How do I transpose a tibble() in R

Question

In R the t() function is really meant for matrices. When I try to transpose my tibble with t() I end up with a matrix. A matrix can't be made into a tibble with tibble(). I end up spending time storing column names as variables and attaching them as I try to re-make a transposed version of my tibble.

Question: What is the simplest way to transpose a tibble where the first column should become the column names of the new tibble and the old column names become the first column of my new tibble.

The only problem with this is that all of my doubles turned into characters. — Alex, Mar 14 '17 at 19:50
`sapply(df[,2:length(df)],as.numeric)` I think this fixed it — Alex, Mar 14 '17 at 19:57
Late for it, but you could convert to numeric with `%>% mutate_at(2:nrow(df), as.numeric)` — BMLopes, Aug 17 '21 at 16:12

score 35 · Accepted Answer · edited May 27 '21 at 13:32

35

As Sotos has mentioned it, you just need to re-declare your matrix as a tibble:

as_tibble(cbind(nms = names(df), t(df)))

edited May 27 '21 at 13:32

micstr

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answered Mar 14 '17 at 16:17

Laurent

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1

This did not work for me and added the original column of rownames as an additional, unwanted row in the resulting tibble, and not as the names of the columns like it should. – Leo Jul 07 '21 at 17:42

score 18 · Answer 2 · edited May 23 '17 at 12:17

18

Solution here: https://stackoverflow.com/a/28917212/3880322

library(dplyr)
library(tidyr)
df %>%
    gather(key = var_name, value = value, 2:ncol(df)) %>% 
    spread_(key = names(df)[1],value = 'value')

edited May 23 '17 at 12:17

Community

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answered Mar 14 '17 at 16:04

Rahul

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Indrajeet Patil · Answer 3 · 2019-07-24T19:29:55.483

14

I was faced with the same problem and tried all the above-mentioned solutions and realized that none of them actually preserve the names of the columns.

Here is, in my opinion, a better way to do the same:

# attaching needed libraries
library(dplyr)
library(data.table)
library(tibble)

# defining the dataframe
df <- cbind.data.frame(x = rnorm(10), y = rnorm(10))

# custom function to transpose while preserving names
transpose_df <- function(df) {
  t_df <- data.table::transpose(df)
  colnames(t_df) <- rownames(df)
  rownames(t_df) <- colnames(df)
  t_df <- t_df %>%
    tibble::rownames_to_column(.data = .) %>%
    tibble::as_tibble(.)
  return(t_df)
}

# using the function
transpose_df(df)
#> # A tibble: 2 x 11
#>   rowname    `1`   `2`     `3`    `4`      `5`   `6`    `7`    `8`    `9`
#>   <chr>    <dbl> <dbl>   <dbl>  <dbl>    <dbl> <dbl>  <dbl>  <dbl>  <dbl>
#> 1 x       -1.38  0.752  1.22    0.296 -0.00298 1.50  -0.719 -0.503 -0.114
#> 2 y        0.618 0.304 -0.0559 -1.27   0.0806  0.156  0.522  0.677  0.532
#> # ... with 1 more variable: `10` <dbl>

Created on 2018-02-17 by the reprex package (v0.2.0).

edited Jul 24 '19 at 19:29

answered Feb 17 '18 at 06:53

Indrajeet Patil

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I like this, but if the input is already tibble, and not data.frame, it fails with `Error in tibble::rownames_to_column(.data = .) : is.data.frame(df) is not TRUE`. Can you fix it to behave as expected with a tibble input? – paulduf Jun 13 '22 at 08:36
I just tried it with a tibble input and there is no issue. Are you sure you are entering a data frame, and not some other data structure? – Indrajeet Patil Jun 13 '22 at 10:12
Indeed, I just learned `purrr:map` returns a list and not a df ... `purrr::map_dfr` is the way to go then. Sorry. I can delete my comment then. – paulduf Jun 13 '22 at 15:25
Thanks, based on the use of `t()`, this worked for me: ```{r} transpose_df <- function(df) { df %>% t() %>% #Tranpose, but function is for matrices. Return Matrix as.data.frame() %>% #Force to be dataframe tibble::rownames_to_column(var = "rowname") %>% #Resave first column from rownames janitor::row_to_names(row_number = 1) #Resave column headers from first row. }``` – phargart Sep 21 '22 at 20:35

How do I transpose a tibble() in R

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