How to split a vector in to sub vectors?

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Jayanta Deb
Jayanta Deb 2017-8-16
评论： Pierre Mathieu 2022-10-10
I have a vector 'v' of size 8812x1,I want to split it into 20(or known length n) sub-vectors. eg. v = {1,2,3,...,8812}
I wish to see:
v1={1,2,3,...,20}
v2={21,22,23,...,41}
v3 = {42,43,44,...,62}
...
v20 = {...} % the last vector doesn't matter if it can have full 20 or not, it can / might get lesser number as its not fully possible to get a whole 20 elements divide equally with the size of the vector but obviously the vector v1 till v19 must have 20 elements.
Then I want to calculate the median from each sub divided vectors.
So it should be like m1 = median(v1), m2 = median(v2)... m20=median(v20)
I would appreciate if it is done with a loop

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采纳的回答

Stephen23
Stephen23 2017-8-16
编辑：Stephen23 2017-8-16
Simpler:
>> v = rand(8812,1);
>> b = 20; % block size
>> n = numel(v);
>> c = mat2cell(v,diff([0:b:n-1,n]));
>> z = cellfun(@median,c);
Or if you really want to use a loop, replace the last line with:
z = NaN(numel(c),1);
for k=1:numel(c)
z(k) = median(c{k});
end
  2 个评论
Md Shahriar Islam
Md Shahriar Islam 2018-10-12
Hi, I would like to know what to do if block size, b is unknown. That is b is dependent on v.

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更多回答（2 个）

José-Luis
José-Luis 2017-8-16
v = rand(8812,1);
n = 20;
dummy = nan(20,ceil(numel(v)./n));
dummy(1:numel(v)) = v;
result = nanmedian(dummy)
ES
ES 2017-8-16
编辑：ES 2017-8-16
v = [1:1:8812];
window_size = 20;
iCount = size(v,2);
iStartIDx = [1:window_size :iCount];
for iLoop=1:length(iStartIDx)-1
segment{iLoop} = v(iStartIDx(iLoop):iStartIDx(iLoop+1)-1);
med{iLoop} = median(segment{iLoop});
end
segment{iLoop+1} = v(iStartIDx(iLoop+1):end);
med{iLoop+1} = median(segment{iLoop+1});
  3 个评论
显示 1更早的评论
Stephen23
Stephen23 2017-8-16
编辑：Stephen23 2017-8-16
@J Smith: the code could be improved:
  • those output arrays should be preallocated.
  • note that square brackets serve no purpose: [1:1:8812] is better written as 1:1:8812 (see the editor warning message and this discussion).
  • using median as a variable name is a bad idea.
José-Luis
José-Luis 2017-8-16
编辑：José-Luis 2017-8-16
This is rather wrong. On top of being incomplete, as pointed out by Jayanta:
There is no pre-allocation, a performance killer.
median is used both as a variable name and as a function. That's bound to be problematic.
Edit: latest comment was valid before answer being edited.

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