Convert a row of a data frame to vector

When you extract a single row from a data frame you get a one-row data frame. Convert it to a numeric vector:

as.numeric(df[1,])

As @Roland suggests, unlist(df[1,]) will convert the one-row data frame to a numeric vector without dropping the names. Therefore unname(unlist(df[1,])) is another, slightly more explicit way to get to the same result.

As @Josh comments below, if you have a not-completely-numeric (alphabetic, factor, mixed ...) data frame, you need as.character(df[1,]) instead.

I recommend unlist, which keeps the names.

unlist(df[1,])
  a   b   c 
1.0 2.0 2.6 

is.vector(unlist(df[1,]))
[1] TRUE

If you don't want a named vector:

unname(unlist(df[1,]))
[1] 1.0 2.0 2.6

Here is a dplyr based option:

newV = df %>% slice(1) %>% unlist(use.names = FALSE)

# or slightly different:
newV = df %>% slice(1) %>% unlist() %>% unname()

If you don't want to change to numeric you can try this.

> as.vector(t(df)[,1])
[1] 1.0 2.0 2.6

Note that you have to be careful if your row contains a factor. Here is an example:

df_1 = data.frame(V1 = factor(11:15),
                  V2 = 21:25)
df_1[1,] %>% as.numeric() # you expect 11 21 but it returns 
[1] 1 21

Here is another example (by default data.frame() converts characters to factors)

df_2 = data.frame(V1 = letters[1:5],
                  V2 = 1:5)
df_2[3,] %>% as.numeric() # you expect to obtain c 3 but it returns
[1] 3 3
df_2[3,] %>% as.character() # this won't work neither
[1] "3" "3"

To prevent this behavior, you need to take care of the factor, before extracting it:

df_1$V1 = df_1$V1 %>% as.character() %>% as.numeric()
df_2$V1 = df_2$V1 %>% as.character()
df_1[1,] %>% as.numeric()
[1] 11  21
df_2[3,] %>% as.character()
[1] "c" "3"

Columns of data frames are already vectors, you just have to pull them out. Note that you place the column you want after the comma, not before it:

> newV <- df[,1]
> newV
[1] 1 2 4 2

If you actually want a row, then do what Ben said and please use words correctly in the future.