# Counting zeros in array

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cem on 6 May 2019
Commented: 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

Image Analyst on 6 May 2019
Just try a for loop.

Adam Danz on 6 May 2019
Edited: Adam Danz on 7 May 2019
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

cem on 7 May 2019
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.
Adam Danz on 7 May 2019
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.
cem on 7 May 2019
I will take a Couple of coffee at extra time;-) In my opinion loop solution nötr readable than second one. Thank you again.

Stephen Cobeldick on 7 May 2019
Edited: 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

Adam Danz on 7 May 2019
Nice. I was certain there was a simpler solution but gave in to time!
Stephen Cobeldick on 7 May 2019
@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...
cem on 7 May 2019
Waow that was super! Thank you Stephen!