Question

I have the following code,

z <- data.frame(a=sample.int(10),b=sample.int(10),c=sample.int(10))
letter <- c("a","c","b") # this will be used as the argument to a function
vec <- unlist(lapply(1:length(letter), 
              function(x) cat(paste("z[[letter[",x,"]]],",sep=""))))
vec[length(vec)] <- paste("z[[letter[",length(vec),"]]]",sep="")

Consequently:

> vec    
[1] "z[[letter[1]]]," "z[[letter[2]]]," "z[[letter[3]]]" 

I want to use vec to order the rows of dataframe z, using the code below,

z.sort <- z[with(z, order(???)),]

How can I get the character vector vec to be evaluated as the arguments to order?
Is there a better way of doing this bearing in mind that letter, which is used to form vec will be an argument to a function?

Desired output would be:

    a  b  c
5   1  1  9
10  2 10  2
1   3  7  1
9   4  2  5
8   5  8  6
2   6  4  3
4   7  9 10
3   8  3  8
6   9  5  7
7  10  6  4

or as dput output:

structure(list(a = 1:10, b = c(1L, 10L, 7L, 2L, 8L, 4L, 9L, 3L, 5L, 6L), c = c(9L, 2L, 1L, 5L, 6L, 3L, 10L, 8L, 7L, 4L)), .Names = c("a", "b", "c"), row.names = c(5L, 10L, 1L, 9L, 8L, 2L, 4L, 3L, 6L, 7L), class = "data.frame")

Answer 1

Here’s what you want (with different random data):

> z[do.call(order, z[,letter]),]
    a  b  c
5   1  2  1
4   2  4  8
1   3  3  9
6   4  6  3
8   5  8  5
10  6  1  4
2   7  5  7
3   8 10  2
9   9  7  6
7  10  9 10

do.call lets us send a list to a function as its arguments, so we can just reorder the columns of z and send them to order using do.call, as a data.frame is just a special kind of list.

Works with no problem in a function:

my.reorder <- function(dat, cols) { dat[do.call(order, dat[,cols]),] }