I want to create a vector out of a row of a data frame. But I don't want to have to row and column names. I tried several things... but had no luck.
This is my data frame:
> df <- data.frame(a=c(1,2,4,2),b=c(2,6,2,1),c=c(2.6,8.2,7.5,3)) > df a b c 1 1 2 2.6 2 2 6 8.2 3 4 2 7.5 4 2 1 3.0 I tried:
> newV <- as.vector(df[1,]) > newV a b c 1 1 2 2.6 But I really want something looking like this:
> newV <- c( 1,2,2.6) > newV [1] 1.0 2.0 2.6 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.
identical(unlist(df[1,], use.names = FALSE), as.numeric(df[1,])) (and btw df still is not a sensible name for a data.frame... ;-))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 
str(as.vector(t(df)[,1])) is num [1:3] 1 2 2.6, i.e. your code does convert the results to a numeric vector ...t(df) R coerces the data frame to a matrix, in this case a numeric matrix because all the elements are numeric. Then [,1] extracts the first column (a numeric vector, because the redundant dimension is automatically dropped). as.vector() just drops the names (which you could also do with unname()).
unname(unlist(x)) solution is a little better (more efficient and more transparent).
as.vector(t(df)[,1]) I love it ! Exactly what I need it !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" 
unfactorizing the columnns one by one?If you are looking for a solution that uses dplyr.
require(dplyr) df <- tibble(a=c(1,2,4,2),b=c(2,6,2,1),c=c(2.6,8.2,7.5,3)) df %>% # create a set of duplicated columns for your source columns mutate(across(c(a, b, c), ~ ., .names = "duplicated_{.col}")) %>% # nest the new/duplicated columns, usable in most cases nest(nested_data = matches("duplicated_")) %>% # or use only the first row of each new data frame (nested) # rowwise is important rowwise %>% mutate(nested_data = list(slice_head(nested_data, n = 1) %>% as.numeric(.))) Note that:
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.
c(t(as.matrix(df)))?