Separating vectors into individual subsets for plotting
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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.
Thanks in advance.
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Azzi Abdelmalek
am 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
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Weitere Antworten (4)
Azzi Abdelmalek
am 5 Aug. 2013
a=[10 20 30 NaN 30 20 10 NaN 10 20 30]
a(isnan(a))=[]
out=reshape(a,3,[])
2 Kommentare
Jan
am 5 Aug. 2013
Bearbeitet: Jan
am 5 Aug. 2013
@Donald: Please post comments in the comment section.
[EDITED, relocated comment] Donald John wrote:
Thanks for this - however in reality my vector has an unknown number of separate ranges. How would I go about separating this data if I do not know how many individual ranges from 10 to 30 there are?
Also I would ideally like to output the result as separate vectors.
Azzi Abdelmalek
am 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
am 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;
3 Kommentare
Donald John
am 5 Aug. 2013
2 Kommentare
Jan
am 5 Aug. 2013
Please post new questions in a new thread. Otherwise the method to accept an answer is not useful anymore.
You would plot the separated data in a FOR loop and yes, the shown method can be applied to matrices also with tiny modifications. Just try it and post again if problems occur.
Andrei Bobrov
am 5 Aug. 2013
Bearbeitet: Andrei Bobrov
am 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});
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