For loop with two conditions

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Saad Alqahtani 2021-7-5

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评论： Saad Alqahtani 2021-7-6

Hi,

I'm trying to run this code where it should stop as soon as one of the conditions has met and give me the index where it stops.

So, here is my code and it's taking so long to run and it also didn't stop utill the end which I'm sure one of those conditions should have been met far away from the end.

isOK1= (Discard2value<=target_value_30_P);
isOK2 =(Discard2value>=t2);
for N = datan(1:end)
	while 1
		if N == (Discard1value) || ((isOK1 && isOK2))
			J = N;
			break
		end
	end
end

I don't know what's missing? Any help would be appreciate it.

Also, I keep getting this error : "array indices must be positive integers or logical values"

Thanks in advance.

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Answer 1

DGM 2021-7-5

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编辑：DGM 2021-7-5

Considering all the comparisons that are being made between unknown variables, I'm not eager to guess at what should be expected. I doubt that any of these loops are necessary, given that all the test values appear to be scalar logicals defined outside the loops. In fact, since nothing changes within the loops, the inner loop serves no purpose and will either pass trivially, or it will get stuck. I also assume that it's unlikely that (presumably) floating point values in datan are exactly 1 or 0, so it's probably never going to do anything but get stuck.

If that doesn't reveal the problem, include enough supporting example data that the problem can be reproduced, or provide enough of the surrounding code that the intent is unambiguous.

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Saad Alqahtani 2021-7-6

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Thanks for the help. I really appreciate it.

So, I'll post the rest of my code so you can have a better idea. My apologies, I just don't want to overwelm you with a long code.

here it is:

MVC= 100; % is a tourque value in Newton Meter 
target_value= 0.30*MVC;
target_lowest_point = 0.70*target_value; % 70 percent of the target value
Discard1= find(datan <=(target_lowest_point),1); % Find the target_lowest_point as soon as it's been found
% datan is the tourque performance in Newton Meter 
Discard1value = (datan (Discard1)); % get the datan value that falls bellow 70% of the target value (first Condition)
target_value_30_Percent = 0.30*target_value; 
Discard2= find(datan <= target_value_30_Percent,1); %Find the target_value_30_Percent as soon as it's been found
Discard2value = (datan (Discard2)); % get the datan value that falls bellow 30% of the target as soon as it's been found (part1 of the second condition) 
Discard2_to_end = ((datan( Discard2 :end))); 
t1 =(3*2048); % I have this just to get at least 3 seconds as soon as the tourque falls bellow 30% of the target value so I can check if it lasts for more than 2 seconds
t2= (2*2048); 
Discard3lasts = ((Discard2_to_end(1:t1))); % Here where it check inside this array if the tourque falls bellow 30% of the target value that lasts for more than 2 seconds (part2 of the second condition) 
isOK1= (Discard2value<=target_value_30_Percent); % Check for part1 of the second condition
isOK2 =(Discard2value>=t2); %Check for part2 of the second condition
for N = datan(1:end)
   while 1
  if N == (Discard1value) || ((isOK1 && isOK2))
     J = N;
       break
  end
   end
end 
Discard3 = find(datan == J);  % I'm planning to only discard anything beyond J 
Discard_length3= length(datan(Discard3:end)); % anything beyond this point is pretty easy for me to do
Discard_length3_in_sec = (Discard_length3)/2048 ;
forceChoppped = datan(1:end-(Discard_length3_in_sec*2048));
timechopped = (0:length(Discard_length3_in_sec)-1)/fsamp;
%the rest is mostly plotting 

Please feel free to ask me for clarification or to modifiy any of the steps. Also, if you have a better way than the For loop approach, I'd really appreciate it. Thanks again.

DGM 2021-7-6

编辑：DGM 2021-7-6

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I did some things at it.

