Anonymous Function - how to impose limits

How to impose max/min limits on anonymous functions i.e. I have to ensure that the current cell does not go beyond this matrix pTransition i.e. bounds are [-2,-2] and [2,2]

Thanks

pTransition = [ 
    0   0   0   0   0
    0  .05  .1 .05  0
    0   .1  .4  .1  0
    0  .05  .1 .05  0
    0   0   0   0   0  ];
getPLocalTransition = @(localCoordinate) ...
    pTransition(localCoordinate(1) + 3, localCoordinate(2) + 3);
% above are the given conditions

%trying to implement the foll function:
calcPTransition = @(x_t, u_t, x_tm1)...
pTransition(x_t(1) + 3, x_t(2) + 3);

[John Chilleri]

Would you mind posting the entire code? I believe that with careful definition of some variables, you can achieve your goal.

[Ken]

Thanks John.

I am trying to define:

calcPTransition(x_t, u_t, x_tm1)

which returns the probability for being in state x_t given the former state x_tm1 and the input u_t

x_t and x_tm1 are 1×2 vectors of integer grid cell coordinates

u_t is a 1×2 vector of integer coordinate offsets defining the expected translation of the robots position after one time step given that input

[John Chilleri]

I see that you add 3 to local coordinates because you're 'translating' [-2,2] to [1,5].

Are your x_t, x_tm1, and u_t also [-2,2], because if they're already [1,5] that would be your problem.

Also, you don't appear to be using u_t in the function.

Lastly, Walter Roberson posted an answer below in case you didn't see it - it might be what you needed.

[Ken]

Yes, x_t, x_tm1, and u_t are [-2,2 ], so before adding the 3, I have to ensure I don't get index exceeds matrix dimensions which I am getting now with: calcPTransition = @(x_t, u_t, x_tm1)... pTransition(x_t(1) + 3, x_t(2) + 3); Errors: Index exceeds matrix dimensions. calcPTransition evaluates to a wrong value for x_t=[1 1], u=[-4 -4], x_tm1=[0 0] : 0.05 (0 expected)

[Image Analyst]

Why not just simply make a regular function? It would only be a few lines and would be far, far less cryptic and hard to read. What's the big deal?

[Ken]

Can you suggest the lines of code?

[Ken]

Image Analyst - if it is so simple please respond

[Image Analyst]

What are you trying to impose limits on? The input or the output? And I don't see why u_t & x_tm1 are being passed in since the function uses x_t only.

pTransition(x_t(1) + 3, x_t(2) + 3);

What is the point of passing in u_t & x_tm1?

Let's say you have an inputMax, inputMin, and outputMax and outputMin. You could limit both input(s) and outputs to those min and max values, but I'm not sure what you want. Do you want to do that? Or you want to clip just the inputs, or just the outputs only? Please clarify.

[Ken]

The input. This is a 5X5 matrix and the intent of limits is to prevent the cells from going outside the matrix. Tried this:

@(M,r,c)min(max([-2, -2], x_t-(x_tm1+u_t)), [2,2]);
calcPTransition = @(x_t, u_t, x_tm1) BoundsIndex2D(pTransition, x_t(1) + 3, x_t(2) + 3);
but get error:
The return value of calcPTransition must be a scalar double but it is a [1 2] double! 
The condition input argument must be a scalar logical. 
The return value of calcPTransition must be a scalar double but it is a [1 2] double! 
The condition input argument must be a scalar logical.

[Ken]

Thanks for replying.

Did not see the 2nd part of the post from Image Analyst, so replying to it now:

Agree, no need of sending u_t and x_tm1. Only need to limit inputs, outputs will take care of themselves. Just want to clip the inputs at -2,-2 on the min and 2,2 on the max, so that I don't go outside the matrix. The matrix centre is at 3,3 hence the limits -2,-2 and 2,2.

