Label segments based on percentage

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Uerm 2019-12-12

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https://ww2.mathworks.cn/matlabcentral/answers/496388-label-segments-based-on-percentage

评论： Uerm 2019-12-23

AL_128.mat

Hi,

I have a variable (AL_128 which I have attached). It is a 1x48 cell where each cell is a Ax20 matrix. The matrices contain the values 0 and 1. For instance, cell 1 is a 225x20 matrix which should be interpret as follows: 225 segments where each segment is of length 20.

What I want to do here is to say that if 30% of the segments contain the label 1, the whole segment should be labelled 1. Therefore, cell 1 will end up being a 225x1 vector. This should be done for all segments and all cells.

How can I do that?

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Answer 1

Guillaume 2019-12-18

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编辑：Guillaume 2019-12-18

Much simpler code:

threshold = 0.3;  %30% threshold
M = cellfun(@(m) mean(m, 1) >= threshold, AL_128, 'UniformOutput', false)

Or if you really want to use a loop over the cell array:

M = cell(size(AL_128));
for cidx = 1:numel(AL_128)
    M{cidx} = mean(AL_128{cidx}, 1) >= threshold;
end

There is certainly no need to loop over the columns of the matrix.

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Uerm 2019-12-23

Thanks a lot, both of you!

In my case it would be

threshold = 0.3;  %30% threshold
M = cellfun(@(m) mean(m, 2) >= threshold, AL_128, 'UniformOutput', false)

as I do the calculation along the rows.

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Answer 2

Image Analyst 2019-12-13

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Your data didn't seem to have any segments where the value was 1, at least in the several cells I looked in, but I think this should work and be pretty easy to follow and understand.

s = load('AL_128.mat')
AL_128 = s.AL_128
% Define output cell array
output = cell(size(AL_128));
for k = 1 : length(AL_128)
	% Get the contents of this cell
	thisCellsContents = AL_128{k}; % This is an N rows - by - 20 columns matrix.
	[rows, columns] = size(thisCellsContents); % Get the number of rows and columns in this matrix.
	% See which rows contain a 1 in any column
	containsA1 = any(thisCellsContents == 1, 2); % This is a N row vector.
	% See if the number of them is 30% or more of the rows
	fractionContaining1 = sum(containsA1) / rows;
	if fractionContaining1 >= 0.3
		output{k} = ones(rows, 1); % Make vector of all 1's
	else
		output{k} = thisCellsContents; % Just the original contents.
	end
end
% Overwrite input, if desired.
AL_128 = output

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Image Analyst 2019-12-15

So "Each row is considered a segment. If a segment contains at least 30% 1's, then the whole segment should be 1." meaning the whole row should be 1, meaning the matrix is the same size, just that certain rows (where more than 30% of the values are 1) should be all 1's.

That's not what I did. I made a column vectors, except in those cases where not more than 30% were 1's and in that case I left the matrix alone. But you also said "should end as a 225x1 vector". So I think your statements are contradictory and I don't know whether you want segments/rows to be set to all 1 in the existing matrix, or if you want a vector where only some elements are 1 (where the row is 30% or more with 1's). I also don't know what you want to do with the existing matrices where the 30% criteria is not met. I just left those alone, but is that what you want?

If you put in an fprintf statement

s = load('AL_128.mat')
AL_128 = s.AL_128
% Define output cell array
output = cell(size(AL_128));
for k = 1 : length(AL_128)
	% Get the contents of this cell
	thisCellsContents = AL_128{k}; % This is an N rows - by - 20 columns matrix.
	[rows, columns] = size(thisCellsContents); % Get the number of rows and columns in this matrix.
	% See which rows contain a 1 in any column
	containsA1 = any(thisCellsContents == 1, 2); % This is a N row vector.
	% See if the number of them is 30% or more of the rows
	fractionContaining1 = sum(containsA1) / rows;
	if fractionContaining1 >= 0.3
		fprintf('Setting cell #%d to a %d-row column vector\n', k, rows);
		output{k} = ones(rows, 1); % Make vector of all 1's
	else
		output{k} = thisCellsContents; % Just the original contents.
	end
end
% Overwrite input, if desired.
AL_128 = output

then you'll see

Setting cell #25 to a 186-row column vector

Setting cell #26 to a 210-row column vector

Setting cell #27 to a 277-row column vector

Setting cell #32 to a 261-row column vector

Setting cell #38 to a 213-row column vector

Setting cell #40 to a 239-row column vector

so I made those cells into vectors.

If on the other hand you want to replace the matrix in each cell with a vector where it's one if more than 30% in that particular row are 1, and zero where less than 30% are 1, then run this code:

s = load('AL_128.mat')
AL_128 = s.AL_128
% Define output cell array
output = cell(size(AL_128));
for k = 1 : length(AL_128)
	% Get the contents of this cell
	thisCellsContents = AL_128{k}; % This is an N rows - by - 20 columns matrix.
	[rows, columns] = size(thisCellsContents); % Get the number of rows and columns in this matrix.
	% See which rows contain a 1 in any column
	num1sPerRow = sum(thisCellsContents == 1, 2); % This is a N row vector.
	% See if the number of them is 30% or more of the columns.
	fractionContaining1 = num1sPerRow / columns;
	if fractionContaining1 >= 0.3
		fprintf('Setting cell #%d to a %d-row column vector\n', k, rows);
		vec = zeros(rows, 1); % Initialize.
		vec(fractionContaining1 > 0.3 * columns) = 1; % Set certain rows to 1.
		output{k} = vec; % Make vector of 0's and 1's
	else
		output{k} = thisCellsContents; % Just the original contents.
	end
end
% Overwrite input, if desired.
% AL_128 = output

You'll see

Setting cell #32 to a 261-row column vector

Setting cell #40 to a 239-row column vector

But again, I'm not sure if you want matrices or vectors. You've said it both ways. And I've given two ways of doing it. If neither is what you want then you'll just have to give an example and explain it. And have some cases where the matrices are a mix of 0s and 1s, not all 0 or 1 like you have now.

Image Analyst 2019-12-16

OK, the picture helped. Try this:

s = load('AL_128.mat')
AL_128 = s.AL_128
% Define output cell array
output = cell(size(AL_128));
for k = 1 : length(AL_128)
	% Get the contents of this cell
	thisCellsContents = AL_128{k}; % This is an N rows - by - 20 columns matrix.
	[rows, columns] = size(thisCellsContents); % Get the number of rows and columns in this matrix.
	% See which rows contain a 1 in any column
	num1sPerRow = sum(thisCellsContents == 1, 2); % This is a N row vector.
	% See if the number of them is 30% or more of the columns.
	fractionContaining1 = num1sPerRow / columns;
	fprintf('The highest fraction of ones any row in matrix #%d has is %.2f.\n', k, max(fractionContaining1));
	% Say what rows need to be set to 1 and which need to be set to zero.
	indexesToSetTo1 = fractionContaining1 > 0.3;
	if any(indexesToSetTo1)
		fprintf('Setting cell #%d to a %d-row column vector\n', k, rows);
		output{k} = indexesToSetTo1; % Make vector of 0's and 1's
	else
		fprintf('Leaving cell #%d alone.\n', k);
		output{k} = thisCellsContents; % Just the original contents.
	end
end
% Overwrite input, if desired.
% AL_128 = output

Uerm 2019-12-18

Thanks a lot! I found that this works too:

function M = threshold(AL_128,n)
for i = 1:length(AL_128)
    for k = 1:size(AL_128{1,i},1)
        M{i}(k,:) = double(mean(AL_128{1,i}(k,:))>=(n/100));
    end
end