counting data points
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I have a data set of 1's and 0's like this 000001111100000111110000011111000001111100000 this data set is called "c"
I need the program to output a count of sections so to speak. In this example they are all even and short but in my data set they are extremely long and of varying lengths. If I could get an output like 5,5,5,5,5,5,5 counting each section it would save me tons of time.
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Image Analyst
le 25 Avr 2012
If you have the "c" as an integer or logical array, and if you have the Image Processing Toolbox, you can do it in one line with regionprops(). Just simply say "measurements = regionprops(~c, 'Area')" Here's a full demo:
% Create sample data.
c = [0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0]
% Measure the zero sections - KEY LINE OF CODE RIGHT HERE!
measurements = regionprops(~c, 'Area')
% Display results.
fprintf('The number of zero sections is %d.\n', length(measurements));
fprintf('The sizes of zero sections are:\n');
allAreas = [measurements.Area]
Plus de réponses (2)
per isakson
le 24 Avr 2012
I learned this trick the other day:
str = '11110111000001111100000111110000011111000001111100000';
cac = regexp( str, '0+', 'split' );
n = cellfun( @(s) length(s), cac )
n =
4 3 5 5 5 5 0
You might want to check for leading and trailing 0. If the string is huge you might need to split it.
--- EDIT without the Image Processing Toolbox ---
c = round( rand( 1,100 ) + 0.1 ); % Sample data
str = sprintf( '%1u', c );
cac = regexp( str, '0+', 'split' );
len = cellfun( @(s) length(s), cac );
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Andrei Bobrov
le 25 Avr 2012
str = '11110111000001111100000111110000011111000001111100000';
x = str -'0';
k = find([true, diff(x)~=0]);
out = [x(k); diff([k;[k(2:end),numel(x)+1]])]
OR
out = diff([find([true, diff(x)~=0]) numel(x)+1]);
ADDED with use regexp
out = cellfun('length',regexp(str,'(0*|1*)','match'))
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