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How to avoid this problem "consider preallocating the memory .for speed" in the below sample code

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aditya sahu
aditya sahu on 14 Sep 2017
Commented: Steven Lord on 14 Sep 2017
The code is like this
m = [];
x=[100,200,300.....]
for k= 1:1:142
if(k >= 142)
break;
end
%
if x(k)== [5];
mm = horzcat(mm,xx);
elseif (x(k+1) >= 20)
bb = 10- lj1 ;
cc = dec2bin(bb,3)-'0';
mm = horzcat(mm,cc);
%
elseif (x(k+2) >= 10)
cc = dec2bin(bb,3)-'0';
mm = horzcat(mm,cc);
%
end
end
how the preallocation of 'mm' can be done to increase the speed of execution. Kindly suggest

  3 Comments

aditya sahu
aditya sahu on 14 Sep 2017
SIR @KSSV This is the right code(piece of sample code from the large code)...and the issue is to preallocate the value of 'mm'. As the loop is very large IN THE ORIGINAL CODE 2 problems are coming 1. is it is killing a lot time for executing 2. memory problem...
dd=imread('x.bmp');
qq=reshape(dd',[100],[])';
qq=double(qq);
qq=qq';
ss=dec2bin(qq)-'0';
ss=ss';
pp=dec2bin(qq)-'0';
xx = [];
bb = [];
cc = [];
mm = [];
lb = [ 0 21 31];
ub = [ 20 30 41];
for k=1:3:105
if(numel(mm) >= 2345)
break;
end
if(k >= 100)
break;
end
xx = pp((k+1),6:8);
mm = horzcat(mm,xx);
stdif1 = abs(qq(k) - qq(k+1));
stdif2 = abs(qq(k+2) - qq(k+1));
if (stdif1 >= lb(1)) && ( stdif1 <= ub(1))
lj1 = lb(1);
bb = stdif1 - lj1 ;
cc = dec2bin(bb,3)-'0';
mm = horzcat(mm,cc);
elseif(stdif1 >= lb(2)) && ( stdif1 <= ub(2))
lj1 = lb(2);
bb = stdif1 - lj1 ;
cc = dec2bin(bb,3)-'0';
mm = horzcat(mm,cc);
else(stdif1 >= lb(3)) && ( stdif1 <= ub(3))
lj1 = lb(3);
bb = stdif1 - lj1 ;
cc = dec2bin(bb,3)-'0';
mm = horzcat(mm,cc);
end
end
mm=reshape(mm,3,[])';
Steven Lord
Steven Lord on 14 Sep 2017
Can you explain the purpose of your code in words? If we know the underlying problem you're trying to solve, we may be able to suggest an alternative approach that avoids growing memory each time through the loop (potentially by avoiding the loop entirely!)

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Answers (2)


KSSV
KSSV on 14 Sep 2017
If you are aware of the dimensions of the allocation preallcoate it using:
x = zeros(m,n) ;
If you are not aware of dimensions, and you know that it is either a row or a column vector:
x = zeros([],n) ; % a row row vector
x = zeros(m,[]) ; % a column vector
If you are not sure of dimensions, allocate it as cell:
x = cell([],n) ;
x = cell(m,[]) ;

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