# Identifying regions in matrix rows

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I have an M x N matrix that contains the indices of columns where an external "validity criterion" is fulfilled, and -1 otherwise. Example:

matrix = zeros (5,8) -1;

matrix(1,1:3) = 1:3;

matrix(3,2:3) = 2:3;

matrix(3,5:7) = 5:7;

matrix(4,2:5) = 2:5;

matrix

Let's call the input above the "regions", i.e., here we have in total 4 regions (always considered row-wise). My objective is to identify, for each row, the largest region, and within that region the central value (rounded up). Currently, I do this as follows:

dummy = zeros(5,1) -1;

jumps = abs(diff([dummy, matrix, dummy], [], 2))

[borders_row, borders_col] = find(jumps>1);

That way, I identify the borders of the regions. Afterwards, I go through each row individually to identify the largest region, as I mentioned, which I do as follows:

first_idx = zeros(1,5);

second_idx = false(1,5);

for i = 1:5

borders_cur = borders_col(borders_row==i).';

ranges = diff(borders_cur);

ranges = ranges(1:2:end);

[width, path_idx] = max(ranges);

idxx = borders_cur(path_idx)-1 + ceil(width/2);

if ~isempty(idxx)

first_idx(i) = idxx;

second_idx(i) = logical(idxx);

end

end

For each row, I consider the current borders and first calculate the distance between them, so that I can judge which region is larger if there is more than one. I skip every second value because those indicate the "invalid spaces" in between the regions. I then pick the bigger region and calculate its central index, which I then record in two arrays (first_idx and second_idx).

I want to get rid of the loop as it's computationally expensive, and generally optimise the code as it's part of a bigger loop, and this is one of the slowest parts of the code currently. Any suggestions?

##### 4 Comments

### Accepted Answer

Bruno Luong
on 18 Sep 2023

Edited: Bruno Luong
on 19 Sep 2023

matrix = zeros (5,8) -1;

matrix(1,1:3) = 1:3;

matrix(3,2:3) = 2:3;

matrix(3,5:7) = 5:7;

matrix(4,2:5) = 2:5;

matrix

dummy = zeros(5,1) -1;

jumps = abs(diff([dummy, matrix, dummy], [], 2));

[borders_row, borders_col] = find(jumps>1);

first_idx = zeros(1,5);

second_idx = false(1,5);

for i = 1:5

borders_cur = borders_col(borders_row==i).';

ranges = diff(borders_cur);

ranges = ranges(1:2:end);

[width, path_idx] = max(ranges);

idxx = borders_cur(2*path_idx-1)-1 + ceil(width/2); % modified by BLU

if ~isempty(idxx)

first_idx(i) = idxx;

second_idx(i) = logical(idxx);

end

end

first_idx

% my method

A = (matrix>0).';

n = size(A,2);

f = false(1, n);

[i,j] = find(diff([f; A; f]));

c = i(1:2:end);

w = i(2:2:end)-c;

r = j(1:2:end);

[rw,is] = sortrows([r,w],[1 -2]);

keep = [true; diff(rw(:,1))>0];

rw = rw(keep,:);

c = c(is(keep));

r = rw(:,1);

w = rw(:,2);

mididx = c-1+ceil(w/2);

first_idx = accumarray(r, mididx, [n,1])'

second_idx = logical(first_idx);

##### 3 Comments

### More Answers (4)

Jon
on 18 Sep 2023

Edited: Jon
on 18 Sep 2023

I wasn't completely clear from your description what output you wanted, but I think this is what you were looking for. Still loops through rows of M, but maybe more efficient that what you had tried. Would have to time it to see.

% Build example input matrix

M = zeros (5,8) -1;

M(1,1:3) = 1:3;

M(3,2:3) = 2:3;

M(3,5:7) = 5:7;

M(4,2:5) = 2:5;

M

% Make matrix that has ones where there are values and 0 otherwise

V = M ~= -1;

% Loop through rows finding midpoint values of longest blocks

[m,n] = size(V);

midVals = NaN(m,1); % preallocate array to hold middle region values

for k = 1:m

% Find indices of beginning and end of each block of values

idx = [true,diff(V(k,:))~=0,true];

% Find length of each block

blklngth = diff(find(idx));

% Make a vector whose elements give length of the block that each

% element in original matrix belongs to

x = repelem(blklngth,blklngth).*V(k,:);

