# Separating vectors into individual subsets for plotting

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Donald John on 5 Aug 2013
I have a vector with a series of data values from 0 to 30. I want to split the vector into separate parts so that each data range from 0 to 30 is seperated and subsequently I can plot the seperate data series. How would I go about doing this?
My vector looks like this:
10 20 30 NaN 30 20 10 NaN 10 20 30
Its on the basis of the NaN that I want to split the vector.

Azzi Abdelmalek on 5 Aug 2013
Using a for loop is faster
if ~isnan(a(end))
a(end+1)=nan;
end
idx2=find(isnan(a));
idx1=[1 idx2(1:end-1)+1];
n=numel(idx1);
out2=cell(n,1);
for k=1:n
out2{k}=a(idx1(k):idx2(k)-1);
end

Azzi Abdelmalek on 5 Aug 2013
a=[10 20 30 NaN 30 20 10 NaN 10 20 30]
a(isnan(a))=[]
out=reshape(a,3,[])
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Azzi Abdelmalek on 5 Aug 2013
a=[10 20 30 NaN 30 20 10 NaN 10 20 30]
if ~isnan(a(end))
a(end+1)=nan
end
idx2=find(isnan(a))
idx1=[1 idx2(1:end-1)+1]
out=arrayfun(@(ii1,ii2) a(ii1:ii2),idx1,idx2-1,'un',0)

nah on 5 Aug 2013
a=[10 20 30 NaN 30 20 10 NaN 10 20 30 0 10 20 30 NaN 30 20 10 NaN 10 10 ]
aaNaNidx = find(isnan(a));
for ix = 2:length(aaNaNidx)
ll = aaNaNidx(ix-1)+1: aaNaNidx(ix);
splitVect{ix,1} = aa(ll);
end
splitVects{1,1} = aa(1:aaNaNidx(1));
splitVects{end+1,1} = aa(aaNaNidx(end)+1:length(aa));
Since you are saying that the vector will be unevenly split, it needs to be stored in an different data structure called Cells;
Jan on 5 Aug 2013
It is a typo and "a" is meant. This might be an effect of the 33 C in my office.

Donald John on 5 Aug 2013
Okay, I have two further questions.
1. Is it possible to plot the new cell structure and how would I do this?
2. Can I do something similar if my data is in fact in a matrix structure, such as below?
650 10
700 20
750 30
NaN NaN
630 30
640 20
650 10
NaN NaN
740 10
720 20
715 30
NaN NaN
700 30
etc etc
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Donald John on 5 Aug 2013
I thought that this followed on from my previous question. To be honest I am still struglling to get this to work, especially now that I am using matrices. Do you have any further advice?

Andrei Bobrov on 5 Aug 2013
Edited: Andrei Bobrov on 5 Aug 2013
z = [nan nan nan nan 1 3 5 nan nan 3 4 nan 6 0 3 nan nan]';
z = z(:);
if ~isnan(z(1)), z = [nan; z]; end;
z1 = z(find(~isnan(z),1,'first')-1:end);
a = zeros(size(z1));
b = ~isnan(z1(:));
a(strfind(~b(:)',[1 0])) = 1;
a1 = cumsum(a);
out = accumarray(a1(b),z1(b),[],@(x){x});