count 1's in binary

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Shasha Glow
Shasha Glow 2014-7-22
评论： Guillaume 2018-2-16
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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采纳的回答

Image Analyst
Image Analyst 2014-7-22
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.
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Joseph Cheng
Joseph Cheng 2014-7-22
编辑：Joseph Cheng 2014-7-22
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).

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更多回答（3 个）

Wayne King
Wayne King 2014-7-22
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{:}
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Shasha Glow
Shasha Glow 2014-7-22
编辑：Shasha Glow 2014-7-22
hi wayne.. actually m not working on short array.... it is very lengthy array which contain more than 5000 binary numbers... so i am not able to run this code because its only work on short arrays... do you have any other idea to count the 1's?

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Laszlo
Laszlo 2016-12-14
编辑：Laszlo 2016-12-14
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 个评论
Image Analyst
Image Analyst 2016-12-14
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)

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Anshuk Uppal
Anshuk Uppal 2018-2-16
编辑：Walter Roberson 2018-2-16
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));
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显示 4更早的评论
Anshuk Uppal
Anshuk Uppal 2018-2-16
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...
Guillaume
Guillaume 2018-2-16
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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