# find() not as expected

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Mike D. on 13 Jul 2019
Commented: Star Strider on 13 Jul 2019
I have a numeric 2D array called A with time in the first column (from zero to 2 minutes every 0.01 seconds). I want to find the rows that match times = 10 : 10 : 100;
When I type find(10==A(:,1), I get 1001
When I type find(20==A(:,1), I get 2001
Why when I type find(times==A(:,1) does it give these values: [1001, 12957, 24913, 36869, ...]?
These values of times (which are 10, 20, 30, ...) are at rows [1001, 2001, 3001, 4001, ...], so that's what I expected.
I know I don't really need a find() in most situations because I can use indexing, but that doesn't work either:
plot(A(A(:,1)==times,3),A(A(:,1)==times,2),'rs')
gives error msg: The logical indices in position 1 contain a true value outside of the array bounds.
because A(:,1)==times gives value of 12957 for the second index which is outside of array A bounds.
##### 2 CommentsShowHide 1 older comment
Image Analyst on 13 Jul 2019
Are they integers or floating point?
See the FAQ for how to test for equality with doubles.

Star Strider on 13 Jul 2019
You have encountered ‘automatic implicit expansion’ introduced in R2016b. The result you get from your equality expressions will be an (Nx10) array, where ‘N’ is the row size of ‘A’ (720001 in my reconstruction here) and 10 is the size of ‘times’.
The only way to get the result you want are either to compare each value of ‘times’ in a loop, or more efficiently to use the accumarray function.
Try this:
A = [0:0.01:7200; rand(size(0:0.01:7200))]'; % Create Matrix
times = 10:10:100; % Create Vector
Out = accumarray((1:numel(times))', times, [], @(x){A(A(:,1)==x,:)})
Reveal = cell2mat(Out)
With my random matrix, this gave the correct results, when I checked them against the actual rows of ‘A’.
You can then plot them as:
figure
plot(Reveal(:,1), Reveal(:,2),'rs')
grid
I named the double matrix ‘Reveal’ to explain what it is. Name it anything you want.
Star Strider on 13 Jul 2019
My pleasure.

R2019a

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