# Interpolate NaN on graph

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Alex Dimko on 27 Oct 2020 at 17:20
Commented: Star Strider on 28 Oct 2020 at 1:11
Hello,
I would like to interpolate the missing data on the graph marked by NaN.
when I try it it does not work.
the code is:
filteredData = data(:,2);
vector = [];
for index = 1:length(filteredData)
if filteredData(index) >= 2048
filteredData(index) = NaN;
vector = [vector index];
end
end
index2 = 2;
while index2 < length(filteredData)
if filteredData(index2) < filteredData(index2-1) - 750 && filteredData(index2) < filteredData(index2+1) - 750
vector = [vector index2];
filteredData(index2) = NaN;
index2 = index2 + 1;
end
if filteredData(index2) > filteredData(index2 - 1) + 750 && filteredData(index2) > filteredData(index2+1) + 750
vector = [vector index2];
filteredData(index2) = NaN;
index2 = index2 + 1;
end
index2 = index2 + 1;
end
vector = sort(vector);
y = interp1(1:length(filteredData),filteredData,vector, "linear");
plot(vector, y, '^r');
format long g;

Star Strider on 27 Oct 2020 at 17:37
If the data have NaN as the value of the dependent variable, and the independent variable is continuous (with no NaN values), then save the original independent variable as a vector, remove the entire row with NaN values from the data, and use the intact (original) independent variable to do the interpolation. The output of interp1 will be the interpolated value of the dependent variable, matching the values of the original independent variable.

Alex Dimko on 28 Oct 2020 at 0:20
by the way filteredData is only one column of data as only the second column was needed and not all 4.
Alex Dimko on 28 Oct 2020 at 1:05
YOU don't need to answer any of the questions. I messed around with the code didbsome searching for the meaning online and copied your code. Modified it for one column and it WORKS. I could not believe it. Thanks again. I will probably encounter more problems on this assignment so I will give you more questions. Thanks!
Star Strider on 28 Oct 2020 at 1:11
how do I delete NaN?
The NaN values are deleted in this assignment:
Dataq = Data(~nanrows,:);
The ‘nanrows’ variable is a logical vector that is true (or 1) where the row has a NaN value and is 0 otherwise. The tilde operator (~) negates that, so this assigns to ‘Dataq’ all the rows that do not have NaN values. See the documentation section on Matrix Indexing for a full explanation.
by the way filteredData is only one column of data as only the second column was needed and not all 4.
If you want to interpolate only the second column, just choose it and the first column (the independent variable) as the ‘Data’ matrix, and use the rest of my code as provided.

KSSV on 27 Oct 2020 at 17:24
Edited: KSSV on 27 Oct 2020 at 17:24

Alex Dimko on 27 Oct 2020 at 17:27
I am requiredto use interp1 by the assignment
KSSV on 27 Oct 2020 at 17:39
Then you have to use like below:
n = 100 ;
x = 1:n ;
t = rand(size(x)) ;
% make some values NaN to fill
y = t ;
idx = sort(randperm(n,20)) ;
y(idx) = NaN ;
% fill nan using interp1
xi = setdiff(x,idx) ;
yi = y(~isnan(y)) ;
y(idx) = interp1(xi,yi,idx) ;
plot(x,t,'r',x,y,'b')
Actually using random data for demo is not good. But you can follow the procedure.