# for loop in an array to substract the values of the array and find a specific value

2 views (last 30 days)
Jose Rego Terol on 1 Feb 2020
Commented: Image Analyst on 2 Feb 2020
I have an array of 200 values. This is the operation I need to program:
The value in the position X minus the value in the position Y EQUALS 0.55. Therefore, I am using a for loop and If statement.
The problem is that I know neither value nor the position of each value. I guess both of them are in the first 50 values of the array. So, how can I code the operation below n times (k-1, k-2, k-3 ...) until I got this number, and disp the 'Here'?
I;
for k =length (I)
x=I(k)-I %Take last value and substract the first value of I, second, k times to the last value.
if x==0.55
disp 'Here'
else
disp 'Not here'
end
end

Guillaume on 1 Feb 2020
Edited: Guillaume on 1 Feb 2020
If you are trying to find the indices X and Y for which I(X) - I(Y) is equal to a given values, this is easily done without a loop with:
[X, Y] = find(I - I.' == seachvalue); %I must be a vector
This will returrn all the XY pairs that match.

Show 1 older comment
Guillaume on 2 Feb 2020
"There is no value for X and Y."
Then there's no values in your array whose difference is exactly equal to the search value.
As pointed out by Image Analyst, if you're comparing floating points values that have been calculated differently or have been rounded then you can't use ==. Instead you must compare against an arbitrary tolerance.
It doesn't change the fact that it can all be done in one line without a loop:
tol = 1e-4; %it's up to you to choose a tolerance value that makes sense for the magnitude of the numbers you want to compare
[X, Y] = find(abs(I - I.' - seachvalue) <= tol)
Jose Rego Terol on 2 Feb 2020
Ok, now I understand. that's right, I do not need a loop.
[X, Y] = find(abs(I - I.' - seachvalue) <= tol)
With tol is hard to find the value. Is there any way to round the result to the nearest value in the vector?
Thanks
Image Analyst on 2 Feb 2020
round() can round the number(s) to any number of decimal points that you want.

Subhadeep Koley on 1 Feb 2020
Try this.
for k = 2:length(I)
x = I(k) - I(k-1);
if x == 0.55
disp('Here');
else
disp 'Not here'
end
end

#### 1 Comment

Jose Rego Terol on 1 Feb 2020
Is not here, but let me ask you what is it doing?
k = 2:length(I)
From the second value to the last.
x = I(k) - I(k-1);
x= the second value - the first one, then the third value - the second one. Is that right?
What happen if the operation is,i.e., the value at the 40 % of the length of the array minus the 20 % of the length of the array? How can I code, try first the sequence you said, then take the 3rd value (k=3:length(I)). When finish, take the third value and substract the first value.
I think it is completely insane for Matlab.

Image Analyst on 1 Feb 2020
Edited: Image Analyst on 1 Feb 2020
First of all read the FAQ : Click here to learn why you shouldn't use == to compare floating point values. You should use ismembertol():
x = zeros(size(I));
for k = 1 : length(I)
% Take last value and substract the first value of I, second, k times to the last value.
% I have no idea what the above means but let's subtract I(1) and see
x(k) = I(k) - I(1);
if ismembertol(x(k), 0.55, 0.004)
fprintf(' ----> Found a match at index %d where I(%d) = %f and x = %f.\n', k, k, I(k), x);
break; % Let's quit when we find a match.
else
fprintf('Found no match at index %d where I(%d) = %f and x = %f.\n', k, k, I(k), x(k));
end
end
histogram(x);

Image Analyst on 1 Feb 2020
Why use a loop? To subtract all values from I(1) and get a new vector with those differences, do
IDiff = I(1) - I;
Then to see which are closer than 0.004 (or whatever tolerance you want) to 0.55, do
indexes = find(abs(IDiff - 0.55) > 0.004);
I don't see how any loop is needed.
Jose Rego Terol on 1 Feb 2020
What happen if I want to do this operation n times?
IDiff = I(1) - I;
IDiff = I(2) - I;
IDiff = I(3) - I;
IDiff = I(4) - I;
IDiff = I(5) - I;
IDiff = I(N) - I;
I need a loop for this reason.
Jose Rego Terol on 1 Feb 2020
I want to continue this operation taking the next value in the case the result is not equal to 0.55.
Do you understand?