I am comparing a value to an array and trying to find out the match. But it is not working though there is a match

function maximumValue = maximumvalue(voltageValues,actTimeStamp, timeStamp,thresholdRowValue)  
actTimestamprow = 0;  
maximumValue=0;  
n=0;  
zeroCrossingTime=0;  
cycle=0;  
analysisArray = [0];  
thresholdRowValue = thresholdRowValue - 1;  
if mod(thresholdRowValue,2) == 0  
    cycle = -1;   
else  
    cycle = 1;  
end  
zeroCrossingTime = timeStamp(thresholdRowValue);  
for i = 1:size(actTimeStamp)  
    if  actTimeStamp(i) = zeroCrossingTime  
        actTimestamprow = i  
    else  
         break;  
    end
end

In this I have a row which matches the value. But the variable actTimestamprow displays zero.

the threshold value is a value passed from another function ( works fine). the actTimeStamp, volatgeValues and timestamp are arrays. Can you tell me why. Any help would be appreciated.

Answer 1

[Image Analyst]

If you're comparing floating point numbers, and not integers, you have to use a tolerance, as described in the FAQ: http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F

And, even if they're integers, the line

if  actTimeStamp(i) = zeroCrossingTime

is the same as

if  actTimeStamp(i)

Why? Well you know if you type a=b, it will spit out the value of a to the command window, because "a" is the result/answer or "ans" as MATLAB calls it. And if zeroCrossingTime is zero, then the if will evaluate to false whereas if zeroCrossingTime is anything other than zero, the if will evaluate as true. Probably not what you want. You probably want

if  abs(actTimeStamp(i) - zeroCrossingTime) < someToleranceValue

[Snehalatha]

Hey this is close to what i want. thank you for the idea. may be this might wor. I'll tell you. Once again thanks for the idea.

[Snehalatha]

This worked. The problem was with comparing floating point numbers. Now it identifies and stops at the cell where there is a match.

Answer 2

[Jan]

Obviously your assumption, that you have "a row which matches the value" is wrong. Perhaps this is the typical floating point problem?

Please format you code properly and provide some input values, such that the problem can be reproduced. Be more precise, because "I have a row" is obvious for you, but the readers do not know, which variable you are talking of.

Are you sure you want to break the "for i" loop, when the first element is not matching?

Did you use the debugger to step through your code line by line, such that you can inspect the values of the variables?

[Snehalatha]

Hi Thanks for telling me the break. I think i didn't notice it. But i removed and recheck it and it doesn't work. Also I ran the code line by line. The problem seems to be in the comparison. my actTimestamprow is always = 0 i.e no match.

Both are double values and I rounded them off to 4 digits. I am attaching an excel with just these 2 columns here so that you can have a look at the values. if u see row B1380 and A118247 match but somehow the comparison doesn't work. i removed the break to display an error msg now to see what happens. It runs through my array and shows the error msg.

Answer 3

[Michael Haderlein]

for i = 1:size(actTimeStamp)

size returns the size of both dimensions. I guess actTimeStamp is something like [1 x N], let's say N=10 (doesn't matter here). In this case, size(actTimeStamp] = [1 10]. If you use this as upper limit of the loop iterator, the first value of the array will be the maximum:

for cnt=1:[4 50]
end
>> cnt

cnt =

       4

You need to use either size(actTimeStamp,2) to address the second dimension (10 in the upper example) or something like length() or numel().

[Snehalatha]

hi it is a (N x1) so i need to access only the first dimension and when i checked it accesses the first dimension and runs through all the rows.

[Michael Haderlein]

Ok, different point: are you sure your code even runs? The line

 if  actTimeStamp(i) = zeroCrossingTime

should immediately produce an error as the equality-check operator is ==, not =.

[Guillaume]

Most people (and probably you as well) don't know the behaviour of the colon operator when given non-scalar inputs.

Save yourself some future trouble and be explicit on the dimension you need, particularly as it's only 2 more characters to write:

for i = 1:size(actTimeStamp, 1)

Your code (this part at least) work for now. In the future you may change your input and this'll be the source of a hard to find bug.

[Snehalatha]

@Michael: I corrected that. It's right in the code but when i was editing it here it must have got deleted but i have used "==" .

When i run the code it runs the line but somehow doesn't recognise the match.

@Guillaumne: this was the first thing i tried when my code didnt work yesterday but it doesn't seem to help. The code runs but doesn't recognise the match.