calculating the time

2012 5 22
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I have a code which runs for 10 iterations in this i want to calculate the time for each iteration and find the minimum time for iteration,please help

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Wayne King
Wayne King 2012-5-22

1 个投票

You can use tic and toc.
t = zeros(1,100);
for n = 1:100
A = rand(n,n);
b = rand(n,1);
tic
x = A\b;
t(n) = toc;
end
[mintime,I] = min(t);

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kash
kash 2012-5-22
Thanks Wayne I have a code for compression ratio
function [CR PSNR]=cpdp(I)
input_image=I;
n=3;
%*****************************************************************************
[Lo_D,Hi_D,Lo_R,Hi_R] = wfilters('haar');
%wavedec2 - Multi-level 2-D wavelet decomposition.
[c,s]=wavedec2(input_image,n,Lo_D,Hi_D);
disp(' the decomposition vector Output is');
disp(c);
%*************************************************************************************
[thr,nkeep] = wdcbm2(c,s,1.5,3*prod(s(1,:)));
[compressed_image,TREED,comp_ratio,PERFL2] =WPDENCMP(thr,'s',n,'haar','threshold',5,1);
re_ima1 = waverec2(c,s,'haar');
re_ima=uint8(re_ima1);
CR=comp_ratio;
Er1= mean2((single(I) - single(re_ima)).^2);
PSNR = 20 * log10 (255 / sqrt(Er1));
I wrote this as a function program where i have 10 images inside for loop i have called this function,now i want to calculate which image has highest CR,plese help
Jan
Jan 2012-5-22
@Wayne: In my opinion your program solves the question sufficiently. +1

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Wayne King
Wayne King 2012-5-22

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Just save the CR values in a vector, then query which one is the max

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kash
kash 2012-5-22
Wayne i tried CR(I)=CR;
max(CR)
BUT NOT GETTING THE RESULT
kash
kash 2012-5-22
ALSO HOW TO DISPLAY THE IMAGE WHICH HAS HIGHEST CR
Jan
Jan 2012-5-22
Dear kash, there is no reason for shouting.
"but not getting the result" is vague. I guess you get a detailed error message instead, because you try to assign the array CR to one of its elements. This is not possible for obvious reasons.
But the questions about the compression ratio do not concern the original question, such that it it recommended to open a new thread. Otherwise it gets impossible to accept the best answer.

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