count 1's in binary
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Shasha Glow
il 22 Lug 2014
Commentato: Guillaume
il 16 Feb 2018
Hi,
This is what i want... I have a binary array
001111000000011100000000011111
from here i have to count the number 1 in such way
result: 0,4,0,3,0,5.... how to get this?
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Image Analyst
il 22 Lug 2014
If you have the Image Processing Toolbox, it's just two real lines of code, a call to regionprops and a line to extract the lengths from what regionprops returns.
% Create sample binary data.
binaryArray = [0 0 1 1 1 1 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1]
% Measure lengths of stretches of 1's.
measurements = regionprops(logical(binaryArray), 'Area');
% Convert from structure to simple array of lengths.
allLengths = [measurements.Area]
% If you want 0's in between for some reason:
out = zeros(1, 2*length(allLengths)+1);
out(2:2:end) = allLengths
5000 numbers is no problem. This code can handle millions of them.
1 Commento
Joseph Cheng
il 22 Lug 2014
Modificato: Joseph Cheng
il 22 Lug 2014
if you don't have the Image Processing Toolbox (or those who find this post trying to do something similar) you can do something like this:
Zs = randi(10,1,10)+1; %number of zeros in a row.
Os = randi(10,1,10)+1; %number of ones in a row.
s = [];
for ind = 1:10
s = [s ones(1,Os(ind)) zeros(1,Zs(ind))]
end
s = strtrim(num2str(s')')
s = ['0' s '0'] %start the values with something you know.
b_bin = logical(s(:)'-'0') %
ds = diff(b_bin) %coincidentally diff of s would work as 1 and 0 strings are 1 value apart.
result = diff(find(abs(ds)==1)); %find transitions
result(2:2:end) = 0 %since forced the start of the array to zero we know the odd indexes are consecutive 1 and even indexes are consecutive 0's (ignoring the leading and trailing zeros).
Più risposte (3)
Wayne King
il 22 Lug 2014
Hi Sasha, I'm presuming your binary number is a character array:
s = '001111000000011100000000011111';
K1 = strfind(s,'1');
F = diff(find([1 diff(K1 - (1:length(K1)))]));
splitvec = mat2cell(K1,1,[F length(K1)-sum(F)]);
NumConsec1 = cellfun(@numel,splitvec);
NumConsec1 gives you the number of consecutive 1's. splitvec is a cell array with the actual indices of those ones, whicy you can see if you enter
splitvec{:}
Laszlo
il 14 Dic 2016
Modificato: Laszlo
il 14 Dic 2016
What about this:
series_length=find(diff(binary_series)==-1)-find(diff(binary_series)==1);
you might have to pad binary_series with 0s at the start and end to ensure switch on and off.
1 Commento
Image Analyst
il 14 Dic 2016
Does not work:
% Create sample binary data.
binary_series = [0 0 1 1 1 1 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 1 1 1 1 1]
% Laslo's code below:
series_length=find(diff(binary_series)==-1)-find(diff(binary_series)==1)
Complete error message:
Matrix dimensions must agree.
Error in test3 (line 4)
series_length=find(diff(binary_series)==-1)-find(diff(binary_series)==1)
Anshuk Uppal
il 16 Feb 2018
Modificato: Walter Roberson
il 16 Feb 2018
A working tested algorithm -
n_ofErrors=flip(find(diff(error_vector)==-1))(1:1:length(find(diff(error_vector)==1))) - flip(find(diff(error_vector)==1));
6 Commenti
Anshuk Uppal
il 16 Feb 2018
You can make it work by truncating the array first and not using the whole expression in a single line. That should solve the error matlab generates. An algorithm is a series of instructions(may be mathematical) that solve a problem. Differences in syntax can occur...
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/187845/image.png)
Guillaume
il 16 Feb 2018
Well, yes. And you can make your algorithm a lot more efficient by performing diff and find only once rather than 3 times each. I also don't understand the purpose of the flip.
transitions = find(diff([0; error_vector(:); 0]));
n_ofErrors = transitions(2:2:end) - transitions(1:2:end)
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