Parsing a vector

Hi,

Given a vectors say a = [1 3 5 NaN 4 5 NaN 8 9] I would like to create new vectors of the following form b = [1 3 5]; c = [4 5]; d = [8 9];

Thanks for your help,

-n

Answer 1

[Jan]

It would not be a good idea to create 'b', 'c', 'd', ... automatically. Better create a cell array:

a = [1 3 5 NaN 4 5 NaN 8 9];
a = [NaN, a, NaN];
sepInd = find(isnan(a));
nSep = length(sepInd) - 1;
b = cell(1, nSep);
for i = 1:nSep
  b{i} = a(sepInd(i) + 1:sepInd(i + 1) - 1);
end

Answer 2

[Paulo Silva]

a = [1 3 5 NaN 4 5 NaN 8 9]
b=[0 find(isnan(a)) numel(a)+1]
c=arrayfun(@(x)a(b(x)+1:b(x+1)-1),1:numel(b)-1,'uni',0)
c(cellfun(@isempty,c))=[]; %remove empty cells caused by consecutive NaN
c{:}

I did this simple benchmark just for fun in MATLAB 2008b

%each loop the NaN are put in a like this
s=10000; %the step of the loop, first value is also equal to s
a=randi([1 100],1,n); %values of the vector a, random integer from 1 to 100
per=randperm(n); %permute the indexes
nnan=randi([1 s]); %get random number of nans
a(per(1:nnan))=NaN; %set the first nnan indexes of a equal to nan

Behaviour of each code varying the number of NaN and keeping the vector with fixed size.

By mistake I had my line of code that removes empty cells commented, it's a little faster but the results might have empty cells that resulted from consecutive NaN values.

%each loop the number of NaN increases by 100 until 10000, 
%locations of the NaN values per loop are randomized with this code  
s=10000;                 %this is the number of columns of vector a
a=randi([1 100],1,s);    %the vector a full of integers from 1 to 100
per=randperm(s);         %generate the permutation of index values
a(per(1:nn))=NaN;        %put NaN on the first nn permuted index values

Same thing but now with my full code, little bit slower.

Another test, now like Fangjun suggested I used cellfun('isempty') instead of cellfun(@isempty) on my code, it does look faster now.

[Jan]

+1: I like embedded graphics.

[Fangjun Jiang]

Interesting! Nice work, Paulo!

My guess is that regexp() or string processing is relatively slow.

Jan told me that cellfun('isempty') is much faster than cellfun(@isempty). Wonder whether that will make a difference in your code.

[Paulo Silva]

Fangjun that part of my code doesn't take much time, actually with or without it doesn't seem to change the execution time, I also tested your suggestion, can't notice much difference.

[Paulo Silva]

ok Fangjun I did more tests and it really is faster with cellfun('isempty') :)

Answer 3

[Fangjun Jiang]

I couldn't resist trying a solution without for-loop. It considered consecutive NaNs.

a = [1 3 5 NaN NaN 4.4 5 NaN 8 9];
b=num2str(a);
c=regexp(b,'(NaN\s+)+','split');
d=cellfun(@str2num,c,'uni',0)
d{1}
d{2}
d{3}