replacing the NAN with value
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hi. i have 2 matrices V & I. some elements of I are zero. then R=V/I returns unavailable values. i need to keep the size of 3 matrices constant. so what do you suggest me for replacing NAN values? does it make sense to replace zeros with positive minimum value of I? thank you for your help
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Answers (4)
Image Analyst
on 27 Jan 2015
Study this snippet. Adapt as needed.
m=[1,3,nan, 5,7,nan,2]
% Find the nans
nanLocations = isnan(m)
% Pick some other value to set the nans to.
alternateValue = 42;
% Do the replacemenet.
m(nanLocations) = alternateValue
In the command window:
m =
1 3 NaN 5 7 NaN 2
nanLocations =
0 0 1 0 0 1 0
m =
1 3 42 5 7 42 2
2 Comments
Image Analyst
on 27 Jan 2015
Well then I certainly wouldn't know. Maybe you just want to do bilinear or bicubic interpolation from surrounding good areas?
Star Strider
on 27 Jan 2015
The NaN values are the result of a 0/0 division, so the same elements of ‘V’ and ‘I’ are both zero in those situations. That is an indeterminate result.
I do not know all of what you are doing, but since the value of ‘R’ (the resistor you are measuring the voltage drop across) will likely not change, I would replace the NaN values with whatever the average value for ‘R’ is in your other calculations.
3 Comments
Star Strider
on 27 Jan 2015
I can’t make any sense out of it.
I have no idea how you generated those data. I also have no idea how you had a zero voltage drop across your resistor with a finite current going through it, unless the resistor became a short circuit, or finite voltage with zero current, unless your resistor became an open circuit.
I completely fail to understand the first column, ‘X’, or whether it is important in understanding the rest of your data.
Something is definitely not correct in whatever you are doing. You probably need to perform your experiment again.
Stephen23
on 27 Jan 2015
What is wrong with keeping the NaN values? Why would you want to get rid of them?
NaN's tell you very important information, and it would be programmatically very poor practice to hide a divide-by-zero with some fake data.
Do not replace them with "some positive minimum value".
If later during your data processing you must use a function that does not accept NaN's (e.g. spline fitting), then you can deal with it then in the most appropriate way for the specific situation. You might need an error message, or to remove them using isnan... but if you have replaced the values with "some positive minimum value" then you have lost information about your data that you cannot retrieve later.
4 Comments
Guillaume
on 27 Jan 2015
Edited: Guillaume
on 27 Jan 2015
Looking at your excel file, I think the first problem is that excel #DIV/0! can mean several things. It can mean NaN which is what matlab interprets it as, or it could mean +Inf or -Inf, which it is in your case.
From a physical perspective, when your switch is open, the resistance is infinite (hence current is 0). So the first thing, I'd do in matlab is recalculate the resistance in matlab. If you just import your excel file with
m = xlsread('42.xlsx');
then
m(:, 5) = m(:, 2) ./ m(:, 3); %calculate resistance
No NaN anymore, but +Inf and -Inf, which is physically correct. That may not help you for further processing, but when you say 'I just want to replace them with an appropriate value', I'm not sure what is more appropriate than infinite. You could replace them with a very big value (but how big is big?), but that would be less correct:
m(~isfinite(m(:, 5), 5) = someverybigvalue; %replace inf by very big R
4 Comments
Guillaume
on 28 Jan 2015
Edited: Guillaume
on 28 Jan 2015
You could always trim all your matrices to the size of the smallest one. Eliminate data at random from the longer ones.
Assuming your 100 matrices are in cell array c:
%c contain matrices, with NaN or Inf in the fifth column
c = cellfun(@(m) m(isfinite(m(:, 5)), :), c, 'UniformOutput', false); %remove NaN/Inf
minheight = min(cellfun(@(m) size(m, 1), c); %get height of smallest matrix
c = cellfun(@(m) m(sort(randperm(size(m, 1), minheight)), :), c, 'UniformOutput', false); %reduce all matrices to the same size by minheight random rows in each
Otherwise you could do some interpolation as Image Analyst suggest. However, that would just be inventing data that does not exists.
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