How to intersect a matrix with a set and keep the rows of the matrix?

I'm trying to intersect an N by M matrix, 'A', with a set (a 1 by X matrix), 'B', but am trying to do so by rows. (e.g. A(1:M) intersect B ).

So I would end up with an N by '__' matrix where the rows are the intersection of the respective row and set 'B', rather than a large set.

Is there a way to do this without using a for loop?

[James Tursa]

Can you provide a short example for what it means to intersect an N x M matrix with a 1 x X vector and end up with a N x M matrix?

[Rion Wilson-Yue]

Sorry I didn't mean ending up with an NxM matrix, I just want the same number of N rows

% 2x5 matrix
A = [
  2  12 6  4  8
  1  3  5  7  9
]

B = [ 1 2 3 4 5 6 ]

C = intersect(A, B);

I want 'C' to be

C = [
  2  6  4
  1  3  5
]

Using the intersect function provided by MATLAB 'C' is

C = [
  1
  2
  3
  4
  5
  6
]

[James Tursa]

Your example conveniently has the same number of elements in each row of the result. What if this is not the case?

[Rion Wilson-Yue]

That would be a problem, wouldn't it... I keep forgetting this is not C. Would it be possible to pad rows with NaN's? (I'm assuming MATLAB functions like min and max ignore NaN's.)

[James Tursa]

There are various ways to do it (padding, cell arrays, etc) ... we just need to know what you want for an output.

[Rion Wilson-Yue]

In the very end I'm looking for the min and max of the intersection of a row and the 1xX array. I want to keep the rows so I can get the min and max for each of them.

So long as I can get that and keep the script fast, format doesn't really matter. And if you can suggest a better way of doing this I'm open to that as well.

Answer 1

[Kelly Kearney]

If you only want to get the min and max of each row, and really want to avoid for loops, here's a solution. This gives the min and max for each row of A, ignoring values not in B, which is I think what you are looking for.

A = [...
  2  12 6  4  8
  1  3  5  7  9
];

B = [ 1 2 3 4 5 6 ];

tf = ismember(A, B);

Amax = A;
Amin = A;
Amax(~tf) = -Inf;
Amin(~tf) = Inf;

maxval = max(Amax, [], 2);
minval = min(Amin, [], 2);

results

>> maxval

maxval =

       6
       5

>> minval

minval =

       2
       1

[Rion Wilson-Yue]

Beautiful! Exactly what I needed, but couldn't seem to convey!

Thanks for your help everyone!

[Kelly Kearney]

Note that for really large matrices, a for loop over the rows may be more efficient than making two extra copies of the A matrix.

[Rion Wilson-Yue]

Interesting... If I don't care about preserving the data could I use NaN instead of +/- Inf and then just replace the elements in 'A'?

[Kelly Kearney]

Yes, that would work. I always forget that min/max ignore NaNs by default.

Answer 2

[Image Analyst]

How about a not-too-fancy but simple, intuitive, and easy to understand "for" loop?

A = randi(20, 2, 15) % Sample data
B = [ 1 2 3 4 5 6 ] % What we're looking for
C = NaN(size(A)); % Preallocate
% Check each row one at a time.
for row = 1 : size(A, 1)
  % Find intersection.
  B_in_A = intersect(A(row, :), B)
  % Stuff them into C
  C(row, 1:length(B_in_A)) = B_in_A;
end
% Display C
C

[Rion Wilson-Yue]

Sorry, I'm trying to keep the script as fast as possible; so I'm trying to avoid 'for' loops.

[Image Analyst]

You're funny. Do you know how fast or slow a for loop even is? I just did one hundred million (yes, 100,000,000) for loop iterations in 0.2 seconds. Just how fast do you need? How many iterations do you plan on doing? Trillions?

The for loop itself is rarely the reason why it's sometimes slightly slower - it's the way the memory is accessed.

[Rion Wilson-Yue]

Yes yes, I know how fast a for loop is. And yes odds are it would, in all probablity, be fine to use a for loop, and even then this smidgen of code would still not be the system's bottle neck. But I'm trying to move from C thinking to MATLAB thinking (using matrix math/functions), and would also like to see how fast I can make this run.

While the difference in time may be insignificant I'd still like to know if this can be done without a for loop. This is why I asked my question (see last line in original question).

[Image Analyst]

Not that I can think of off the top of my head. The problem is that you want your answers row-by-row, and each row has a different number of numbers, so all I can think of is a row-by-row solution.