i need to code speed ..anyone can help me? Getting rid of the loop would be a good start

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pipor
pipor 2023-9-4
评论： pipor 2023-9-4
Ran in:
v=[1 0 0 1];
d=[0.3 0 0 0.4]
d = 1×4
0.3000 0 0 0.4000
n="lp";
count=1;
gg=find(v)
gg = 1×2
1 4
clear b
g={'Cycle n.',count}
g = 1×2 cell array
{'Cycle n.'} {[1]}
j=1; %
j = 1
for h=1:numel(gg)
b{j,h}={[g,gg(h),d(gg(h))]};
end
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pipor
pipor 2023-9-4
编辑：pipor 2023-9-4
because I'm working on a laptop and on a lot of data it happened that I got the error about the memory..hence thinking about not making three-dimensional matrices
however I am modifying the code..maybe I make it clearer
Bruno Luong
Bruno Luong 2023-9-4
编辑：Bruno Luong 2023-9-4
But you have very bad start. This data organization is awful. We warn you, with such data you'll have speed, memory, and coding problems later.

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采纳的回答

Bruno Luong
Bruno Luong 2023-9-4
Ran in:
clear
v=[1 0 0 1];
d=[0.3 0 0 0.4];
count=1;
gg=find(v);
g={'Cycle n.',count};
j=1;
clear b
for h=1:numel(gg)
b{j,h}={[g,gg(h),d(gg(h))]};
end
% you want horrible code to store horrible data, here we go
bb = num2cell(num2cell([ repmat(g,[numel(gg) 1]) num2cell([ gg' d(gg)' ])], 2)');
isequal(bb,b)
ans = logical
1
  1 个评论
pipor
pipor 2023-9-4
you want horrible code to store horrible data, here we go :D
thank but i write :"however I am modifying the code..maybe I make it clearer"
however I accept your work

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