Undesired behaviour of bsxfun and colon-operator

I have a problem using bsxfun for a certain user-defined function. A simplified version of it looks like this:

bsxfun(@(x,y)x+sum(0:y),(1:3)',6:9)

ans =

    22    29    37    46
    23    30    38    47
    24    31    39    48

This does exactly what I want. But if i use it with a single number of x it gives me:

bsxfun(@(x,y)x+sum(0:y),1,6:9)

ans =

    22

While I would like to get

ans =

    22    29    37    46

Any ideas to build a workaround for this problem?

[Stephen23]

"Undesired behaviour of bsxfun and colon-operator"

You might consider this to be undesired behavior, but the cause is that your function is not truly vectorized, as required by the bsxfun documentation:

"fun must support scalar expansion, such that if A or B is a scalar, then C is the result of applying the scalar to every element in the other input array.

A simple example shows that this is not the case when x is scalar and y is non-scalar:

>> F = @(x,y)x+sum(0:y);
>> F([1;2;3],5) % Vectorized.
ans =
   16
   17
   18
>> F(6,[7,8,9]) % Not vectorized!
ans =  34

This is due to how the colon operator interprets non-scalar arguments, and bsxfun changing the orientations of the arguments that it provides to your function.

Answer 1

[Stephen23]

A robust workaround using ndgrid and arrayfun and exactly the same function:

>> [X,Y] = ndgrid(1:3,6:9);
>> arrayfun(@(x,y)x+sum(0:y),X,Y)
ans =
   22   29   37   46
   23   30   38   47
   24   31   39   48

>> [X,Y] = ndgrid(1,6:9);
>> arrayfun(@(x,y)x+sum(0:y),X,Y)
ans =
   22   29   37   46

Answer 2

[Matt J]

bsxfun(@(x,y)x+y.*(y+1)/2,1,6:9)

[Matt J]

Better:

f=@(z) z.*(z+1)./2;

bsxfun(@plus,(1:3).', f(6:9))

Answer 3

[Matt J]

bsxfun(@(x,y)x+arrayfun(@(z)sum(0:z),y),1,6:9)