statistics - How to print the name of current row when using apply in R?

Question

For example, I have a matrix k

> k
  d e
a 1 3
b 2 4

I want to apply a function on k

> apply(k,MARGIN=1,function(p) {p+1})
a b
d 2 3
e 4 5

However, I also want to print the rowname of the row being apply so that I can know which row the function is applied on at that time.

It may looks like this:

apply(k,MARGIN=1,function(p) {print(rowname(p)); p+1})

But I really don't do how to do that in R. Does anyone has any idea?

Answer 1

Here's a neat solution to what I think you're asking. (I've called the input matrix mat rather than k for clarity - in this example, mat has 2 columns and 10 rows, and the rows are named abc1 through to abc10.)

In the code below, the result out1 is the thing you wanted to calculate (the outcome of the apply command). The result out2 comes out identically to out1 except that it prints out the rownames that it is working on (I put in a delay of 0.3 seconds per row so you can see it really does do this - take this out when you want the code to run full speed obviously!)

The trick I came up with was to cbind the row numbers (1 to n) onto the left of mat (to create a matrix with one additional column), and then use this to refer back to the rownames of mat. Note the line x = y[-1] which means that the actual calculation within the function (here, adding 1) ignores the first column of row numbers, which means it's the same as the calculation done for out1. Whatever sort of calculation you want to perform on the rows can be done this way - just pretend that y never existed, and formulate your desired calculation using x. Hope this helps.

set.seed(1234)
mat = as.matrix(data.frame(x = rpois(10,4), y = rpois(10,4)))
rownames(mat) = paste("abc", 1:nrow(mat), sep="")
out1 = apply(mat,1,function(x) {x+1})
out2 = apply(cbind(seq_len(nrow(mat)),mat),1,
             function(y) {
                           x = y[-1]
                           cat("Doing row:",rownames(mat)[y[1]],"
")
                           Sys.sleep(0.3)
                           x+1
                          }
            )

identical(out1,out2)