Replace elements of a matrix by a smaller matrix which contains circular shaped values margins

Hello,

I'm having problems to efficiently replace some values of a result matrix.

The process is taking place ~50k times, so its very time intensive (result matrix: m,n > 1000).

Each time a much smaller matrix replaces a corresponding area in the result matrix.

The problem is, that the values in the small matrix form "the shape of a circle" and represent the presence of an attribute.

Like this:

A = result matrix (in reality: much bigger):

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

1 1 1 1 1

B = small matrix (in reality: more elements in matrix, so that the value margins form the shape of a circle):

0 0 0 0

0 5 5 0

0 5 5 0

0 0 0 0

By performing "A(1:4, 1:4) = B" I get the result C:

0 0 0 0 1

0 5 5 0 1

0 5 5 0 1

0 0 0 0 1

1 1 1 1 1

I want the result:

1 1 1 1 1

1 5 5 1 1

1 5 5 1 1

1 1 1 1 1

1 1 1 1 1

I managed to get the result by first deleting the specific area of the result to "0" (only the ones that are not "0" in the small matrix "B") and then adding only the values of the small matrix that are not "0" by using two for-loops element-wise (50k times).

Unfortunately this takes very long....

Is there a more efficient way to do so?

Thank you very much in advance!

Answer 1

[Bruno Luong]

C = A;
sB = size(B);
idx = {1+(1:sB(1)),1+(1:sB(2))}; % adjust offset 1 and 1 if B to be moved at other place
C(idx{:}) = B

or depends on desired result

C(idx{:})= max(C(idx{:}), B)

Answer 2

[KALYAN ACHARJYA]

A=[1 1 1 1 1;1 1 1 1 1;1 1 1 1 1;1 1 1 1 1;1 1 1 1 1];
B=[0 0 0 0;0 5 5 0;0 5 5 0;0 0 0 0];
[r c]=find(B>1);
A(r,c)=B(r,c)

Result:

A =
     1     1     1     1     1
     1     5     5     1     1
     1     5     5     1     1
     1     1     1     1     1
     1     1     1     1     1

[Image Analyst]

Krassor's "Answer" to Kalyan moved here (to be a "Comment") since it's not an answer to the original question.

Thank you for your answer!

At first that looked exactly like what I wanted...

But when implementing, the result didn't seem right:

For example:

A=[1 1 1 1 1 1 1;1 1 1 1 1 1 1;1 1 1 1 1 1 1;1 1 1 1 1 1 1;1 1 1 1 1 1 1; 1 1 1 1 1 1 1;1 1 1 1 1 1 1];
B=[0 5 0; 5 5 5;0 5 0]
[r, c]=find(B>1)
A(1+r,1+c)=B(r,c)
B(r,c)

results in:

B =

   0   5   0
   5   5   5
   0   5   0

r =

   2
   1
   2
   3
   2

c =

   1
   2
   2
   2
   3

A =

   1   1   1   1   1   1   1
   1   0   5   0   1   1   1
   1   5   5   5   1   1   1
   1   0   5   0   1   1   1
   1   1   1   1   1   1   1
   1   1   1   1   1   1   1
   1   1   1   1   1   1   1

B =

   5   5   5   5   5
   0   5   5   5   0
   5   5   5   5   5
   0   5   5   5   0
   5   5   5   5   5

As you can see, matrix A contains "0"s again.

I also don't understand the output r, c of the find-function.

Neither the result of B(r, c) and why A replaces the original "B" matrix (3x3) and not the B(r,c) (5x5)...

I'm very confused right now :)

Answer 3

[Krassor]

@Bruno Luong:

By doing this, I just replace the area in "A" by "B" with all the "0"s...

This is what I try to avoid.

I tried the following:

A=[1 1 1 1 1;1 1 1 1 1;1 1 1 1 1;1 1 1 1 1;1 1 1 1 1];
B=[0 5 0; 5 5 5;0 5 0]

[r c]=find(B)

A(r(1):r(size(r)), c(1):c(size(c)))=B(r(1):r(size(r)), c(1):c(size(c)))

Result:

B =

   0   5   0
   5   5   5
   0   5   0

r =

   2
   1
   2
   3
   2

c =

   1
   2
   2
   2
   3

A =

   1   1   1   1   1
   5   5   5   1   1
   1   1   1   1   1
   1   1   1   1   1
   1   1   1   1   1

Basically I think the "find"-approach is the right one... If I replace each element of the "rows" and "columns"-vector one by one (for example: A(r(1), c(1)) = B(r(1), c(1)) ... ) the result is perfect.

Only I want to avoid using a loop.

[Bruno Luong]

replace the last stetement of my code above with

C(1+(1:sB(1)),1+(1:sB(2))) = max(C(1+(1:sB(1)),1+(1:sB(2))), B)

;

people who suggests you using FIND did not care about what it actually do. It obviously actually do something very wrong about indexing.

[Krassor]

Thank you very much, that almost got the desired result.

In fact, it's not completely what I wanted, but for my purpose maybe even better.

The simple example with "A" and "B" is not telling the whole story.

In reality "A" can contain results from previous iterations.

These values can be greater than the values in "B".

Originally I wanted to replace all values in "A" with the "newest" ones from "B".

Now I replace the values in "A" with the biggest ones (across all iterations) from "B".

In my case this is an alternative interpretation and might be at least as correct as my original intention.

Anyway, just for my interest, is there a solution for the original problem?

[Bruno Luong]

But the first version is literally replacing sub-region A with B, inluding 0s.

If you want replacing only at the places where B > 0 you need to tell us (otherwise I consider there is nothing special to differentiate value 0 or 5 in B).

[Krassor]

Your last solution is replacing at the places where B > A.

What I want is, replacing at the places where B ~= 0.

You are right, I didn't specify that in first place....

The replacement of the places where B = 0 is not wanted.

[Bruno Luong]

[i,j,b] = find(B);
A(sub2ind(size(A),i+1,j+1)) = b

[Krassor]

That looks like the solution!

Perfect!

Thank you! :)