how to calculate the maximum of different parts of a vector without using a for loop?
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Hi!
I have a vector H, which is separated in different events of known size, i.e. s=[2,5,7,2,...] and I want to find the maximum that corresponds to each event. How can I do it without using a for loop (I want to reduce the computational time)?
Using a for loop would be:
maxH(1) = max(H(1:s(1)));
for i=2:length(s)
maxH(i) = max(H( 1+sum(s(1:i-1)) : sum(s(1:i-1)) + s(i) ));
end
Thanks!
[EDITED, Jan, code formatted]
Accepted Answer
Teja Muppirala
on 31 Aug 2012
Although I don't recommend programming like this in MATLAB, sometimes writing out the logic out explicitly can make it run extremely fast. MATLAB has become very good at loops you know:
maxH = zeros(numel(s),1);
V = 1;
thismax = -inf;
thiss = s(1);
cnt = 1;
N = numel(H);
for n = 1:N
if cnt > thiss
maxH(V) = thismax;
thismax = H(n);
V = V+1;
cnt = 1;
thiss = s(V);
elseif H(n) > thismax
thismax = H(n);
end
cnt = cnt+1;
end
maxH(V) = thismax;
4 Comments
Jan
on 31 Aug 2012
@Teja: Runtime rules. As long as ACCUMARRAY is not multithreaded and/or you run the code on less than 12 cores, your solution is the most efficient. Therefore I see good reasons to recommend such programming techniques even in Matlab. +1
More Answers (4)
Andrei Bobrov
on 30 Aug 2012
Edited: Andrei Bobrov
on 30 Aug 2012
try this is code
for condition: numel(H) == sum(s)
maxH = cellfun(@max,mat2cell(H(:),s(:),1));
OR
idx = zeros(size(H));
idx(cumsum(s) - s +1) = 1;
maxH = accumarray(cumsum(idx(:)),H(:),[],@max);
1 Comment
Jan
on 30 Aug 2012
Edited: Jan
on 30 Aug 2012
maxH = zeros(1, length(s)); % Pre-allocate!!!
maxH(1) = max(H(1:s(1)));
cs = cumsum(s); % Avoid repeated summation
for ii = 2:length(s)
maxH(ii) = max(H(1 + cs(ii - 1) : cs(ii)));
end
[EDITED] Timings:
s = 1 + randi(10, 1, 10000);
H = rand(1, sum(s));
% Original:
tic;
maxH(1) = max(H(1:s(1)));
for i=2:length(s)
maxH(i) = max(H( 1+sum(s(1:i-1)) : sum(s(1:i-1)) + s(i) ));
end
toc % 0.533000 seconds
% Andrei's CELLFUN:
tic;
maxH = cellfun(@max,mat2cell(H(:),s(:),1));
toc % 0.080097 seconds
% Andrei's ACCUMARRAY:
tic;
idx = zeros(size(H));
idx(cumsum(s) - s +1) = 1;
maxH = accumarray(cumsum(idx(:)),H(:),[],@max);
toc % 0.021015 seconds
% FOR-loop:
tic;
maxH = zeros(1, length(s)); % Pre-allocate!!!
maxH(1) = max(H(1:s(1)));
cs = cumsum(s); % Avoid repeated summation
for ii = 2:length(s)
maxH(ii) = max(H(1 + cs(ii - 1) : cs(ii)));
end
toc % 0.033937 seconds
0 Comments
Robert Cumming
on 30 Aug 2012
In your example above the biggest speed gain would be achieved by pre-allocating the maxH variable
maxH = zeros(length(s),1);
note: For reference - removing the for loop will not necessarily speed things up....
0 Comments
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