% don't know what datan range or size is
% this is just a test vector
datan = 100*rand(10000,1);
MVC= 100;
target_value= 0.30*MVC;
target_lowest_point = 0.70*target_value;
target_value_30_Percent = 0.30*target_value; % ... why is this even a variable?  
% it literally takes up more space and is less clear than the code itself
% magic numbers that depend on some array size or resolution factor 
% and will otherwise cause the code to break
% is fsamp = 2048? if so, use it.
t1 = 3*2048;
t2 = 2*2048; 
% this could be compacted and clarified, but eh.
Discard1 = find(datan <= target_lowest_point,1);
Discard1value = datan(Discard1);
Discard2 = find(datan <= target_value_30_Percent,1);
Discard2value = datan(Discard2);
Discard2_to_end = datan(Discard2:end); 
Discard3lasts = Discard2_to_end(1:t1);
isOK1 = Discard2value<=target_value_30_Percent;
isOK2 = Discard2value>=t2;
% i have no idea what this means, but it should do the same thing as the loop
% or rather, it does what i assume the loop was _supposed_ to do
Jidx = (datan == Discard1value) | (isOK1 && isOK2);
Discard3 = find(Jidx);
Discard_length3 = length(datan(Discard3:end)); 
Discard_length3_in_sec = (Discard_length3)/2048 ;
forceChoppped = datan(1:end-(Discard_length3_in_sec*2048));
timechopped = (0:length(Discard_length3_in_sec)-1)/fsamp; % let me guess.  fsamp is 2048?  
% if so, you should make it a parameter and use it

I only deleted the comments to make my own remarks visible. I'm not suggesting you remove yours. Having plenty of comments is good.

Saad Alqahtani 2021-7-6

This works perfect. Thank you so much!

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Answer 2

Paul Hoffrichter 2021-7-5

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>> this code where it should stop as soon as one of the conditions has met

But the break just gets you out of the while loop when one of the two conditions is met. You need to add a test after the while loop to see if you broke out, in which case you can add another break do get out of the for-loop.

But within the while loop, here are your constants:

N (is being tested, but not changing)
Discard1value (is being tested, but not changing)
isOK1, isOK2 (are being tested but not changing)

Within the for-loop, only one variable N changes.

If the first time you hit the if N... test, you have a false condition, then you will never break out - you will be in an infinite loop since all the variables in the if test are constant within the body of the if-statement.

isOK1= (Discard2value<=target_value_30_P);
isOK2 =(Discard2value>=t2);
for N = datan(1:end)
   while 1
      if N == (Discard1value) || ((isOK1 && isOK2))
         J = N;
         break
      end
   end % END while
end % END for

>> it's taking so long to run and it also didn't stop utill the end

Since you are not reporting an infinite loop, then one of the conditions in the if-test must be true the first time you hit it. If the isOK test true, then it remains true for all values of N.

I advise you to put a breakpoint at the if-test, and step through a few loops to better understand your two loops.

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Image Analyst 2021-7-6

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This would be some more robust code, which breaks out of both the for loop and the while loop as well as having a failsafe to prevent an infinite loop

isOK1= (Discard2value<=target_value_30_P);
isOK2 =(Discard2value>=t2);
abort = false;
loopCounter = 1; % Failsafe
maxIterations = 1000; % A lot more than you EVER expect to have.
for N = datan(1:end)
	while ~abort && loopCounter < maxIterations
		if N == (Discard1value) || ((isOK1 && isOK2))
			J = N;
			abort = true;
			break; % out of while loop
		end
		% Now update conditions N, Discard1value, isOK1, and/or isOK2 somehow
		% or else it will never hit the break.
		% TODO : update code, then increment the loop counter.
		loopCounter = loopCounter + 1; % Failsafe
	end
	% You get here when the while loop breaks OR
	% if you hit the max number of iterations.
	if abort
		break; % Out of for loop.
	end
end

Saad Alqahtani 2021-7-6

Thanks for the help.

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For loop with two conditions

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