% Find the index of the middle (rounding up) of the longest block

% and assign middle value to output vector

maxBlkLngth = max(x);

if maxBlkLngth > 0

idxMid = round(mean(find(x == max(x))));

midVals(k) = M(k,idxMid);

end

end

midVals

##### 12 Comments

Jon
on 20 Sep 2023

Mathieu NOE
on 18 Sep 2023

hello

I tried to put some code together and ended with that ; maybe interesting (?)

it will detect the largest region for each row and store (and display - see red crosses ) the center position of each region

I haven't rounded these values so add it if you need it

matrix = zeros (5,8) -1;

matrix(1,1:3) = 1:3;

matrix(3,2:3) = 2:3;

matrix(3,5:7) = 5:7;

matrix(4,2:5) = 2:5;

matrix

%% main code

mat = (matrix>0); % creat logical array

for k = 1:5

[begin,ends] = find_start_end_group(mat(k,:));

ll = ends-begin+1;

if isempty(ll)

index_pos(k) = NaN;

else

if numel(ll)>1 % more than one region in this row

[v,id] = max(ll);

index_pos(k) = 0.5*(begin(id)+ends(id)); % % center position (index) of largest region

else

index_pos(k) = 0.5*(begin+ends); % % center position (index) of largest region

end

end

end

% plot

cmap = [0 0 0; parula(128)];

matrix(matrix<0) = NaN;

imagesc(matrix)

colormap(cmap);

caxis([-1 10]);

colorbar('vert');

hold on

for k = 1:5

plot(index_pos(k),k,'r+','markersize',25);

end

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

function [begin,ends] = find_start_end_group(ind)

% This locates the beginning /ending points of data groups

% Important : ind must be a LOGICAL array

D = diff([0;ind(:);0]);

begin = find(D == 1);

ends = find(D == -1) - 1;

end

##### 0 Comments

Fifteen12
on 18 Sep 2023

I've also put some work into this, but I'm unclear with what you want as your output, so I've got it to a good stopping place and I'll let you finish it with what you want to do. This script basically identifies the first and last index for every region in an inputted matrix using vectorized calls.

function [first, last] = findRegion(in)

valid = find((in + 1)');

neighbors = [0; valid];

regions = [valid; numel(in) + 1] - [-1; valid]; %Spaces between each valid index (first cell inflated for easier processing, last cell is handled later)

starting_indices = valid(regions > 1);

temp = logical([regions > 1; 0]);

ending_indices = valid(temp(2:end));

if (starting_indices(end)) == numel(in)

ending_indices(end+1) = numel(in);

end

%% Separate regions on different rows

rows = transpose(0:width(in):numel(in));

row_pairs = [floor((starting_indices - 1) ./ width(in)), floor((ending_indices - 1) ./ width(in))];

wrapped_regions = logical(row_pairs(:, 1) - row_pairs(:, 2));

first = sort([starting_indices; rows(wrapped_regions) + 1]);

last = sort([ending_indices; rows(wrapped_regions)]);

end

I wasn't sure if the largest region was the one with the most values or the one with the highest aggregated value, but I think you can probably calculate either from here.

##### 0 Comments

Image Analyst
on 19 Sep 2023

If you have the Image Processing Toolbox, it's pretty easy:

M = [...]

1 2 3 -1 -1 -1 -1 -1

-1 -1 -1 -1 -1 -1 -1 -1

-1 2 3 -1 5 6 7 -1

-1 2 3 4 5 -1 -1 -1

-1 -1 -1 -1 -1 -1 -1 -1]

[rows, columns] = size(M); % Get dimensions.

% Process each row one at a time.

for row = 1 : rows

mask = M(row, :) ~= -1;

props{row} = regionprops(mask, 'PixelIdxList')

% Now, not sure how to define the "central value rounded up"

end

So I'm just not sure how you define central value. What would be the central value of a vector like this: [3,6,1,9]?

##### 2 Comments

Bruno Luong
on 20 Sep 2023

Edited: Bruno Luong
on 20 Sep 2023

@Image Analyst 0.1-0.2 second is time with data of size 1000 x 10000 as bellow, not with OP toy example. Your code take 4 seconds which is more than 20 time slower than our codes (without tollbox).

M = repmat(1:1000, 10000, 1);

M(randi(end, 1000000, 1)) = -1;

tic

for row = 1 : size(M,1)

mask = M(row, :) ~= -1;

props{row} = regionprops(mask, 'PixelIdxList');

% Now, not sure how to define the "central value rounded up"

end

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