Merge the contents of neighboring non-zero cells

Hi i have only one cell array and, what i would like to do is to merge the contents of neighboring non-zero cells.

So, in the figure all the contents of the groups with the blue outline will be "merged" into one cell.

i.e. in the final cell array in the col n.8 row n.6 it will be [9;9;9;9;1;1;.....9;2;1;1;2;1;......2;1;9;2;9;9;.....].

Thanks

Answer 1

[Image Analyst]

Why do you need to to this? I'm thinking you think you need to do this to accomplish some task, but you really don't. That's why I was wanting to know the use case -- what you're going to do once you have this.

It's possible but not easy. You'd have to get a map of what cells are non-zero, then label that map with bwlabel() to determine what cells are contiguous. Then you could identify the upper left cell (if that's what you want) of each region, and replace it with the concatenation of all the other cells inside that contiguous region. Then replace the other cells with 0 (I guess).

So it is possible but a bit of work so I need to know if it's really necessary.

Here's start, but let's hope you don't really need this:

[rows, columns] = size(ca); % Get size of the cell array variable, ca.
binaryImage = true(rows, columns);
for col = 1 : columns
    for row = 1 : rows
        cellContents = ca{row, col};
        if isequal(cellContents, 0)
            binaryImage(row, col) = false;
        end
    end
end
[labeledImage, numRegions] = bwlabel(binaryImage);
for k = 1 : numRegions
    thisRegion = ismember(labeledImage, k);
    [r, c] = find(thisRegion);
    % Collect contents of all cells in this region and concatenate them.
    % You do this.  Loop over all cells defined by r and c getting contents and concatenating them.
    % Then store result into ca(r(1), c(1));
    % You do this.  Something like ca{r(1), c(1)} = concatenatedResults.
    % Other cells set to zero if you want.
end

[LL]

Hi @Image Analyst, I'm working on a study concerning the microcalcifications of breast cancer, my goal is to show that a type of calcium called Whitlockite tends to form aggregates located on the edge of another form of calcium called hydroxyapatite.

So to do this I thought is best to look around the whitlokite pixel (defined by the number 2 in the cell array); leave out the "isolated" whitlockite pixel and study only the "aggregates" pixels.

I then created an array of cells with the district of this type of pixel ("aggregates"), and to know if they are at the border of hydroxyapatite (defined by the number 1 in the cell array) I thought of merging the individual contents of neighboring pixels into a single pixel containing the information of the whole whitlockite aggregate.

I don't know if I have been clear, but I really need a hand to be able to overcome this obstacle, maybe you have a better idea?

Thank you.

[Image Analyst]

I still don't think you're approaching this the right way. I'd use image analysis instead of this weird cell array way. What do the different labels (0, 1, 2, and 9) mean?

Anyway if you haven't finished the code and need me to do it, you'll have to attach your cell array in a .mat file so I have something to work with.

[LL]

I gladly accept advice :)

So, the number represent the component of the microcalcifications, in this case we have only 1 (Hydroxyapatite), 2(Whitlockite) and 9 (Non defined that is tissue, necrosis ..) but generally it's possibile to find also: 3(calcium carbonate), 7(calcite) and 8(calcium oxalate) but in this study the focus is on:

• the number of whitlockite aggregates
• the position of the aggregates with respect to hydroxyapatite(for this reason I had thought about merging the contents of the cells)

In the cell array the zeros represent anything that is not whitlockite, as the study is focused on this component.

Thank you so much!!!!

[Image Analyst]

I think it would be much easier to have up to 9 separate binary images that are maps of where that material is. Then you can see what materials are next to what other materials using functions like graycomatrix() or imdilate() and imreconstruct().

[LL]

@Image Analyst I talked to my boss and he prefers merging the cells, could you please help me?

[Image Analyst]

Do you want every cell in the contiguous region to have the same contents?

I'm preparing to leave for a trip to Washington DC tomorrow so I'm not sure how much time I'll have to finish the code. I'll be in DC through Tuesday and won't be able to answer many Answers posts during that time. In the last 5 hours, while you were working on completing the code I gave already, what do you have so far? Or did you not even bother trying?

