Counting zeros in array

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?

[Image Analyst]

Just try a for loop.

Answer 1

[Adam Danz]

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]

I really thank you, it works perfectly. Also I wonder is there any solution way without using loop?

[Adam Danz]

I updated my answer with a non-loop method. What makes this tricky is that the last number in your example is a zero but it's counted as if it were not a zero. That's built in to both solutions but makes the method a little more klunky, but functional.

[cem]

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]

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]

I will take a Couple of coffee at extra time;-) In my opinion loop solution nötr readable than second one. Thank you again.

Answer 2

[Stephen23]

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]

Nice. I was certain there was a simpler solution but gave in to time!

[Stephen23]

@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]

Waow that was super! Thank you Stephen!