Getting a row from a data frame as a vector in R

You can't necessarily get it as a vector because each column might have a different mode. You might have numerics in one column and characters in the next.

If you know the mode of the whole row, or can convert to the same type, you can use the mode's conversion function (for example, as.numeric()) to convert to a vector. For example:

> state.x77[1,]
Population     Income Illiteracy   Life Exp     Murder    HS Grad      Frost 
   3615.00    3624.00       2.10      69.05      15.10      41.30      20.00 
      Area 
  50708.00 
> as.numeric(state.x77[1,])
[1]  3615.00  3624.00     2.10    69.05    15.10    41.30    20.00 50708.00

This would work even if some of the columns were integers, although they would be converted to numeric floating-point numbers.

There is a problem with what you propose; namely that the components of data frames (what you call columns) can be of different data types. If you want a single row as a vector, that must contain only a single data type - they are atomic vectors!

Here is an example:

> set.seed(2)
> dat <- data.frame(A = 1:10, B = sample(LETTERS[1:4], 10, replace = TRUE))
> dat
    A B
1   1 A
2   2 C
3   3 C
4   4 A
5   5 D
6   6 D
7   7 A
8   8 D
9   9 B
10 10 C
> dat[1, ]
  A B
1 1 A

If we force it to drop the empty (column), the only recourse for R is to convert the row to a list to maintain the disparate data types.

> dat[1, , drop = TRUE]
$A
[1] 1

$B
[1] A
Levels: A B C D

The only logical solution to this it to get the data frame into a common type by coercing it to a matrix. This is done via data.matrix() for example:

> mat <- data.matrix(dat)
> mat[1,]
A B 
1 1

data.matrix() converts factors to their internal numeric codes. The above allows the first row to be extracted as a vector.

However, if you have character data in the data frame, the only recourse will be to create a character matrix, which may or may not be useful, and data.matrix() now can't be used, we need as.matrix() instead:

> dat$String <- LETTERS[1:10]
> str(dat)
'data.frame':   10 obs. of  3 variables:
 $ A     : int  1 2 3 4 5 6 7 8 9 10
 $ B     : Factor w/ 4 levels "A","B","C","D": 1 3 3 1 4 4 1 4 2 3
 $ String: chr  "A" "B" "C" "D" ...
> mat <- data.matrix(dat)
Warning message:
NAs introduced by coercion 
> mat
       A B String
 [1,]  1 1     NA
 [2,]  2 3     NA
 [3,]  3 3     NA
 [4,]  4 1     NA
 [5,]  5 4     NA
 [6,]  6 4     NA
 [7,]  7 1     NA
 [8,]  8 4     NA
 [9,]  9 2     NA
[10,] 10 3     NA
> mat <- as.matrix(dat)
> mat
      A    B   String
 [1,] " 1" "A" "A"   
 [2,] " 2" "C" "B"   
 [3,] " 3" "C" "C"   
 [4,] " 4" "A" "D"   
 [5,] " 5" "D" "E"   
 [6,] " 6" "D" "F"   
 [7,] " 7" "A" "G"   
 [8,] " 8" "D" "H"   
 [9,] " 9" "B" "I"   
[10,] "10" "C" "J"
> mat[1, ]
     A      B String 
  " 1"    "A"    "A" 
> class(mat[1, ])
[1] "character"

Data.frames created by importing data from a external source will have their data transformed to factors by default. If you do not want this set stringsAsFactors=FALSE

In this case to extract a row or a column as a vector you need to do something like this:

as.numeric(as.vector(DF[1,]))

or like this

as.character(as.vector(DF[1,]))

How about this?

library(tidyverse)
dat <- as_tibble(iris)
pulled_row <- dat %>% slice(3) %>% flatten_chr()

If you know all the values are same type, then use flatten_xxx.

Otherwise, I think flatten_chr() is safer.