how velocize it? (it's possibile to vectorize it?)

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pipin
pipin el 30 de Jul. de 2023
Editada: pipin el 31 de Jul. de 2023
Ntradess =[2 99 1 8 15 67 74 51 58 40
3 80 7 14 16 73 55 57 64 41
3 80 7 20 22 54 56 63 70 47
3 81 7 21 3 60 62 69 71 28
3 81 9 2 9 61 68 75 52 34
4 82 9 83 90 42 49 26 33 65
4 82 9 89 91 48 30 32 39 66
5 85 10 95 97 29 31 38 45 72
6 85 11 96 78 35 37 44 46 53
7 85 12 77 84 36 43 50 27 58
8 86 13 80 84 38 45 51 30 58
8 86 13 82 84 38 45 51 31 58
8 86 13 83 84 38 45 51 31 58
9 87 13 84 84 38 45 51 31 58];
RP_bin=[0 0 1 0 0 1 0 0 1 0 0 1 0 0]';
period=2;
minTrades=2;
g=find(RP_bin>0);
Ntradess_=Ntradess(g,:);
[r,c]=size(Ntradess_);
MinNtrad=zeros(r,c);
for x=1:c %loop colonne
for i=period+1:r
a=Ntradess_(i,x); %n:trade del idx attuale
if a>minTrades
a=find(flip(Ntradess_(1:i-1,x))<=(a-minTrades),1);
if ~isempty(a)
MinNtrad(i,x)= i-a; %MEMORIZZO la posizione della schiera per avere in mintrades richiesti
end
end
end
end
  4 comentarios
Walter Roberson
Walter Roberson el 30 de Jul. de 2023
You could save some coding effort if you were to use
a = find(Ntradess_(1:i-1,x))<=(a-minTrades), 1, 'last');
if ~isempty(a)
MinNtrad(i,x) = a;
end
The change in logic might make it easier to figure out how to vectorize.
pipin
pipin el 30 de Jul. de 2023
thank

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Bruno Luong
Bruno Luong el 30 de Jul. de 2023
Editada: Bruno Luong el 31 de Jul. de 2023
I don't know the memory requirement would go up if I vectorize the outer loop in a non-toy case. So I leave it for now.
Ntradess =[2 99 1 8 15 67 74 51 58 40
3 80 7 14 16 73 55 57 64 41
3 80 7 20 22 54 56 63 70 47
3 81 7 21 3 60 62 69 71 28
3 81 9 2 9 61 68 75 52 34
4 82 9 83 90 42 49 26 33 65
4 82 9 89 91 48 30 32 39 66
5 85 10 95 97 29 31 38 45 72
6 85 11 96 78 35 37 44 46 53
7 85 12 77 84 36 43 50 27 58
8 86 13 80 84 38 45 51 30 58
8 86 13 82 84 38 45 51 31 58
8 86 13 83 84 38 45 51 31 58
9 87 13 84 84 38 45 51 31 58];
RP_bin=[0 0 1 0 0 1 0 0 1 0 0 1 0 0]';
period=2;
minTrades=2;
g=find(RP_bin>0);
Ntradess_=Ntradess(g,:);
[r,c]=size(Ntradess_);
MinNtrad=zeros(r,c);
for x=1:c %loop colonne
for i=period+1:r
a=Ntradess_(i,x); %n:trade del idx attuale
if a>minTrades
a=find(flip(Ntradess_(1:i-1,x))<=(a-minTrades),1);
if ~isempty(a)
MinNtrad(i,x)= i-a; %MEMORIZZO la posizione della schiera per avere in mintrades richiesti
end
end
end
end
MinNtrad
MinNtrad = 4×10
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 2 2 1 0 0 2 2 1 3 2 3 1 3 3 3 3 0 3
[r,c]=size(Ntradess_);
MinNtrad = zeros(r,c);
Mask = flip(triu(true(r),1),1);
for x = 1:c
nx = Ntradess_(:,x);
[v,j] = max(Mask .* (nx'-flip(nx,1) >= minTrades), [], 1);
j = r+1-j;
j(v(:)==0 | nx(:)<=minTrades) = 0;
MinNtrad(:,x) = j;
end
MinNtrad(1:period,:) = 0;
MinNtrad
MinNtrad = 4×10
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 2 2 2 1 0 0 2 2 1 3 2 3 1 3 3 3 3 0 3
  2 comentarios
Bruno Luong
Bruno Luong el 30 de Jul. de 2023
Editada: Bruno Luong el 31 de Jul. de 2023
Full vectorize
warning: potential runout of memory
[r,c] = size(Ntradess_);
Mask = (1:r)' > (r:-1:1); % flip(triu(true(r),1),1);
D = reshape(Ntradess_, [1 r c]) - reshape(flip(Ntradess_, 1), [r 1 c]);
B = Mask .* (D >= minTrades);
[V,J] = max(B, [], 1);
MinNtrad = r+1-reshape(J, [r c]);
V = reshape(V, [r c]);
MinNtrad(V==0 | Ntradess_<=minTrades) = 0;
MinNtrad(1:period,:) = 0;
pipin
pipin el 31 de Jul. de 2023
Editada: pipin el 31 de Jul. de 2023
congrats for the vectorization....but i don't use it it requires a lot of free memory

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