how to speed ...i need very fast code

q=6062; %%I want to call this code and pass it a different 'q' each time
matrix=matri(1:q,:);
[~,c]=size(matrix);

COR=zeros(c,c);

  tic
    gg=1:c;       
    a=matrix(:,gg);
    for yy=1:c
        b=matrix(:,yy);
        cc=sum((a>0 & b>0)|(a<0 & b<0),1);
        cc1=sum(a~=0 & b~=0,1);
        COR(yy,gg)=round(cc./cc1*1000)/1000;   %round(100 * cc./cc1)/100; %%array multidimension 
    end 
toc

Answer 1

[Star Strider]

Putting the ‘a’ conditional tests outside the loop (and still within the tic-toc block) sppeds it up a bit —

LD = load('matlab_matri.mat')

LD = struct with fields:
    matri: [6165×351 double]

matri = LD.matri;

q=6062; %%I want to call this code and pass it a different 'q' each time
matrix=matri(1:q,:);
[~,c]=size(matrix);

COR=zeros(c,c);

  tic
    gg=1:c;       
    a=matrix(:,gg);
    for yy=1:c
        b=matrix(:,yy);
        cc=sum((a>0 & b>0)|(a<0 & b<0),1);
        cc1=sum(a~=0 & b~=0,1);
        COR(yy,gg)=round(cc./cc1*1000)/1000;   %round(100 * cc./cc1)/100; %%array multidimension 
    end 
toc

Elapsed time is 3.543419 seconds.

COR

COR = 351×351

1.0000 0.5210 0.5090 0.5000 0.5000 0.5180 0.5010 0.5180 0.5150 0.5030 0.4950 0.4820 0.4870 0.4990 0.5020 0.4840 0.5020 0.5100 0.4980 0.4950 0.5140 0.5190 0.4920 0.5110 0.5070 0.5110 0.4900 0.5270 0.5200 0.5080 0.5210 1.0000 0.4900 0.4890 0.4810 0.5120 0.4730 0.5660 0.4860 0.4940 0.4740 0.5540 0.5320 0.5310 0.5290 0.5150 0.5120 0.5160 0.4970 0.5190 0.5190 0.4750 0.5250 0.5360 0.4580 0.5290 0.5120 0.5440 0.4720 0.5060 0.5090 0.4900 1.0000 0.5100 0.5160 0.5320 0.5320 0.5430 0.4900 0.5170 0.5620 0.4810 0.4930 0.4880 0.4990 0.4770 0.4640 0.4930 0.4940 0.4860 0.5120 0.5000 0.4950 0.5160 0.5060 0.5300 0.5040 0.4980 0.4820 0.4790 0.5000 0.4890 0.5100 1.0000 0.6350 0.5780 0.5720 0.4980 0.5090 0.5310 0.5430 0.5050 0.5050 0.4890 0.5110 0.5250 0.4900 0.4900 0.5000 0.4730 0.5010 0.5100 0.5090 0.5280 0.4930 0.4590 0.4940 0.4980 0.4440 0.5220 0.5000 0.4810 0.5160 0.6350 1.0000 0.5260 0.6010 0.6720 0.4570 0.5190 0.5480 0.5070 0.4980 0.4890 0.4760 0.5030 0.4590 0.5040 0.4970 0.4980 0.5180 0.4940 0.4940 0.4840 0.4790 0.4800 0.5470 0.4980 0.4310 0.5110 0.5180 0.5120 0.5320 0.5780 0.5260 1.0000 0.5210 0.4540 0.5660 0.5380 0.4030 0.5250 0.4960 0.4860 0.4870 0.4980 0.5090 0.4890 0.4800 0.5170 0.4890 0.5280 0.4910 0.4840 0.5210 0.4690 0.5380 0.4780 0.5560 0.5030 0.5010 0.4730 0.5320 0.5720 0.6010 0.5210 1.0000 0.5930 0.4810 0.5490 0.4970 0.5080 0.4930 0.4950 0.4940 0.5180 0.4810 0.4860 0.5090 0.4960 0.4810 0.4950 0.5130 0.5240 0.5090 0.5360 0.5090 0.5010 0.4850 0.5140 0.5180 0.5660 0.5430 0.4980 0.6720 0.4540 0.5930 1.0000 0.2900 0.4920 0.5990 0.5670 0.4830 0.4860 0.5000 0.5140 0.5080 0.4900 0.5130 0.5120 0.4940 0.4690 0.4670 0.5190 0.4570 0.4330 0.5070 0.5160 0.5000 0.5010 0.5150 0.4860 0.4900 0.5090 0.4570 0.5660 0.4810 0.2900 1.0000 0.5280 0.4470 0.4890 0.5200 0.5020 0.5140 0.5150 0.5260 0.5140 0.5020 0.5260 0.4950 0.5030 0.5190 0.5050 0.4850 0.5150 0.4770 0.4970 0.5380 0.5150 0.5030 0.4940 0.5170 0.5310 0.5190 0.5380 0.5490 0.4920 0.5280 1.0000 0.4810 0.5040 0.5000 0.5120 0.5150 0.4830 0.4950 0.4860 0.4960 0.4980 0.4890 0.5130 0.5040 0.5030 0.5290 0.4920 0.5180 0.5090 0.5220 0.5030

