R: applying function over matrix and keeping matrix dimensions

Question

So I want to apply a function over a matrix in R. This works really intuitively for simple functions:

> (function(x)x*x)(matrix(1:10, nrow=2))  [,1] [,2] [,3] [,4] [,5] [1,]    1    9   25   49   81 [2,]    4   16   36   64  100 

...but clearly I don't understand all of its workings:

> m = (matrix(1:10, nrow=2)) > (function(x) if (x %% 3 == 0) { return(NA) } else { return(x+1) })(m)      [,1] [,2] [,3] [,4] [,5] [1,]    2    4    6    8   10 [2,]    3    5    7    9   11 Warning message: In if (x == 3) { :   the condition has length > 1 and only the first element will be used 

I read up on this and found out about Vectorize and sapply, which both seemed great and just like what I wanted, except that both of them convert my matrix into a list:

> y = (function(x) if (x %% 3 == 0) { return(NA) } else { return(x+1) }) > sapply(m, y)  [1]  2  3 NA  5  6 NA  8  9 NA 11 > Vectorize(y)(m)  [1]  2  3 NA  5  6 NA  8  9 NA 11 

...whereas I'd like to keep it in a matrix with its current dimensions. How might I do this? Thanks!

Answer 1

@Joshua Ulrich (and Dason) has a great answer. And doing it directly without the function y is the best solution. But if you really need to call a function, you can make it faster using vapply. It produces a vector without dimensions (as sapply, but faster), but then you can add them back using structure:

# Your function (optimized) y = function(x) if (x %% 3) x+1 else NA  m <- matrix(1:1e6,1e3) system.time( r1 <- apply(m,1:2,y) ) # 4.89 secs system.time( r2 <- structure(sapply(m, y), dim=dim(m)) ) # 2.89 secs system.time( r3 <- structure(vapply(m, y, numeric(1)), dim=dim(m)) ) # 1.66 secs identical(r1, r2) # TRUE identical(r1, r3) # TRUE 

...As you can see, the vapply approach is about 3x faster than apply... And the reason vapply is faster than sapply is that sapply must analyse the result to figure out that it can be simplified to a numeric vector. With vapply, you specified the result type (numeric(1)), so it doesn't have to guess...

UPDATE I figured out another (shorter) way of preserving the matrix structure:

m <- matrix(1:10, nrow=2) m[] <- vapply(m, y, numeric(1)) 

You simply assign the new values to the object using m[] <-. Then all other attributes are preserved (like dim, dimnames, class etc).

Answer 2

One way is to use apply on both rows and columns:

apply(m,1:2,y)      [,1] [,2] [,3] [,4] [,5] [1,]    2   NA    6    8   NA [2,]    3    5   NA    9   11 

You can also do it with subscripting because == is already vectorized:

m[m %% 3 == 0] <- NA m <- m+1 m      [,1] [,2] [,3] [,4] [,5] [1,]    2   NA    6    8   NA [2,]    3    5   NA    9   11