Answer 1

[Image Analyst]

See if this is what you want:

function test()
% Call GetPLocalTransition with -2, 99 (both outside)
calcPTransition = GetPLocalTransition(-2, 99);
% Call GetPLocalTransition with 2, 4 (both inside)
calcPTransition = GetPLocalTransition(2, 4);
% Call GetPLocalTransition with 2, 15 (row inside, column outside)
calcPTransition = GetPLocalTransition(2, 15);

% Row and col can be anything.  If outside the edges of pTransition,
% then they will be clipped to the edge location
function output = GetPLocalTransition(row, col)
pTransition = [ 
    0   0   0   0   0
    0  .05  .1 .05  0
    0   .1  .4  .1  0
    0  .05  .1 .05  0
    0   0   0   0   0  ];
[rows, columns] = size(pTransition);
% Clip to being inside the array.
if row < 1
  fprintf('Row of %d clipped to row #1\n', row);
  row = 1;
end
if col < 1
  fprintf('Column of %d clipped to column #1\n', col);
  col = 1;
end
if row > rows
  fprintf('Row of %d clipped to row #%d\n', row, rows);
  row = rows;
end
if col > columns
  fprintf('Column of %d clipped to column #%d\n', col, columns);
  col = columns;
end
% Row and col are definitely inside now.
output = pTransition(row, col);
fprintf('pTransition(%d, %d) = %.2f\n', row, col, pTransition(row, col));

[Ken]

Tried it but get error "Not enough input arguments"

They are looking for a one-line command; the problem is stated verbatim below:

pTransition = [ 0 0 0 0 0 0 .05 .1 .05 0 0 .1 .4 .1 0 0 .05 .1 .05 0 0 0 0 0 0 ]; getPLocalTransition = @(localCoordinate) ... pTransition(localCoordinate(1) + 3, localCoordinate(2) + 3); %%%%%%%%%%% PLEASE DON'T CHANGE ANYTHING ABOVE THIS LINE %%%%%%%%%%%

calcPTransition = @(x_t, u_t, x_tm1) ???; % maybe use getPLocalTransition

[Image Analyst]

Can't be. I just copied and pasted that from my test.m and it worked fine. You must have changed something.

And who is this "they" that you are talking about? I thought you needed it for yourself. Why would your question be stated verbatim in a "problem"? It's not homework, or else you would have tagged it as homework. Nor did you say it was homework anywhere I remember. So who is "they"? Your employer??? Why would they impose a one line requirement instead of just getting the job done regardless of how many lines?

http://www.mathworks.com/matlabcentral/answers/13205#answer_18099

[Ken]

Tried it in Matlab - fine. This is a problem I attempted from an Auto. Mobile Robots course.

Thanks a lot!

Answer 2

[Walter Roberson]

If you mean in the sense that if the index is out of bounds then instead the closest in-bounds index should be used, then:

BoundsIndex2D = @(M,r,c) M( max( min(r, size(M,1), 1 ), max( min(c, size(M,2), 1 );

calcPTransition = @(x_t, u_t, x_tm1) BoundsIndex2D(pTransition, x_t(1) + 3, x_t(2) + 3);

However, in the cases where you might want something like this, such as running a minimization where the probe location is to index an array, then typically the indices would end up not being constrained to integers, and then you would have the difficulty that you are trying to index at a fractional location. In the cases where you can be sure that the indices are integer you can typically also be sure that the indices will be in range.

[Ken]

Thanks Walter. Tried but get: Error: Line: 11 Column: 85 Unbalanced or unexpected parenthesis or bracket.

[Walter Roberson]

BoundsIndex2D = @(M,r,c) M( max( min(r, size(M,1)), 1 ), max( min(c, size(M,2)), 1) );

[Ken]

Fixed the bracket problem but but now get: Could not eval "calcPTransition([0, 0], [0, 0], [0, 0])" : MIN with two matrices to compare and a working dimension is not supported.

[Walter Roberson]

Please check your bracket positions. I had them wrong before but the version I posted most recently corrected them.

[Ken]

Thanks Walter - so far the closest solution with least errors, still get one error:

calcPTransition evaluates to a wrong value for x_t=[1 1], u=[-4 -4], x_tm1=[0 0] : 0.05 (0 expected)

Used the following script:

pTransition = [ 
    0   0   0   0   0
    0  .05  .1 .05  0
    0   .1  .4  .1  0
    0  .05  .1 .05  0
    0   0   0   0   0  ];
getPLocalTransition = @(localCoordinate) ...
    pTransition(localCoordinate(1) + 3, localCoordinate(2) + 3);
%%%%%%%%%%%PLEASE DON'T CHANGE ANYTHING ABOVE THIS LINE %%%%%%%%%%%

BoundsIndex2D = @(M,r,c) M( max( min(r, size(M,1)), 1 ), max( min(c, size(M,2)), 1) );
calcPTransition = @(x_t, u_t, x_tm1) BoundsIndex2D(pTransition, x_t(1) + 3, x_t(2) + 3);

[Ken]

Any suggestions?

