Outer division (vectors to matrix) #2

[wageningen]

In the R user meeting of Nov 16th, we had a question why

 (1:5) %*% t(2:6)
     [,1] [,2] [,3] [,4] [,5]
[1,]    2    3    4    5    6
[2,]    4    6    8   10   12
[3,]    6    9   12   15   18
[4,]    8   12   16   20   24
[5,]   10   15   20   25   30

but

(1:5) %/% t(2:6)
     [,1] [,2] [,3] [,4] [,5]
[1,]    0    0    0    0    0

??
I don't know, but the solution (given by another participant) is:

outer((1:5),(2:6),"/")
     [,1]      [,2] [,3] [,4]      [,5]
[1,]  0.5 0.3333333 0.25  0.2 0.1666667
[2,]  1.0 0.6666667 0.50  0.4 0.3333333
[3,]  1.5 1.0000000 0.75  0.6 0.5000000
[4,]  2.0 1.3333333 1.00  0.8 0.6666667
[5,]  2.5 1.6666667 1.25  1.0 0.8333333

[bbrede]

Regarding the % operators: They signal a special binary operator. %/% means integer division, %% modulo.
In your example: (1:5) %/% t(2:6) the integer division will be applied and the t() will make the output a matrix, because the second argument is treated as a column vector.

That %*% means matrix multiplication, but %/% integer division is a bit unfortunate in R. But R is not a mathematical language (like Matlab), matrix multiplication is seen as special so get's a special syntax.

BTW: you can define a special binary operator yourself:
%s% <- function(x,y) x + y
3 %s% 4

[almeidaxan]

Just out of curiosity, the same result can also be achieved without the "outer" function, by simply doing

(1:5) %*% t(1/(2:6))