Effacer les filtres
Effacer les filtres

counting data points

9 vues (au cours des 30 derniers jours)
Sonia Wiemann
Sonia Wiemann le 24 Avr 2012
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.
  1 commentaire
Andrei Bobrov
Andrei Bobrov le 25 Avr 2012
question a duplicate

Connectez-vous pour commenter.

Réponse acceptée

Image Analyst
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]
  1 commentaire
Sonia Wiemann
Sonia Wiemann le 26 Avr 2012
Thanks a lot! This will save loads of time counting thousands of zeros!

Connectez-vous pour commenter.

Plus de réponses (2)

per isakson
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 );

Andrei Bobrov
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'))

Catégories

En savoir plus sur Data Type Conversion dans Help Center et File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by