tic
        agt0 = matrix>0;
        alt0 = matrix<0;
        ane0 = matrix~=0;

        for yy = 1:c
            cc=sum((agt0 & agt0(:,yy))|(alt0 & alt0(:,yy)),1);
            cc1=sum(ane0 & ane0(:,yy),1);
            COR(:,yy) = cc./cc1;
        end
        COR = round(COR,4);
toc

Elapsed time is 1.354815 seconds.

COR

COR = 351×351

1.0000 0.5214 0.5091 0.5004 0.5000 0.5175 0.5008 0.5180 0.5149 0.5029 0.4950 0.4825 0.4866 0.4991 0.5015 0.4843 0.5015 0.5105 0.4979 0.4950 0.5139 0.5189 0.4918 0.5114 0.5071 0.5105 0.4899 0.5267 0.5199 0.5084 0.5214 1.0000 0.4899 0.4886 0.4810 0.5116 0.4734 0.5665 0.4858 0.4944 0.4737 0.5544 0.5316 0.5314 0.5294 0.5146 0.5117 0.5158 0.4974 0.5185 0.5191 0.4747 0.5247 0.5361 0.4582 0.5287 0.5117 0.5441 0.4718 0.5063 0.5091 0.4899 1.0000 0.5104 0.5159 0.5322 0.5320 0.5432 0.4901 0.5170 0.5622 0.4811 0.4930 0.4883 0.4993 0.4773 0.4639 0.4934 0.4944 0.4857 0.5116 0.5000 0.4949 0.5161 0.5056 0.5304 0.5044 0.4980 0.4819 0.4786 0.5004 0.4886 0.5104 1.0000 0.6349 0.5780 0.5720 0.4984 0.5089 0.5313 0.5430 0.5051 0.5052 0.4885 0.5109 0.5252 0.4895 0.4898 0.4996 0.4734 0.5015 0.5104 0.5094 0.5282 0.4931 0.4593 0.4944 0.4979 0.4444 0.5220 0.5000 0.4810 0.5159 0.6349 1.0000 0.5259 0.6009 0.6725 0.4570 0.5192 0.5476 0.5069 0.4976 0.4890 0.4761 0.5030 0.4594 0.5040 0.4975 0.4978 0.5176 0.4943 0.4941 0.4838 0.4789 0.4800 0.5473 0.4985 0.4305 0.5109 0.5175 0.5116 0.5322 0.5780 0.5259 1.0000 0.5210 0.4540 0.5655 0.5380 0.4029 0.5251 0.4961 0.4855 0.4871 0.4982 0.5087 0.4888 0.4797 0.5167 0.4890 0.5276 0.4913 0.4835 0.5209 0.4685 0.5381 0.4777 0.5563 0.5030 0.5008 0.4734 0.5320 0.5720 0.6009 0.5210 1.0000 0.5931 0.4810 0.5494 0.4968 0.5082 0.4934 0.4948 0.4937 0.5185 0.4807 0.4862 0.5094 0.4961 0.4811 0.4954 0.5132 0.5243 0.5086 0.5358 0.5086 0.5015 0.4847 0.5144 0.5180 0.5665 0.5432 0.4984 0.6725 0.4540 0.5931 1.0000 0.2897 0.4920 0.5989 0.5667 0.4829 0.4862 0.5000 0.5144 0.5081 0.4904 0.5133 0.5117 0.4943 0.4694 0.4669 0.5195 0.4573 0.4330 0.5066 0.5159 0.5000 0.5010 0.5149 0.4858 0.4901 0.5089 0.4570 0.5655 0.4810 0.2897 1.0000 0.5284 0.4468 0.4890 0.5205 0.5022 0.5140 0.5147 0.5258 0.5141 0.5022 0.5261 0.4949 0.5029 0.5187 0.5053 0.4851 0.5148 0.4771 0.4972 0.5375 0.5152 0.5029 0.4944 0.5170 0.5313 0.5192 0.5380 0.5494 0.4920 0.5284 1.0000 0.4811 0.5037 0.5000 0.5121 0.5149 0.4826 0.4946 0.4856 0.4957 0.4976 0.4886 0.5129 0.5043 0.5033 0.5292 0.4918 0.5183 0.5092 0.5216 0.5033

EDIT — Slight tweak to create all the conditionals together.

