Separating vectors into individual subsets for plotting

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.

Answer 1

[Azzi Abdelmalek]

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

Answer 2

[Azzi Abdelmalek]

a=[10 20 30 NaN 30 20 10 NaN 10 20 30]
a(isnan(a))=[]
out=reshape(a,3,[])

[Jan]

@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]

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)

Answer 3

[nah]

    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]

Do not forget the pre-allocation:

idx = [1, find(isnan(a)), length(a) + 1];
n   = length(idx) - 1;
V   = cell(1, n);
for ix = 1:n
  V{ix} = a(idx(ix)+1:idx(ix + 1));  % EDITED, "aa" -> "a"
end

What should happen, if the last element of a is NaN?

[Azzi Abdelmalek]

What is aa?

[Jan]

It is a typo and "a" is meant. This might be an effect of the 33 C in my office.

Answer 4

[Donald John]

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

[Jan]

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.

[Donald John]

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?

Answer 5

[Andrei Bobrov]

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});