Counting zeros in array

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cem
cem il 6 Mag 2019
Commentato: cem il 7 Mag 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?

Risposta accettata

Adam Danz
Adam Danz il 6 Mag 2019
Modificato: Adam Danz il 7 Mag 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 Commenti
Adam Danz
Adam Danz il 7 Mag 2019
Modificato: Adam Danz il 7 Mag 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
cem il 7 Mag 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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Più risposte (1)

Stephen23
Stephen23 il 7 Mag 2019
Modificato: Stephen23 il 7 Mag 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 Commenti
Stephen23
Stephen23 il 7 Mag 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
cem il 7 Mag 2019
Waow that was super! Thank you Stephen!

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