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cem
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Counting zeros in array

Asked by cem
on 6 May 2019
Latest activity Commented on by cem
on 7 May 2019
Hi everyone,
I have an array such as;
input = [1 -2 -1 -1 -1 0 0 -1 0 0 0 3 0 0 4 0 0 0 0 0]
By counting zeros and determining the value after zero, I want to create a new dimentional array such as;
output = [0 1; 0 -2; 0 -1; 0 -1; 0 -1; 2 -1; 3 3; 2 4; 4 0]
Each row of "output" array determines that [number of zeros before non zero element non zero element].
For example, [2 4] represents that there are 2 zeros before "4".
How can I create the "output" array based on this rule?

  1 Comment

Just try a for loop.

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2 Answers

Answer by Adam Danz
on 6 May 2019
Edited by Adam Danz
on 7 May 2019
 Accepted Answer

Loop method
input = [1 -2 -1 -1 -1 0 0 -1 0 0 0 3 0 0 4 0 0 0 0 0];
output = nan(numel(input),2);
for i = 1:numel(input)
if input(i)==0
continue
end
output(i,:) = [max(cumsum(input(1:i)==0)),input(i)];
input(1:i) = 1; %make sure all previous 0s are overwritten
end
% if input ended in 0, count the consecutive 0s minus 1 (which matches the example)
if input(end)==0
output(end,:) = [sum(input==0)-1,0];
end
% get rid of leftover output rows
output(isnan(output(:,1)),:) = [];
Without a loop
inputTemp = [1,input(1:end-1),1]; %make sure last digit is non-zero (for now)
cs = cumsum(inputTemp==0); %cumulative sum of 0-counts
zeroCounts = diff(cs(inputTemp~=0)); %count consecutive zeros
nonZeros = [input(input~=0 & 1:numel(input)<numel(input)),input(end)];
output = [zeroCounts', nonZeros'];
Result for both methods
output =
0 1
0 -2
0 -1
0 -1
0 -1
2 -1
3 3
2 4
4 0

  5 Comments

Thank you for both solutions, Elapsed time of first solution takes 0.009498 seconds, and the other version without loop code takes 0.007459 seconds at the same computer.
What are you going to do with all of that extra time? :D
Readability is always important, too. Especially if other people will work with your code some day. That being said, I'm not sure which of my proposals is more readable.
I will take a Couple of coffee at extra time;-) In my opinion loop solution nötr readable than second one. Thank you again.

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Answer by Stephen Cobeldick on 7 May 2019
Edited by Stephen Cobeldick on 7 May 2019

Simpler:
>> V = [1,-2,-1,-1,-1,0,0,-1,0,0,0,3,0,0,4,0,0,0,0,0];
>> X = find([V(1:end-1),1]);
>> Z = [diff([1,X+1])-1;V(X)].'
Z =
0 1
0 -2
0 -1
0 -1
0 -1
2 -1
3 3
2 4
4 0

  3 Comments

Nice. I was certain there was a simpler solution but gave in to time!
@Adam Danz: thank you. Together diff and find are great for these kind of things, but there appears to be no shortcut: I just sit and puzzle them out the hard way...
Waow that was super! Thank you Stephen!

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