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how to speed ...i need very fast code

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Luca Re
Luca Re 2023-8-30
评论： Luca Re 2023-8-31
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

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采纳的回答

Star Strider
Star Strider 2023-8-30
编辑：Star Strider 2023-8-30
Ran in:
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.
.
  2 个评论
Luca Re
Luca Re 2023-8-30
I saw that the vectorization can not be done ..ne get times of 1/10 from my ..patience I accept this last

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更多回答（2 个）

Bruno Luong
Bruno Luong 2023-8-30
编辑：Bruno Luong 2023-8-30
Ran in:
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
  4 个评论
显示 2更早的评论
Bruno Luong
Bruno Luong 2023-8-31
编辑：Bruno Luong 2023-8-31
Ran in:
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;

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David Hill
David Hill 2023-8-30
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
  1 个评论
Luca Re
Luca Re 2023-8-30
编辑：Luca Re 2023-8-30
hi,
Elapsed time is 4.489091 seconds. My Code
Elapsed time is 4.406116 seconds. Your Code
time is similar

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