fix loop

[Omar Latif]

this loop should give me for -1 0 and for 1 should give me
a 1, can anyone help me. My code is at the bottom
thanks

p=[-1.0000 1.0000 -1.0000 -1.0000 -1.0000 -1.0000 1.0000
1.0000 1.0000 1.0000 1.0000 -1.0000 ];

k=1;
i=0;
D=[];
for i=1:length(p)
a=p(i);
if a==-1.0000

D(1,k)=0;

else

D(1,k)=1;


end

k=k+1;
end
D

[Roger Stafford]

"Omar Latif" <omar-...@hotmail.com> wrote in message <g5tmdc$c3c
$1...@fred.mathworks.com>...

At first glance it should work as you asked, but it depends on how you
obtained the numbers in p. If you had round off errors arriving at the minus
ones, they might not be exactly minus ones and then the loop wouldn't work.

However, it's a awkward way to do the job in any case. You could simply
write:

D = zeros(1,length(p));
D(p>0) = 1;

Roger Stafford

[Roger Stafford]

"Roger Stafford" <ellieandr...@mindspring.com.invalid> wrote in
message <g5toob$si3$1...@fred.mathworks.com>...

In the interest of brevity you could also use the one-liner:

D = +(p>0);

Roger Stafford

[Matt Fig]

Even shorter:

D = p>0

Because

all(p>0==+(p>0))

ans =

1

Or was it just stylistic?

[Roger Stafford]

"Matt Fig" <spam...@yahoo.com> wrote in message <g5ub6f$66c
$1...@fred.mathworks.com>...

I assumed Omar would like to have a numeric type result rather than logical.
(On my own version 4a it makes no difference. It never heard of logical
types.)

Roger Stafford

[Matt Fig]

> > all(p>0==+(p>0))
> >
> > ans =
> > 1
> >
> > Or was it just stylistic?
>
> I assumed Omar would like to have a numeric type result
rather than logical.
> (On my own version 4a it makes no difference. It never
heard of logical
> types.)
>
> Roger Stafford

Thanks Roger, I had never seen that conversion before. I
have found that sometimes it is confusing as to when Matlab
enforces the "This is a logical variable, you cannot use it
as a double" (or the opposite)rule. For example:

p=[-1.0000 1.0000 -1.0000 -1.0000 -1.0000 -1.0000 1.0000
1.0000 1.0000 1.0000 1.0000 -1.0000 ];

p1 = p>0; % logical
p2 = +(p>0); % numeric
% Yet Matlab has no problem with the following:
p1+2.4
% Nor this
p2+true(size(p2))

So the types mix easily and intuitively.

[Roger Stafford]

"Matt Fig" <spam...@yahoo.com> wrote in message <g5vkva$ruo
$1...@fred.mathworks.com>...

> Thanks Roger, I had never seen that conversion before. I
> have found that sometimes it is confusing as to when Matlab
> enforces the "This is a logical variable, you cannot use it
> as a double" (or the opposite)rule. For example:
>
> p=[-1.0000 1.0000 -1.0000 -1.0000 -1.0000 -1.0000 1.0000
> 1.0000 1.0000 1.0000 1.0000 -1.0000 ];
>
> p1 = p>0; % logical
> p2 = +(p>0); % numeric
> % Yet Matlab has no problem with the following:
> p1+2.4
> % Nor this
> p2+true(size(p2))
>
> So the types mix easily and intuitively.

You do have to be careful in certain cases. For example, suppose p =
[5,10]. Then

p(p>0)

produces [5,10] whereas

p(+(p>0))

gives [5,5] as a result.

I guess one lesson here is don't rely on "==" completely for distinguishing
different types.

Roger Stafford