thanks for the very nice video…

Rushikesh Borse

The way you’ve written your code, the value alpha only depends on whether or not the last element of d is less than 1.5. Anyway, using “logical indexing” (you can search the documentation for this term for more information if you’re interested) you can operate on a subset of the elements all at once.

d = 1:0.1:2;
% Create a "logical mask"
smallEnough = d < 1.5;
% Raise all elements of d to the 3rd power
k = d.^3;
% Square those elements that are "small enough" using the mask
k(smallEnough) = d(smallEnough).^2;

If you want alpha to be a vector the same size as d you can do what I did above (predefine it to be a vector of the same size as d with a “default” value then fill in certain elements to override that default) or you can do something slightly different:

alpha = zeros(size(d));
alpha(smallEnough) = 2;
alpha(~smallEnough) = 4;

Note that I negated the smallEnough mask to set elements of alpha corresponding to those elements of d that were NOT smallEnough to 4. You can also AND (&) or OR (|) those logical masks together to form more complicated masks for use with logical indexing.

d = 1:0.1:2
for h = 1:length(d)
if d(h)(1,:)&it 1.500
alpha = 2
else
alpha = 4
end
k(h)(1,:) = d(h)(1,:)^2
end

Thank you, your help will be much appreciated.

Your code simplified looks like this:

While 1
   do something
end

The boolean in the while statement is ALWAYS true, that is why you never leave the loop.

Doug

n=2;
for i=1:n,
disp(‘types of things’)
disp(’1 – thing one’)
disp(’2 – thing two’)
disp(’3 – thing 3′)
disp(’4 – thing 4′)
opcomb=input(strcat (‘what’s the number ‘,num2str(i),’thing:’,32));
while ((opcomb ~=1)|(opcomb~=2)|(opcomb~=3)||(opcomb~=4))
switch opcomb
case 1
combst(i)= 1;
case 2
combst(i)=2;
case 3
combst(i)=3;
case 4
combst(i)=4;
otherwise
disp(‘wrong answer’)
opcomb=input(strcat (‘what’s the number ‘,num2str(i),’thing:’,32));
end
end
end

what its doing is after choosing in the first opcomb and then entering the switch it doesnt “come out”…i’m at a loss as to where my problem is, if it’s in the switch or in the conditions i set in whyle.
thank you in advance

I don’t know, but one quick test is worth a thousand expert opinions! :)

I would first get the code working and then profile to find the bottle necks. They are rarely where you expect them.

Doug