.

[Luca Re]

I saw that the vectorization can not be done ..ne get times of 1/10 from my ..patience I accept this last

[Star Strider]

Thank you!

Answer 2

[Bruno Luong]

You want REALLY FAST code?

Watch this, almost 100 time faster

load('matlab_matri.mat')
q=6062; %%I want to call this code and pass it a different 'q' each time
matrix=matri(1:q,:);
[~,c]=size(matrix);

tic
COR=zeros(c,c); % you have to count this as well
gg=1:c;
a=matrix(:,gg);
for yy=1:c
    b=matrix(:,yy);
    cc=sum((a>0 & b>0)|(a<0 & b<0),1);
    cc1=sum(a~=0 & b~=0,1);
    COR(yy,gg)=round(cc./cc1*1000)/1000;   %round(100 * cc./cc1)/100; %%array multidimension
end
toc

Elapsed time is 2.403425 seconds.

tic
s = sign(matrix);
b = double(matrix~=0);
cor = round((1 + (s'*s)./(b'*b))*500)/1000;
toc

Elapsed time is 0.025416 seconds.

% Does it match?
isequaln(cor,COR)

ans = logical
   1

[Luca Re]

excellent work thanks

[Luca Re]

there is a small correction i need to do to my original code : COR(yy,gg)=round(cc./(cc1+0.00000001)*1000)/1000; in round(100 * cc./cc1)/100; I added a 0.000001 to avoid division with zero (which would lead to the result ==Nan) Where can I put this in your code?

cor = round((1 + (s'*s)./((b'+0.0000001)*b))*500)/1000;

is correct this correction?

[Bruno Luong]

No it is not correct.

Correct one is this

% cor = round((1 + (s'*s)./((b'*b)+eps(0.5)))*500)/1000;

This will return identical numerical result, excepted when working on colum dot-product that contain all 0s.

The reason is the minimum strictly positive of (b'*b) is 1, adding eps(0.5) to it does not change the value, unless it is 0 where it protect the denominator to vanishe, as showed here

load('matlab_matri.mat')
q=6062; %%I want to call this code and pass it a different 'q' each time
matrix=matri(1:q,:);
[~,c]=size(matrix);

s = sign(matrix);
b = double(matrix~=0);
cor = round((1 + (s'*s)./(b'*b))*500)/1000;

s = sign(matrix);
b = double(matrix~=0);
cornan = round((1 + (s'*s)./((b'*b)+eps(0.5)))*500)/1000;

any(isnan(cor),'all') % nan is present

ans = logical
   1

any(isnan(cornan),'all') % nan disappears

ans = logical
   0

% Does it match? 0 is perfectly match
max(abs(cor-cornan),[],'all')

ans = 0

Note that in my case cornan contain 1 for correlation of two colums of matrix that do not share 1s, and not 0 as with your code. It is somewhat an arbitrary choice, since the correlation is undefined. Without protection MATLAB NaN result is actually more "correct" IMO since it reflects this fact of arbitrary choice.

If it bother you, just do not protect, then simply add this at the end:

cor(isnan(cor)) = 0; 

Alternatively you can do this to return 0 for degenerated case

s = sign(matrix);
b = double(matrix~=0);
btb = b'*b;
cornan = round(((btb + (s'*s))./(btb+eps(0.5)))*500)/1000;

[Luca Re]

Thank you

Answer 3

[David Hill]

q=6062;
matrix=matri(1:q,:);
c=size(matrix,2);
COR=zeros(c);

a=matrix;
for yy=1:c
  b=matrix(:,yy);
  COR(yy,:)=round(sum((a>0 & b>0)|(a<0 & b<0),1)./sum(a~=0 & b~=0,1),3);
end 

[Luca Re]

hi,

Elapsed time is 4.489091 seconds. My Code

Elapsed time is 4.406116 seconds. Your Code

time is similar