[LL]

I tried to merge the contents of the contiguous cells and the result is this (contenuto.mat), i inialize contenuto like a zeros matrix(:,numAgg) ; I do not understand why it does not carry out the cycle on all 4 numAgg but only on the fourth.

Also, I can't find a way to have the contiguous cell contents automatically inserted into my cell array cont.

for k = 1 : numAgg

regioni = ismember(etichettaIm, k);

end

for k=1:numAgg

[r, c] = find(regioni);

for i=1:c

for y=1:r

cont=CIRC_agg(r(1),c(1)),CIRC_agg(r(2),c(2)),CIRC_agg(r(3),c(3)),CIRC_agg(r(4),c(4)),CIRC_agg(r(5),c(5))];

contenuto=cat(1,cont{:});

end

end

end

[Image Analyst]

You didn't answer my first question.

Tell me what you want in the cell array locations (6,8), (6,9), and (7,9).

Do all have the same concatenated vector? Or are some set to 0? Or are some set to null/empty/[]???

[LL]

In (6,8) i would like [9;9;9;9;1;1;.....9;2;1;1;2;1;......2;1;9;2;9;9;.....] so the merge of the three contigent cells, in the other two cells (6,9) and (7,9) only 0; the same practice for the other aggregate of non zero cells because they don't have the same concatenated vector.

[LL]

Hi @Image Analyst, I thought I could check the LabelIm array, especially for the first loop I initialize number = 1 then I check all cells = 1, find their rows and columns and get their value from CIRC_agg when I have all the contents of the contingent cells I have concatenated them and studied the contents to verify if the aggregates are internal, external or boundary of hydroxypatite.

I show you the code.

[etichettaIm, numAgg] = bwlabel(binaryIm);

contenuto=zeros();

numero=1

for k=1:size(etichettaIm)

regione=ismember(etichettaIm,numero);

[r, c]=find(regione);

for y=1:r

for i=1:c

%%%i would like to cycling but i don't know how i tried with cont=[CIRC_agg(r(y),c(i))]; without results

cont=[CIRC_agg(r(1),c(1)),CIRC_agg(r(2),c(2)),CIRC_agg(r(3),c(3))];

contenuto=cat(1, cont{:});

for j=1:length(contenuto)

idrox=contenuto(find(contenuto==2));

perc_idrox=(length(idrox)*100)/length(contenuto);

%%%%%%%%%%%%

%%%EXTERNAL%%%

%%%%%%%%%%%%

if length(idrox)==0

WHI_ext_agg=0;

WHI_ext_agg=WHI_ext_agg+1;

%%%%%%%%%%%%

%%%INTERNAL%%%

%%%%%%%%%%%%

elseif perc_idrox>65

WHI_int_agg=0;

WHI_int_agg=WHI_int_agg+1;

%%%%%%%%%%%%%%

%%%BOUNDARIES%%%

%%%%%%%%%%%%%%

else

WHI_bound_agg=0;

WHI_bound_agg= WHI_bound_agg+1;

end

numero=numero+1;

%numero=numero+1;

if numero>numAgg

fprintf('Non ci sono più aggregati da controllare')

end

end

end

end

end

Answer 2

[Awais Saeed]

Remove the cells that has 0 in it.

c = {25,0,'anyString','aChar',[12 5 48 7];0,0,'anyString','bChar',[10 01]}

c = 2×5 cell array
    {[25]}    {[0]}    {'anyString'}    {'aChar'}    {[12 5 48 7]}
    {[ 0]}    {[0]}    {'anyString'}    {'bChar'}    {[     10 1]}

nonzero_indices = cellfun(@(x) ~isequal(x, 0), c);
M = c(nonzero_indices)

M = 7×1 cell array
    {[       25]}
    {'anyString'}
    {'anyString'}
    {'aChar'    }
    {'bChar'    }
    {[12 5 48 7]}
    {[     10 1]}

[LL]

Sorry @Awais Saeed, I need that the final array has the same size of the initial one.

So when i find a cell non-zero, i need to check around this cell to find and merge the contents of the other non-zero cells and put inside these one the 0 or NaN.