Question about :
I have a string vector foo:
>foo = c("1x2","3x4","5x6","7x8","9x10")
I split the individual strings over the “x” and stick the result in goo:
>goo = strsplit(foo, "x")
>goo
[[1]]
[1] "1" "2"
[[2]]
[1] "3" "4"
[[3]]
[1] "5" "6"
[[4]]
[1] "7" "8"
[[5]]
[1] "9" "10"
How do I extract the first and second ‘column’ from this list? (I want (1,3,5,7,9) and (2,4,6,8,10))
Use sapply to serially 'extract' using "[[":
sapply(goo, "[[" , 1)
[1] "1" "3" "5" "7" "9"
I've always thought that should be the result, but I probably don't understand the issues.
> result <- do.call(rbind, goo)
> result
[,1] [,2]
[1,] "1" "2"
[2,] "3" "4"
[3,] "5" "6"
[4,] "7" "8"
[5,] "9" "10"
> result[, 1] # column 1
[1] "1" "3" "5" "7" "9"
> result[, 2] # column 2
[1] "2" "4" "6" "8" "10"
The easiest way is to wrap the results in an sapply statement
odd_results <- sapply(goo, function(x) x[1])
and
even_results <- sapply(goo, function(x) x[2])
You'll need to specify the extraction of the list elements by position.
# load data
foo <- c("1x2","3x4","5x6","7x8","9x10")
# split foo by 'x'
foo.list <- strsplit( x = foo, split = "x", fixed = TRUE )
# store the first and second elements of each list in a data frame
foo.df <-
data.frame(
First_Element = unlist( lapply( X = foo.list, FUN = "[[", FUN.VALUE = 1 ) )
, Second_Element = unlist( lapply( X = foo.list, FUN = "[[", FUN.VALUE = 2 ) )
)
For the sake of variety, you could use
goo <- unlist(strsplit(foo, "x"))
a <- goo[seq(1, length(goo), 2)]
b <- goo[seq(2, length(goo), 2)]
Which yields
[1] "1" "3" "5" "7" "9"
and
[1] "2" "4" "6" "8" "10"
respectively.
