Compare two vector arrays and if they match set vector array 2 = nan?

So i have two arrays the same size:

time = [0,0,0,0,2,3,4,5,5,6,7,8]';

solution = [1,2,0,0,1,1,2,2,2,0,0,0];

What would be the best way to compare the two and if both are the same make solution 0 = nan;

So the final outcome i would like is solution = [1,2,nan,nan,1,1,2,2,2,0,0,0];

would i use a function such as:

if strfind(time,solution)==0

Do something?

Thanks in adavnce!

Answer 1

[Walter Roberson]

time =  [0,0,0,0,2,3,4,5,5,6,7,8]

time = 1×12

0 0 0 0 2 3 4 5 5 6 7 8

solution = [1,2,0,0,1,1,2,2,2,0,0,0]

solution = 1×12

1 2 0 0 1 1 2 2 2 0 0 0

outcome = solution;
outcome(solution == 0 & time == 0) = nan

outcome = 1×12

1 2 NaN NaN 1 1 2 2 2 0 0 0

It is not clear what you want to have happen in positions where the values are equal, but the values are not 0.

[MKM]

Thanks for this. Sorry, I only wanted the solution's '0s' to = nan WHEN the positions of time & solution both = 0, else ignore and leave the solution's 0 untouched. Any other position that are equal i.e. when time is 2 and solution is 2 should not be changed.

Would this work if it was in a statement:

if outcome(solution == 0 & time == 0)

outcome == 0 = nan;

Thanks

[MKM]

if any(time==0 & solution == 0)

solution(time ==0 & solution==0)=nan;

else

return

end

Used this solution in the end, so the solution is - update where time and solution 0s are equal and if not return to script

[Walter Roberson]

solution(time ==0 & solution==0) = nan;
if ~any(isnan(solution)); return; end

Answer 2

[Arif Hoq]

Try this...

time =  [0,0,0,0,2,3,4,5,5,6,7,8];
solution = [1,2,0,0,1,1,2,2,2,0,0,0];
[idx]=find(time==solution);
solution(idx)=[NaN NaN]

solution = 1×12

1 2 NaN NaN 1 1 2 2 2 0 0 0

[Stephen23]

FIND is not required, logical indexing is simpler and more efficient.

It would be better to allocate a scalar NaN.

[Arif Hoq]

@Stephen Thank you very much for the clue