How to vectorize?

Hi, is it possible to avoid the loop? Thanks.

x=(1:3:r);

Unrecognized function or variable 'r'.

[r,c]=size(Tr);
E1=zeros(r,c);
for i=1:c
  E1(x,i)=E(Tr(r,i),i);
end    

>> size(Tr)

ans =

21 6

[DGM]

What is z? What is Tr?

What is the value of r before it's redefined? Is it the same?

[pipin]

excuse me.. z=r...i change it in code

Tr is a matrix

[pipin]

[r,c]=size(Tr);
x=(1:3:r);
E1=zeros(r,c);
for i=1:c
  E1(x,i)=E(Tr(r,i),i);
end    

[Image Analyst]

Not seeing it. Again, what is Tr and E (not E1, but E)?

Does your for loop even work? What is the desired output?

If you have any more questions, then attach your data and missing code to read it in after you read this:

TUTORIAL: How to ask a question (on Answers) and get a fast answer

[pipin]

thanks your help I made some mistakes while creating the code because it's a bit laborious. Here's the correct version

%input:
row=10;
columns=5;
step=3;
%%*************
E = round(rand(row,columns)*10)
Tr = randi(size(E,1),row,size(E,2))
[r,c]=size(Tr);
x=(1:step:r);
% Your method
E1=zeros(r,c);
for i=1:c
    Tr(x,i)
    E(Tr(x,i),i)
  E1(x,i)=E(Tr(x,i),i);
end    
E1

Answer 1

[Bruno Luong]

Just to pull out the right answer (my previous answer applied on a WRONG calculation specified by OP)

% vectorized method
c = size(Tr,2);
m = size(E,1);
x = 1:step:m;
E1 = zeros(size(Tr));
E1(x,:) = E(Tr(x,:) + (0:c-1)*m);

[pipin]

hello.. why do you say that the previous answer was wrong? on the isequal test it was correct

(what is OP?)

[Bruno Luong]

OP is Original Poster, it's you who wrote : "excuse me.. z=r...i change it in code" ... remember?

My answer (not the comments below it) is wrong because the question is wrong.

Answer 2

[Bruno Luong]

If you want obscure code by avoiding for-loop

E = rand(10,5)

E = 10×5

0.4433 0.5126 0.5907 0.7564 0.9740 0.9908 0.4916 0.9634 0.8613 0.0549 0.9769 0.2085 0.4765 0.2052 0.7497 0.0450 0.8816 0.9243 0.6475 0.1040 0.8076 0.1900 0.4109 0.6581 0.2044 0.4685 0.1936 0.7925 0.7647 0.4082 0.9868 0.9336 0.2478 0.0403 0.7413 0.1741 0.5068 0.1629 0.0827 0.0434 0.1120 0.7868 0.0827 0.4696 0.7350 0.4545 0.5871 0.0055 0.1273 0.2868

Tr = randi(size(E,1),9,size(E,2))

Tr = 9×5

9 10 5 2 4 9 10 9 9 2 1 10 4 6 2 2 8 2 3 4 8 2 7 9 8 5 7 2 4 5 5 5 7 9 5 5 10 10 4 3 9 8 3 3 7

[r,c]=size(Tr);
x=(1:3:r);

[r,c]=size(Tr);
x=(1:3:r);
% Your method
E1=zeros(r,c);
for i=1:c
  E1(x,i)=E(Tr(r,i),i);
end    
E1

E1 = 9×5

0.1120 0.5068 0.4765 0.2052 0.7413 0 0 0 0 0 0 0 0 0 0 0.1120 0.5068 0.4765 0.2052 0.7413 0 0 0 0 0 0 0 0 0 0 0.1120 0.5068 0.4765 0.2052 0.7413 0 0 0 0 0 0 0 0 0 0

% vectorized method
E1=zeros(r,c);
i=1:c;
E1(x(:) + (i-1)*r) = repmat(E(Tr(r,i) + (i-1)*size(E,1)),length(x),1);
E1

E1 = 9×5

0.1120 0.5068 0.4765 0.2052 0.7413 0 0 0 0 0 0 0 0 0 0 0.1120 0.5068 0.4765 0.2052 0.7413 0 0 0 0 0 0 0 0 0 0 0.1120 0.5068 0.4765 0.2052 0.7413 0 0 0 0 0 0 0 0 0 0

[Bruno Luong]

PS: I have a feeling that is NOT what you expect to get.

