# How to use if statement for than one variables?

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Ivan Mich el 4 de Jul. de 2021
Editada: Sulaymon Eshkabilov el 4 de Jul. de 2021
I would like to set one if statement. I am using the following code:
filename2= 'OutputFile1L.xlsx';
a=d2(:,1);
for ii=1:numel(a)
if a(ii)==0
b(ii)==1 & c(ii)==0
elseif a(ii)==1
b(ii)==0 & c(ii)==0
elseif a==2
b(ii)==0 & c(ii)==0
else
b(ii)==nan & c(ii)==nan
end
but command window shows me
Undefined function or variable b
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Sulaymon Eshkabilov el 4 de Jul. de 2021
Editada: Sulaymon Eshkabilov el 4 de Jul. de 2021
Note that for loop and "if" based calcs are very slow and inefficient.
Anyhow if indeed you wish to fix your code, then check these corrected errs with ==, & |, , e.g:
for ii=1:numel(a)
if a(ii)==0
b(ii)=1; c(ii)=0;
elseif a(ii)==1 | a(ii)==2
b(ii)=0; c(ii)=0;
else
b(ii)=nan; c(ii)=nan;
end
end
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Image Analyst el 4 de Jul. de 2021
Editada: Image Analyst el 4 de Jul. de 2021
For loops are not always slower than vectorized code - that's a myth from long ago. I ran your code both ways:
d2 = rand(1000, 2);
a=d2(:,1);
% Vectorized method:
tic
Idx=find(a(:)==0 | a(:)==1 | a(:)==2);
Idx2=find(a~=0 & a~=1 & a~=2);
b(Idx)=0;
b(Idx2)=nan;
c=b;
elapsedTime1 = toc
% For loop method.
tic
for ii=1:numel(a)
if a(ii)==0 | a(ii)==1 | a(ii)==2
b(ii)=0;
else
b(ii)=nan;
end
end
c=b;
elapsedTime2 = toc
I find:
elapsedTime1 =
4.8e-05
elapsedTime2 =
1.5e-05
so your for loop code takes less than a third of the time of your vectorized code.
Of course the two codes don't do exactly the same thing. If I make them more efficient and do the comparison 1000 times, with this code:
clc; % Clear the command window.
close all; % Close all figures (except those of imtool.)
clear; % Erase all existing variables. Or clearvars if you want.
workspace; % Make sure the workspace panel is showing.
format long g;
format compact;
% Initialize variables:
a = rand(1000, 1);
b = nan(size(a));
numTrials = 1000;
vecWon = 0;
for t = 1 : numTrials
% Find where a is 0, 1, or 2 and set it = 0 there.
% Try it vectorized first.
vectorizedStartTime = tic;
linearIndexes = find(a(:)==0 | a(:)==1 | a(:)==2);
b(linearIndexes)=0;
elapsedTimeVectorized = toc(vectorizedStartTime);
% Now try it with a for loop
b = nan(size(a)); % Reset
forStartTime = tic;
for k = 1 : numel(a)
if a(k)==0 || a(k)==1 || a(k)==2
b(k)=0;
else
b(k)=nan;
end
end
elapsedTimeFor = toc(forStartTime);
fprintf('At iteration %d\n Vectorized took %.7f seconds.\n For loop took %.7f seconds.\n', ...
k, elapsedTimeVectorized, elapsedTimeFor);
% See who was faster.
if elapsedTimeVectorized < elapsedTimeFor
vecWon = vecWon + 1;
end
end
vecPercentage = 100 * vecWon / numTrials;
forPercentage = 100 * (numTrials - vecWon) / numTrials;
fprintf('--------------------------------------------------\n');
fprintf('Vectorized won %3d times out of %d = %.1f%%.\nFor loop won %3d times out of %d = %.1f%%.\n', ...
vecWon, numTrials, vecPercentage, (numTrials - vecWon), numTrials, forPercentage);
We see
--------------------------------------------------
Vectorized won 182 times out of 1000 = 18.2%.
For loop won 818 times out of 1000 = 81.8%.
Now if we improve it even further for the vectorized code by using logical indexes rather than linear indexes:
logicalIndexes = a==0 | a==1 | a==2;
b(logicalIndexes) = 0;
we see that vectorized wins the majority of the time, but not always:
--------------------------------------------------
Vectorized won 903 times out of 1000 = 90.3%.
For loop won 97 times out of 1000 = 9.7%.
Sulaymon Eshkabilov el 4 de Jul. de 2021
@Image Analyst Very Good point and discussion! "Not Always" cases are small size simulation problems which require insignifcantly small amount of sim time.

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### Más respuestas (2)

Yongjian Feng el 4 de Jul. de 2021
You meant this?
if a(ii)==0
b(ii)=1;
c(ii)=0;
elseif a(ii)==1
b(ii)=0;
c(ii)=0;
elseif a==2
b(ii)=0;
c(ii)=0;
else
b(ii)=nan;
c(ii)=nan;
end
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Sulaymon Eshkabilov el 4 de Jul. de 2021
Editada: Sulaymon Eshkabilov el 4 de Jul. de 2021
There are a couple of ERRs in Feng's proposed code:
for ... % is missing
...
if ..
elseif a==2 % MUST be a(ii)==2
...
end
end

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Sulaymon Eshkabilov el 4 de Jul. de 2021
Editada: Sulaymon Eshkabilov el 4 de Jul. de 2021
There are several errs in your loop code that is not advised. Just because it is slow and inefficient.
Here is an easy sol isntead of loop based code:
...
a=d2(:,1);
Idx0=find(a(:)==0);
Idx1=find(a(:)==1 | a(:)==2);
Idx2=find(a~=0 & a~=1 & a~=2);
b(Idx0)=1; c(Idx0)=0;
b(Idx1)=0; c(Idx1)=1;
b(Idx2)=nan; c(Idx2)=nan;
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