Get a row in data.frame as a vector where each element is a string

Question

I have issues getting the output I want from a data.frame. The structure of the data is the following:

head(test)

    A  T  L   B  E  X   D
4  no no no yes no no yes
7  no no no  no no no  no
11 no no no  no no no  no
12 no no no yes no no yes
17 no no no  no no no  no
27 no no no  no no no  no

The output I get:

test[1,]
   A  T  L   B  E  X   D
4 no no no yes no no yes

The output I want:

[1] "no"  "no"  "no"  "yes" "no"  "no"  "yes"

Or simply, the output I want is a vector where each element is a string value of that column in the df. I could do a for loop or some stupid thing like that, but I belive there should be a much simpler way that I am missing.

I have tried:

as.character(test[1,])
[1] "1" "1" "1" "2" "1" "1" "2"

Not sure what I am missing here?

Ronak Shah · Accepted Answer · 2019-09-05T10:49:06.077

Use unlist and then as.character

as.character(unlist(test[1, ]))
#[1] "no"  "no"  "no"  "yes" "no"  "no"  "yes"

test[1, ] is still a dataframe and applying as.character on data frame doesn't work. We use unlist to make dataframe to vector and then use as.character to convert it into character.

data

test <- structure(list(A = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = "no", class = "factor"), 
T = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = "no", class = "factor"), 
L = structure(c(1L, 1L, 1L, 1L, 1L, 1L), .Label = "no", class = "factor"), 
B = structure(c(2L, 1L, 1L, 2L, 1L, 1L), .Label = c("no", 
"yes"), class = "factor"), E = structure(c(1L, 1L, 1L, 1L, 
1L, 1L), .Label = "no", class = "factor"), X = structure(c(1L, 
1L, 1L, 1L, 1L, 1L), .Label = "no", class = "factor"), D = structure(c(2L, 
1L, 1L, 2L, 1L, 1L), .Label = c("no", "yes"), class = "factor")), 
class = "data.frame", row.names = c("4", "7", "11", "12", "17", "27"))

Get a row in data.frame as a vector where each element is a string

1 Answers1