Values from multiple dataframe columns into one vector

Question

I have a dataframe df that has many cols and say 100 rows.

How do I take all the level values from the columns with names "alpha", "gamma" and "zeta" and store the 300 of them in a single vector?

Answer 1

You have an accepted answer, but here's what I think is happening: You have a combination of factor and character columns. In that case, unlist doesn't work directly, but if they were all factor or if they were all character, there would be no problem:

Some sample data:

mydf <- data.frame(A = LETTERS[1:3], B = LETTERS[4:6], C = LETTERS[7:9],
                   D = LETTERS[10:12], E = LETTERS[13:15])
df <- mydf
df$E <- as.character(df$E)
colsOfInterest <- c("A", "B", "E")

Case 1, all columns are factors

unlist(mydf[colsOfInterest], use.names = FALSE)
# [1] A B C D E F M N O
# Levels: A B C D E F M N O

Case 2, column E = characters, other columns factors

unlist(df[colsOfInterest], use.names = FALSE)
# [1] "1" "2" "3" "1" "2" "3" "M" "N" "O"

unlist(lapply(df[colsOfInterest], as.character), use.names = FALSE)
# [1] "A" "B" "C" "D" "E" "F" "M" "N" "O"

---

For a problem at the scale described here, the benchmarks show that converting to character first and using unlist is actually the fastest approach if you don't care for retaining factors. Note that the result of fun1() won't be correct if some columns are factors and some are characters. Here's a benchmark on a 100 row data.frame:

library(microbenchmark)    
microbenchmark(fun1(), fun2(), fun3())
# Unit: microseconds
#    expr      min        lq    median       uq      max neval
#  fun1()  572.606  587.3595  595.4845  606.175 3439.055   100
#  fun2()  327.570  334.6265  341.2550  350.449 3443.758   100
#  fun3() 1037.020 1055.6215 1064.1745 1086.197 3929.981   100

Of course, here we're talking microseconds, but the results scale too.

For reference, here's what was used for benchmarking. Change "nRow" and "nCol" if you want to test on a different sized data.frame extracting different numbers of columns.

nRow <- 100
nCol <- 30
set.seed(1)
mydf <- data.frame(matrix(sample(LETTERS, nRow*nCol, replace = TRUE), nrow = nRow))
colsOfInterest <- sample(nCol, sample(nCol*.7, 1))
length(colsOfInterest)
# [1] 17

library(microbenchmark)    
fun1 <- function() unlist(mydf[colsOfInterest], use.names = FALSE)
fun2 <- function() unlist(lapply(mydf[colsOfInterest], as.character), use.names = FALSE)
fun3 <- function() as.vector(as.matrix(mydf[colsOfInterest]))
microbenchmark(fun1(), fun2(), fun3())

Answer 2

I've found that converting to a matrix first makes getting to levels a bit easier.

as.vector(as.matrix(df[,c("alpha", "gamma", "zeta")]))

Of course, you could have just done stringsAsFactors=FALSE when you read the data in initially.