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Label segments based on percentage

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Uerm
Uerm 2019 年 12 月 12 日
コメント済み: Uerm 2019 年 12 月 23 日
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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Guillaume
Guillaume 2019 年 12 月 18 日
編集済み: Guillaume 2019 年 12 月 18 日
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
Uerm 2019 年 12 月 23 日
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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Image Analyst
Image Analyst 2019 年 12 月 13 日
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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Uerm
Uerm 2019 年 12 月 15 日
Unfortunately, the second chunk did not do that. I attach the original variable (AL_128) and the result of running mode(AL_128,2), which is called M.mat. As you can see, all the cells end being vectors containing 0 and 1. Please look at cell 25-26 as they contain a lot of 1's and 0's.
Does it make sense?
When the criteria is met, the label is 1. If not, the label is 0.
I have also attached an image for better understanding. The image shows an example where the segment length is 10.
Image Analyst
Image Analyst 2019 年 12 月 16 日
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
Uerm 2019 年 12 月 18 日
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

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