numerics - Finding a fit to a multi-dimensioned function

I have a model function $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$, and a bunch of data points for which I'd like Mathematica to fit for me. Unfortunately FindFit seems to only deal with functions $\mathbb{R}^n\rightarrow\mathbb{R}$. I guess I could make my own square difference from the data to the model and use NMinimize on that, but I wondering if there was an easier way?

Edit - toy example with answers

$g\left(x,y\right)=\left$

g[{x_, y_}, {a_, b_, c_, d_}] := {a b x + (2 c + d) y, (2 a + b) x + c d y};
points = Flatten[Table[{{x, y}, g[{x, y}, {1, 2, 3, 4}]},
  {x, -1, 1, 0.5}, {y, -1, 1, 0.5}], 1];

Finding a fit for each component separately doesn't work:

data1 = points /. {{x_, y_}, {u_, v_}} -> {x, y, u};
data2 = points /. {{x_, y_}, {u_, v_}} -> {x, y, v};
FindFit[data1, g[{x, y}, {a, b, c, d}][[1]], {a, b, c, d}, {x, y}]
FindFit[data2, g[{x, y}, {a, b, c, d}][[2]], {a, b, c, d}, {x, y}]

{a -> 0.729723, b -> 2.74077, c -> 3.35659, d -> 3.28681}

{a -> 1.48555, b -> 1.02891, c -> 2.81936, d -> 4.25629}

Jens suggestion works well:

data3 = points /. {{x_, y_}, {u_, v_}} -> Sequence[{x, y, 0, u}, {x, y, 1, v}];
FindFit[data3, g[{x, y}, {a, b, c, d}].{1 - s, s}, {a, b, c, d}, {x, y, s}]

{a -> 1., b -> 2., c -> 3., d -> 4.}

My original suggestion (which is similar to user840's links)

error = Plus @@ (points /.
    {{x_, y_}, {u_, v_}} -> Norm[g[{x, y}, {a, b, c, d}] - {u, v}]^2);

NMinimize[error, {a, b, c, d}]

Gives a different, but still correct, solution (or at least a close approximation thereof)

{1.41754*10^-13, {a -> 1.00004, b -> 1.99993, c -> 2., d -> 6.}}

Answer

Without a specific example, I can only make general suggestions here.

The first thing that comes to mind is that you could fit each of the two components of your vector field independently. I.e., if $\vec{f}: \mathbb{R}^2\to\mathbb{R}^2$ is split up into $\vec{f} = \{f_x, f_y\}$ with $f_{1,2}: \mathbb{R}^2\to\mathbb{R}$, then FindFit would work on each of these component functions.

If you don't want the fits to be determined independently, it could still be possible to use FindFit by introducing an auxiliary variable $s$ that labels the two component functions above, $f_s$, and then provide the fitting data with this variable s included. Here, s can only have two discrete values (I borrowed the idea from spin-half quantum mechanics). Let's choose s = 0 for the function $f_x$ and s = 1 for $f_y$. So you'd have data in the form

data = {{x1, y1, 0, fx1}, {x1, y1, 1, fy1}, ...}

where fx1 is the first value of $f_x$ and fy1 the first value of $f_y$, both at the point x1, y1.

Your model could then look something like this:

model[x_, y_, s_] := modely[x, y]*s + modelx[x, y]*(1 - s)

where modely[x, y] is the model for $f_y$ and modelx[x, y] is the model for $f_x$. The variables in the FindFit call would be x, y, s, and the parameters in the models modelx, modely could be the same (or different).

There are probably many other reasonable ways to do such a fit, but these are some ideas.