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

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Jose Rego Terol 2020-2-1

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评论： Image Analyst 2020-2-2

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

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Answer 1

Guillaume 2020-2-1

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编辑：Guillaume 2020-2-1

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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.

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Jose Rego Terol 2020-2-2

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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 2020-2-2

round() can round the number(s) to any number of decimal points that you want.

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Answer 2

Subhadeep Koley 2020-2-1

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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

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Jose Rego Terol 2020-2-1

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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.

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Answer 3

Image Analyst 2020-2-1

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编辑：Image Analyst 2020-2-1

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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);

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Jose Rego Terol 2020-2-1

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Hi Image Analyst,

You did not undestand this line

% Take last value and substract the first value of I, second, k times to the last value.

Let me explain it with the code you provided me:

for k = 1 : length(I)	
        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);

Here, the first value is subtracted from the first value. If the condition is not true (ismembertol x ~= 0.55), the second value is subtracted from the first value. If the condition is not true (ismembertol x ~= 0.55), the third value is subtracted from the first value. This rutine is run N times until ismembertol x = 0.55, right?

This is what I tried to explain in this comment. Unfortunaly, I did not explain properly. My fault.

Now, let's back to the code.

In this first round, the value was not found. Therefore, I want to check this operation but using the second value:

x(k) = I(k) - I(2);

The point is that I do not know which value is the one that x=0.55. There are 200 values. Can't change it manually.

I was wondering how the code can go back to x(k) and take I(2) if there was no match with I(1):

n=1
for k = 1 : length(I)	
        x(k) = I(k) - I(n);
	  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
   %If there is no match, go back to x(k) and take I(2). Do the same operation (I(n+1)) 
   % until x(k) matches 0.55.

Jose Rego Terol 2020-2-1

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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 2020-2-1

https://www.youtube.com/watch?v=VQ3vg9DSHxo

I want to continue this operation taking the next value in the case the result is not equal to 0.55.

Do you understand?

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

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