[Walter Roberson]

Ensuring that the indices do not go beyond bounds is a quite different matter than calculating for a specific purpose. My adjustment makes sure that you do not get any out of bounds error by instead accessing the location closest to the border. Depending on how the function is to be used, it could be that what you want to do is not suitable for an anonymous function.

Or in some situations a way around the problem might be to return 0 instead of an actual value, perhaps by using

BoundsIndex2D = @(M,r,c) (r >= 1 & r <= size(M,1) & c >=1 & c <= size(M,2)) .* M( max( min(r, size(M,1)), 1 ), max( min(c, size(M,2)), 1) );

[Ken]

Thanks Walter, get the same error. Some more info on this problem:

Indeed we have to take care of not calling getPLocalFunction with out of bounds indices, however I would like to point out that this can happen not only when the norm of the displacement vector is > 2, but in general when either the first or the second component of the displacement vector are less than -2 or more than 2. Intuitively, this is given by the fact that pTransition indices go from 1 to 5 in both dimensions, so you need to take care both of the lower and upper bounds for your getPLocalTransition call. Now, how could you embed this logic in an anonymous function? You could just enforce the upper and lower bounds with the min and max MATLAB functions. For example: max([-2, -2], x_t-(x_tm1+u_t)) would return a new displacement vector, where the lowest possible value for every component is -2. Then you could apply the upper bound in a similar fashion, and finally you'll be able to safely pass the new displacement vector to your getPLocalTransition call.

[Walter Roberson]

Well, you can follow what they suggest there of using max() and min() .

[Ken]

But not sure how to implement it. i.e. use 2 if loops, one for max, one for min?

[Walter Roberson]

min(max([-2, -2], x_t-(x_tm1+u_t)), [2,2])

[Ken]

Tried but error:

•Could not eval "calcPTransition([0, 0], [0, 0], [0, 0])" : Undefined function or variable 'x_t'. 
Undefined function or variable 'x_t'. 
  pTransition = [ 
      0   0   0   0   0
      0  .05  .1 .05  0
      0   .1  .4  .1  0
      0  .05  .1 .05  0
      0   0   0   0   0  ];
  getPLocalTransition = @(localCoordinate) ...
      pTransition(localCoordinate(1) + 3, localCoordinate(2) + 3);
  %%%%%%%%%%%PLEASE DON'T CHANGE ANYTHIN.. 
  BoundsIndex2D = @(M,r,c)min(max([-2, -2], x_t-(x_tm1+u_t)), [2,2])

    calcPTransition = @(x_t, u_t, x_tm1) BoundsIndex2D...
    (pTransition, x_t(1) + 3, x_t(2) + 3);

[Ken]

Since I was getting errors, I init'ed:

x_t = [0,0];
x_tm1 = [0,0];

then added the 2 lines:

@(M,r,c)min(max([-2, -2], x_t-(x_tm1+u_t)), [2,2]);
calcPTransition = @(x_t, u_t, x_tm1) BoundsIndex2D(pTransition, x_t(1) + 3, x_t(2) + 3);

but get error:

The return value of calcPTransition must be a scalar double but it is a [1 2] double! 
The condition input argument must be a scalar logical. 

u_t = [0,0];

Answer 3

[Eva Barceló Michans]

Can someone help me please?

Is regarding the same exercise Ken share.

I put this code:

pTransition = [ 
      0   0   0   0   0
      0  .05  .1 .05  0
      0   .1  .4  .1  0
      0  .05  .1 .05  0
      0   0   0   0   0  ];
  getPLocalTransition = @(localCoordinate) ...
      pTransition(localCoordinate(1) + 3, localCoordinate(2) + 3);
  %%%%%%%%%%%PLEASE DON'T CHANGE ANYTHIN..
  x_t = [0,0];
x_tm1 = [0,0];
u_t = [0,0];

  BoundsIndex2D = @(M,r,c)min(max([-2, -2], x_t-(x_tm1+u_t)), [2,2])
    calcPTransition = @(x_t, u_t, x_tm1) BoundsIndex2D...
    (pTransition, x_t(1) + 3, x_t(2) + 3);


plotPredictionUpdate(calcPTransition);

The error Matlab is:

Unable to perform assignment because the size of the left side is 1-by-1 and the size of the right side is 1-by-2.

I don't know how to solve it.

Thank you

[Image Analyst]

Instead of using anonymous functions, just make them regular functions, which are SO much easier to debug.

[Eva Barceló Michans]

The problem is that I am doing an online course and I have to do it with anonymous function.