[pipin]

thanks your help I made some mistakes while creating the code because it's a bit laborious. Here's the correct version

%input:

row=10;
columns=5;
step=3;

%%*************


E = round(rand(row,columns)*10)

Tr = randi(size(E,1),row,size(E,2))

[r,c]=size(Tr);
x=(1:step:r);



% Your method
E1=zeros(r,c);
for i=1:c
    Tr(x,i)
    E(Tr(x,i),i)
  E1(x,i)=E(Tr(x,i),i);
end    

E1

[Bruno Luong]

"I made some mistakes while creating the code because it's a bit laborious. "

I knew it, not it is not because laborious, it's because you are simply not serious on what you are doing.

row=10;
columns=5;
step=3;
%%*************
E = round(rand(row,columns)*10)

E = 10×5

3 10 7 6 1 1 6 1 1 1 8 2 4 3 5 7 9 6 6 5 9 8 1 7 5 4 10 3 6 1 7 9 8 5 3 2 3 8 4 1 10 1 3 0 2 2 5 2 8 2

Tr = randi(size(E,1),row,size(E,2))

Tr = 10×5

3 7 9 10 2 4 3 2 5 8 4 10 9 3 10 10 3 5 7 6 5 6 1 10 7 3 5 4 6 1 2 10 8 5 1 8 1 8 1 2 7 7 10 1 10 1 9 1 7 2

[r,c]=size(Tr);
x=(1:step:r);
% Your method
E1=zeros(r,c);
for i=1:c
    E1(x,i)=E(Tr(x,i),i);
end    
E1

E1 = 10×5

8 9 3 8 1 0 0 0 0 0 0 0 0 0 0 2 2 1 5 1 0 0 0 0 0 0 0 0 0 0 1 5 8 7 1 0 0 0 0 0 0 0 0 0 0 3 1 7 5 1

% vectorized method
E1=zeros(r,c);
i=1:c;
E1(x(:) + (i-1)*r) = E(Tr(x,i) + (i-1)*size(E,1));
E1

E1 = 10×5

8 9 3 8 1 0 0 0 0 0 0 0 0 0 0 2 2 1 5 1 0 0 0 0 0 0 0 0 0 0 1 5 8 7 1 0 0 0 0 0 0 0 0 0 0 3 1 7 5 1

[pipin]

:D

very good

[pipin]

i try it with

row=500000;

columns=500;

step=30;

Elapsed time is 20.148787 seconds. (mYcODE)

Elapsed time is 18.746811 seconds. (VECTORIZED CODE)

Why do you think vectorization doesn't improve execution time?

i use:

Processore Intel(R) Core(TM) i5-7200U CPU @ 2.50GHz 2.71 GHz

RAM installata 8,00 GB (7,88 GB utilizzabile)

Tipo sistema Sistema operativo a 64 bit, processore basato su x64

[Bruno Luong]

Why do you think it does?

This is the timining of the Online server, your PC does not have anough memory IMO. Each array E, E1, and Tr take 2 Gb, so a total of 6 Gb of free RAM is needed.

row=500000;
columns=500;
step=300;
%%*************
E = round(rand(row,columns)*10);
Tr = randi(size(E,1),row,size(E,2));

tic
[r,c]=size(Tr);
x=(1:step:r);
% Your method
E1=zeros(r,c);
for i=1:c
    E1(x,i)=E(Tr(x,i),i);
end    
toc

Elapsed time is 0.607406 seconds.

tic
% vectorized method II
[r,c]=size(Tr);
x=(1:step:size(E,1));
E1=zeros(r,c);
E1(x,:) = E(Tr(x,:) + (0:c-1)*size(E,1));
toc

Elapsed time is 0.633170 seconds.

[pipin]

20 sec (my method) vs 18 sec (vectorized) are similar

often with vectorization I see advantages of 1:10 as speed compared to loops

likely because this pc has few cores and is not benefited by parallel computing

[pipin]

ok thank

[Bruno Luong]

"often with vectorization I see advantages of 1:10 as speed compared to loops"

Your information is surely NOT up-to-date and too generic to have any real value.

[pipin]

hi, i'm using your method vectorized

% vectorized method II
[r,c]=size(Tr);
x=(1:step:size(E,1));
E1=zeros(r,c);
E1(x,:) = E(Tr(x,:) + (0:c-1)*size(E,1));
toc


I have seen that Tr can also assume the value zero which would lead to an error in the function (you cannot search for a zero index)
is it possible to do the search of the indexes excluding if the value is zero?
i solve it:
 [r,c]=size(MinTrade_Tab_Short);
 a=MinTrade_Tab_Short+ (0:c-1)*r;
 a2=a+(~a); %tolgo lo zero (per evitare errore riga sucessiva)
 E1 = E(a2);
 E2=E1.*(a>0);
 
 it's good for you?
   

[Bruno Luong]

No your solution is NOT good as it only catches Tr == 0 at column 0. For other column a2 is not 0 so you won't correct the overflowed row indexing.

[Bruno Luong]

Code to handle the case Tr == 0

c = size(Tr,2);
m = size(E,1);
x = 1:step:m;
E1 = zeros(size(Tr));

Trx = Tr(x,:);
iE = Trx + m*(0:c-1);
tmp = E(max(iE,1));
tmp(Trx == 0) = 0;
E1(x,:) = tmp;

it does not handle still the case Tr > m or Tr < 0.

If you havve data that are not valid for proper indexing it will be a mess to deal with.

[pipin]

Correct..thank you!