Counting zeros in array

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

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Accepted Answer

Adam Danz
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
  5 Comments
cem
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.

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More Answers (1)

Stephen
Stephen on 7 May 2019
Edited: Stephen 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
cem
cem on 7 May 2019
Waow that was super! Thank you Stephen!

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