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- Transposing data frames 1 answer
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.
As Sotos has mentioned it, you just need to re-declare your matrix as a tible:
as_tibble(cbind(nms = names(df), t(df)))
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')
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).
来源:https://stackoverflow.com/questions/42790219/how-do-i-transpose-a-tibble-in-r