Identifying time for prolong changes in a sequence

Hello

Please I have a sequence for instance

010001100000111110000100011001

And I want to identify the ones that are more than 2(onces) and the time

for this example is 5 seconds

Also another example is when I have more than one occurances of 1 greater than 2

000001110001111001000110001

Then the time are [ 3sec 4sec]

I have used the code A = seconds(x)

It only shows me a total number of 1 in seconds.

Is there a way I can get about this.

Thanks for your help in advance

Answer 1

[Image Analyst]

Try this (requires the Image Processing Toolbox):

binarySequence = [0 1 0 0 0 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 0 0 0 1 1 0 0 1];
props = regionprops(logical(binarySequence), 'Area');
allLengths = [props.Area]
% Count number of regions longer than some number:
maxLength = max(allLengths)
for k = 1 : maxLength
	longerThanN = sum(allLengths == k);
	fprintf('There are %d sequences of %d.\n', longerThanN, k);
end

You get:

There are 3 sequences of 1.

There are 2 sequences of 2.

There are 0 sequences of 3.

There are 0 sequences of 4.

There are 1 sequences of 5.

[Tino]

Thanks a lot for your time and effort.

What if I want an sequence of the output. For instance if I want to list the (There are 1 sequences of 5) to become

000000000000111110000000000000

or

There are 3 sequences of 1 to become

010000000000000000000100000001

thanks for your time in advance

[Image Analyst]

You can use bwlabel and ismember. Write back if you can't figure it out.

[Tino]

Hi

I cannot figure out the bwlabel and ismember.

Can you assist me on this if possible?

Thanks

[Image Analyst]

Actually you can just ask regionprops() for the locations:

binarySequence = [0 1 0 0 0 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 0 0 0 1 1 0 0 1];
% Label the data
props = regionprops(logical(binarySequence), 'Area', 'PixelIdxList');
allLengths = [props.Area]
% Count number of regions longer than some number:
maxLength = max(allLengths)
for k = 1 : maxLength
	% Get the labels for this length
	labels = allLengths == k;
	longerThanN = sum(labels);
	fprintf('There are %d sequences of %d.\n', longerThanN, k);
	% Get 1's wherever the run has a length of k.
	locations = zeros(size(binarySequence));
	for k2 = 1 : length(labels)
		if labels(k2)
			locations(props(k2).PixelIdxList) = 1;
		end
	end
	% Report to the command window:
	locations
end

If you want locations in a 2-D array, you can do this:

binarySequence = [0 1 0 0 0 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 0 0 0 1 1 0 0 1];
% Label the data
props = regionprops(logical(binarySequence), 'Area', 'PixelIdxList');
allLengths = [props.Area]
% Count number of regions longer than some number:
maxLength = max(allLengths)
locations = zeros(maxLength, length(binarySequence));
for k = 1 : maxLength
	% Get the labels for this length
	labels = allLengths == k;
	longerThanN = sum(labels);
	fprintf('There are %d sequences of %d.\n', longerThanN, k);
	% Get 1's wherever the run has a length of k.
	for k2 = 1 : length(labels)
		if labels(k2)
			locations(k, props(k2).PixelIdxList) = 1;
		end
	end
end
% Report to the command window:
locations

and you get

locations =

Columns 1 through 23

0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0

0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0

Columns 24 through 30

0 0 0 0 0 0 1

0 0 1 1 0 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0

0 0 0 0 0 0 0

Note the length is the row number. So row 2 has 1's wherever those locations are in a run of 2 digits long, namely 1's in columns 6 & 7, and 26 & 27.

[Tino]

Thank you very much sir.

Very grateful.

Another question if it is possible? what of identifying numbers that have conditions

for instance if 1000011111000001111000000110000000111100

if 1 > 3

The answer becomes

0000001111100000111100000000000000111100

Thank you for your help so far

[Tino]

Another question if it is possible? what of identifying numbers that have conditions

for instance if 1000011111000001111000000110000000111100

if 1 > 3

The answer becomes

0000001111100000111100000000000000111100

Thank you for your help so far

[Image Analyst]

binarySequence2 = bwareaopen(binarySequence, 4); % Keep regions >3, meaning 4 or longer.

[Tino]

Thank you very much

[Tino]

Hi Image analyst.

If I wanted only onces to show (and there group occurences in the sequence) in the result instead of both 1 and zero

for instance if I have 1111100001111000000111111

I will like the results to be

There are 1 sequences of 4 = second occurences

There are 2 sequences of 5 = first and third occurences

I will appreciate it if you answer the question for me as you ve done in the past

Best wishes

[Image Analyst]

Try this:

binarySequence = [0 1 0 0 0 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 1 0 0 0 1 1 0 0 1]
% Measure the data
props = regionprops(logical(binarySequence), 'Area', 'PixelIdxList');
allLengths = [props.Area]
% Count number of regions longer than some number:
maxLength = max(allLengths)
for k = 1 : maxLength
	% Get the labels for this length
	labels = allLengths == k;
	longerThanN = sum(labels);
	fprintf('There are %d sequences of %d.\n', longerThanN, k);
	indexes{k} = find(labels);
	for k2 = 1 : length(indexes{k})
		fprintf('    Sequence %d\n', indexes{k}(k2));
	end
	% Get 1's wherever the run has a length of k.
	locations = zeros(size(binarySequence));
	for k2 = 1 : length(labels)
		if labels(k2)
			locations(props(k2).PixelIdxList) = 1;
		end
	end
	% Report to the command window:
	%locations
end

There are 3 sequences of 1.

Sequence 1

Sequence 4

Sequence 6

There are 2 sequences of 2.

Sequence 2

Sequence 5

There are 0 sequences of 3.

There are 0 sequences of 4.

There are 1 sequences of 5.

Sequence 3