498

Another variant of Adam's idea:

as_tibble(sapply(vec, \(x) character()))

# A tibble: 0 x 3
# ... with 3 variables: a <chr>, b <chr>, c <chr>

Upvotes: 3

Martin Morgan
Martin Morgan

Reputation: 46856

A variant on Adam's solution is to create a named list of appropriate types, e.g.,

n = length(vec)
tmpl = setNames(rep(list(character()), n), head(letters, n))

This reflects the underlying structure of a data.frame or tibble -- a named list of equal-length vectors. Then

as.data.frame(tmpl, stringsAsFactors = FALSE)
tibble::as_tibble(tmpl)

Upvotes: 1

user10917479
user10917479

Reputation:

You can create a named vector, vec, where the first argument sets the type of column you want. The rep("", 3) line says I want three character columns. Then the second argument is the vector of column names.

Use dplyr::bind_rows to convert this into tibble with one row. Then [0, ] selects zero rows, leaving it empty.

With this method, you can control the data type for each column easily.

library(dplyr)

vec <- setNames(rep("", 3), letters[1:3])
bind_rows(vec)[0, ]

# A tibble: 0 x 3
# ... with 3 variables: a <chr>, b <chr>, c <chr>

You can also use as_tibble if you transpose the named vector. I guess I use bind_rows because I usually have dplyr loaded but not tibble.

library(tibble)

vec <- setNames(rep("", 3), letters[1:3])
as_tibble(t(vec))[0, ]

# A tibble: 0 x 3
# ... with 3 variables: a <chr>, b <chr>, c <chr>

If you know all of the columns are of a single type (e.g., character), you can do something like this.

vec <- letters[1:3]
df <- bind_rows(setNames(rep("", length(vec)), vec))[0, ]

Upvotes: 5

Humpelstielzchen
Humpelstielzchen

Reputation: 6441

You can do

library(tibble)

tb <- as_tibble(matrix(nrow = 0, ncol = length(vec), dimnames = list(NULL, vec)))

tb
# A tibble: 0 x 3
# ... with 3 variables: a <lgl>, b <lgl>, c <lgl>

Upvotes: 3

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