How to use if statement for than one variables?

I would like to set one if statement. I am using the following code:

filename2= 'OutputFile1L.xlsx';

[d2,tex]= xlsread(filename2);

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

What is the wrong? could you please help me?

Answer 1

[Sulaymon Eshkabilov]

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

[Image Analyst]

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]

@Image Analyst Very Good point and discussion! "Not Always" cases are small size simulation problems which require insignifcantly small amount of sim time.

Answer 2

[Yongjian Feng]

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

[Sulaymon Eshkabilov]

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

Answer 3

[Sulaymon Eshkabilov]

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;