Preallocating str for speed

I have this code. I want it to run faster. And I keep getting the warning to preallocate str for speed.
clc
str=[];
for T='956754674275'
if T=='0'
str = [str sprintf('A')];
elseif T=='1'
str = [str sprintf('B')];
elseif T=='2'
str = [str sprintf('C')];
elseif T=='3'
str = [str sprintf('D')];
elseif T=='4'
str = [str sprintf('E')];
elseif T=='5'
str = [str sprintf('F')];
elseif T=='6'
str = [str sprintf('G')];
elseif T=='7'
str = [str sprintf('H')];
elseif T=='8'
str = [str sprintf('I')];
elseif T=='9'
str = [str sprintf('J')];
end
end
a=str
Please how do I preallocate str for speed. Please help.

5 Comments

just do
a = '956754674275';
str = a;
Then
ct = 0;
for T='956754674275'
ct = ct+1;
if T=='0'
str(ct) = 'A'; % following you just do in this form
elseif T=='1'
str = [str sprintf('B')];
elseif T=='2'
str = [str sprintf('C')];
elseif T=='3'
str = [str sprintf('D')];
elseif T=='4'
str = [str sprintf('E')];
elseif T=='5'
str = [str sprintf('F')];
elseif T=='6'
str = [str sprintf('G')];
elseif T=='7'
str = [str sprintf('H')];
elseif T=='8'
str = [str sprintf('I')];
elseif T=='9'
str = [str sprintf('J')];
end
end
Please it gives an output
'956754674275JFGHFEGHECHF'
instead of 'JFGHFEGHECHF'
Thank you Wan Ji. how do I accept this comment as an answer.
Stephen23
Stephen23 on 11 Aug 2021
Edited: Stephen23 on 11 Aug 2021
"...how do I accept this comment as an answer."
Or take look at dpb's much better approach.
(assuming that you want neat, simple, very efficient code using basic character code operations)
@Stephen Cobeldick@Ernest Adamtey, Yes, @dpb 's answer is pretty efficient and faster!

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Answers (3)

correspondingChars = 'ABCDEFGHIJ';
str = correspondingChars(arrayfun(@str2double,T,'UniformOutput',true));
No loop needed here -- using relationship to ASCII coding and that MATLAB char() strings are just arrays, write
T='956754674275';
% the engine
C='A'-'0'; % difference offset from '0' to 'A' in code--MATLAB returns 17
str=char(T+C); % convert to char() from double of input + offset
As long as the substitution pattern is ordered sequentially, the above works no matter what the initial coding point is simply by changing the target reference in computing the offset constant, C. If it could be a variable, then create a lookup table and index into it instead.
Above produces--
>> T='956754674275';
>> C='A'-'0'
C =
17
>> char(T+C)
ans =
'JFGHFEGHECHF'
>> all(ans==str)
ans =
logical
1
>>
According to @dpb's answer, here I provide a more general answer!
T='956754674275';
encode = 'ABCDEFGHIJ';
str = encode(double(T)-double('0')+1)
then
str =
'JFGHFEGHECHF'

1 Comment

Yes, that's the lookup table version mentioned.
NB: that the above can be more succinctly written as
str=encode(T-'0'+1);
or
str=encode(T-'/');

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Asked:

on 11 Aug 2021

Commented:

dpb
on 12 Aug 2021

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