Count the occurence of a number in between other numbers

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Ashane Fernando
Ashane Fernando am 12 Feb. 2019
Kommentiert: Jos (10584) am 22 Feb. 2019
I need to extract the number of zeros of the vector highlighted below.
x = [ 1 0 0 1 1 1 0 1 1 0 0 0 0 0 1 1 0 1 0]
My expected outcome is,
y = [2 1 5 1 1]
I need to get the number of 0s in between the 1s. I tried several "for" and "while" loops. But wasnt able to get this result.
Appreciate if anyone can help me out.

Akzeptierte Antwort

Stephen23
Stephen23 am 22 Feb. 2019
Bearbeitet: Stephen23 am 22 Feb. 2019
This is simpler and actually works for all horizontal vectors (unlike the accepted answer):
>> x = [ 1 0 0 1 1 1 0 1 1 0 0 0 0 0 1 1 0 1 0];
>> D = diff([false,x==0,false]);
>> find(D<0)-find(D>0)
ans =
2 1 5 1 1
For a slightly faster version you can call find once:
>> F = find(diff([false,x==0,false]));
>> F(2:2:end)-F(1:2:end)
ans =
2 1 5 1 1
EDIT: uses Jan's logical vector suggestion.
  5 Kommentare
Stephen23
Stephen23 am 22 Feb. 2019
"...like I stated those cases would be satisfied with one simple if condition ."
Please show how that would work.
Ashane Fernando
Ashane Fernando am 22 Feb. 2019
Hi Stephen,
Thank you for pointing out the limitation of the previously accepted answer.
I think your answer looks very straight forward and easy.

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Weitere Antworten (4)

Jan
Jan am 21 Feb. 2019
Another solution:
x = [ 1 0 0 1 1 1 0 1 1 0 0 0 0 0 1 1 0 1 0];
y = [1, x, 1];
strfind(y, [0,1]) - strfind(y, [1,0])

Jan
Jan am 12 Feb. 2019
Bearbeitet: Jan am 12 Feb. 2019
[B, N] = RunLength(x);
Result = N(B == 0)
If you do not have a C-compiler installed, use RunLength_M from the same submission.
Without RunLength:
x = [ 1 0 0 1 1 1 0 1 1 0 0 0 0 0 1 1 0 1 0]
d = [true, diff(x) ~= 0]; % TRUE if values change
b = x(d); % Elements without repetitions
n = diff(find([d, true])); % Number of repetitions
Result = n(b == 0)

madhan ravi
madhan ravi am 12 Feb. 2019
Bearbeitet: madhan ravi am 12 Feb. 2019
index=find(x==0); % edited after Jan’s comment
idx=find(diff(index)~=1);
R=[idx(1) diff(idx) numel(index)-idx(end)]
  13 Kommentare
Stephen23
Stephen23 am 22 Feb. 2019
Bearbeitet: Stephen23 am 22 Feb. 2019
"Just a simple if condition would satisfy those needs."
There are several problems with that approach:
  1. How do you know that your IF covers all of the special cases? Can you prove that you have thought of every special case?
  2. Extra complexity makes code harder to understand and maintain, and is more libale to bugs.
  3. It just shifts the problem: okay, so you can detect a special case, but then how do you deal with it? Solution: by writing code that handles that special case... in which case, you still have to solve the same problem.
  4. It really would not be that simple. Make a list of the cases for which your code fails: how would you deal with them using one IF statement?
madhan ravi
madhan ravi am 22 Feb. 2019
Yes it turns out lots of conditions are required :D.

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Jan
Jan am 22 Feb. 2019
Bearbeitet: Jan am 22 Feb. 2019
See Jos solution (link):
D = diff(find(diff([false, x==0, false])));
R = D(1:2:end);
A short test (yes, I know that timeit is more accurate):
A = ones(1, 1e5);
A(randperm(1e5, 5e4)) = 0;
tic % My strfind method
for k = 1:100
D = [false, A==0, false];
R = strfind(D, [true, false]) - strfind(D, [false,true]);
end
toc % 0.182 seconds
tic % My general RunLength code:
for k = 1:100
d = [true, diff(A) ~= 0];
b = A(d);
n = diff(find([d, true]));
Result = n(b == 0);
end
toc % 0.220 seconds
tic % Stephen's 2 FIND
for k = 1:100
D = diff([0, A==0, 0]);
R = find(D<0) - find(D>0);
end
toc % 0.190 seconds
tic % Stephen's 1 FIND (with modification)
for k = 1:100
D = find(diff([false, A==0, false])); % 10% faster than [0, A==0, 0] !
R = D(2:2:end) - D(1:2:end);
end
toc % 0.161 seconds, was 0.172 with [0, A==0, 0]
tic % Jos
for k = 1:100
D = diff(find(diff([false, A==0, false])));
R = D(1:2:end);
end
toc % 0.161 seconds
[false, A==0, false] creates a logical vector, while [0, A==0, 0] is a double vector, which is 8 times larger.
I'm going to create a C-MEX function for this task, which avoids the time consuming explicit creation of [false, A==0, false].
  1 Kommentar
Jos (10584)
Jos (10584) am 22 Feb. 2019
Nice, Jan!
I think this shows the power of the matlab community: we all can learn from it! :-)

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