Thread Subject: Basic Matlab Question

I have a 32 x 1300 2D array, lets call it S. I know I can pick off a particular row with the following argument: A = S(1,:); which gives me a 1 x 1300 array from the first
row of the 2D array. I want every row assigned to a new indexed array, so I would start off with

for n = 1:rows
A(n) = S(n,:);
end

This doesn't work because of a dimensional mismatch. I'm not really trying to define
a dimension but simply declare a unique identifier. How am I able to assign rows
such that I can put them in A(1), A(2), etc., all with the dimension 1 x 1300.

Thanks!

On Nov 16, 3:46 pm, "James Carter" <jim...@msn.com> wrote:
> I have a 32 x 1300 2D array, lets call it S.  I know I can pick off a particular row with the following argument: A = S(1,:); which gives me a 1 x 1300 array from the first
> row of the 2D array.  I want every row assigned to a new indexed array, so I would start off with
>
> for n = 1:rows
> A(n) = S(n,:);
> end
>
> This doesn't work because of a dimensional mismatch.   I'm not really trying to define
> a dimension but simply declare a unique identifier.   How am I able to assign rows
> such that I can put them in A(1), A(2), etc., all with the dimension 1 x 1300.  
>
> Thanks!

So essentially you want A to be a 32x1 vector? You're trying to
access single elements of A, yet you're trying to put a 1x1300 length
vector into that single element.

You would either need to do A(n,:) = S(n, :) (which is pointless
because then A=S), or you need to make A either a structure or cell
array. That would A could be a 32x1 cell array, where each cell is
then a 1x1300 vector.

"James Carter" <jimctr@msn.com> wrote in message <hdso7p$sh5$1@fred.mathworks.com>...
> I have a 32 x 1300 2D array, lets call it S. I know I can pick off a particular row with the following argument: A = S(1,:); which gives me a 1 x 1300 array from the first
> row of the 2D array. I want every row assigned to a new indexed array, so I would start off with
>
> for n = 1:rows
> A(n) = S(n,:);
> end
>
> This doesn't work because of a dimensional mismatch. I'm not really trying to define
> a dimension but simply declare a unique identifier. How am I able to assign rows
> such that I can put them in A(1), A(2), etc., all with the dimension 1 x 1300.
>
> Thanks!

How about a cell array:
 for n = 1:rows
 A{n} = S(n,:);
 end

Use a cell array instead.
A{n} = S(n,:)

or easier:


A = [magic(3),magic(3)]
B = mat2cell(A,ones(1,size(A,1)),size(A,2))

"Matt Fetterman" <mattinjersey@yahoo.com> wrote in message <hdsp40$mg5$1@fred.mathworks.com>...
> "James Carter" <jimctr@msn.com> wrote in message <hdso7p$sh5$1@fred.mathworks.com>...
> > I have a 32 x 1300 2D array, lets call it S. I know I can pick off a particular row with the following argument: A = S(1,:); which gives me a 1 x 1300 array from the first
> > row of the 2D array. I want every row assigned to a new indexed array, so I would start off with
> >
> > for n = 1:rows
> > A(n) = S(n,:);
> > end
> >
> > This doesn't work because of a dimensional mismatch. I'm not really trying to define
> > a dimension but simply declare a unique identifier. How am I able to assign rows
> > such that I can put them in A(1), A(2), etc., all with the dimension 1 x 1300.
> >
> > Thanks!

When I attempt your suggestion using the Cell Array A{n} I get the following error

??? Cell contents assignment to a non-cell array object.

>
> How about a cell array:
> for n = 1:rows
> A{n} = S(n,:);
> end

"James Carter" <jimctr@msn.com> wrote in message <hdso7p$sh5$1@fred.mathworks.com>...
> I have a 32 x 1300 2D array, lets call it S. I know I can pick off a particular row with the following argument: A = S(1,:); which gives me a 1 x 1300 array from the first
> row of the 2D array. I want every row assigned to a new indexed array, so I would start off with
>
> for n = 1:rows
> A(n) = S(n,:);
> end
>
> This doesn't work because of a dimensional mismatch. I'm not really trying to define
> a dimension but simply declare a unique identifier. How am I able to assign rows
> such that I can put them in A(1), A(2), etc., all with the dimension 1 x 1300.
>
> Thanks!

By definition A will be multidimensional since you have A(x) and each A(x) has a length of 1300.

If you need 1 x 1300 vectors you could use a unique name, i.e. A1=..., A2=...

for n = 1:rows
  q=['A' num2str(n) '= S(', num2str(n), ',:);'];
  eval(q);
end

or maybe use a cell array where each cell contains a 1x1300 array

c=cell(32,1);
for n = 1:rows
  c{n}=S(n,:);
end

"Doug " <nospam@thx.com> wrote in message <hdsqh1$flv$1@fred.mathworks.com>...
> "James Carter" <jimctr@msn.com> wrote in message <hdso7p$sh5$1@fred.mathworks.com>...
> > I have a 32 x 1300 2D array, lets call it S. I know I can pick off a particular row with the following argument: A = S(1,:); which gives me a 1 x 1300 array from the first
> > row of the 2D array. I want every row assigned to a new indexed array, so I would start off with
> >
> > for n = 1:rows
> > A(n) = S(n,:);
> > end
> >
> > This doesn't work because of a dimensional mismatch. I'm not really trying to define
> > a dimension but simply declare a unique identifier. How am I able to assign rows
> > such that I can put them in A(1), A(2), etc., all with the dimension 1 x 1300.
> >
> > Thanks!

That did the trick. Thanks
>
> By definition A will be multidimensional since you have A(x) and each A(x) has a length of 1300.
>
> If you need 1 x 1300 vectors you could use a unique name, i.e. A1=..., A2=...
>
> for n = 1:rows
> q=['A' num2str(n) '= S(', num2str(n), ',:);'];
> eval(q);
> end
>
> or maybe use a cell array where each cell contains a 1x1300 array
>
> c=cell(32,1);
> for n = 1:rows
> c{n}=S(n,:);
> end

DON'T use eval. Use mat2cell like I showed above.


http://matlabwiki.mathworks.com/MATLAB_FAQ#How_can_I_create_variables_A1.2C_A2.2C....2CA10_in_a_loop.3F

"Doug " <nospam@thx.com> wrote in message
news:hdsqh1$flv$1@fred.mathworks.com...
> "James Carter" <jimctr@msn.com> wrote in message
> <hdso7p$sh5$1@fred.mathworks.com>...
>> I have a 32 x 1300 2D array, lets call it S. I know I can pick off a
>> particular row with the following argument: A = S(1,:); which gives me a
>> 1 x 1300 array from the first
>> row of the 2D array. I want every row assigned to a new indexed array,
>> so I would start off with
>>
>> for n = 1:rows
>> A(n) = S(n,:);
>> end
>>
>> This doesn't work because of a dimensional mismatch. I'm not really
>> trying to define
>> a dimension but simply declare a unique identifier. How am I able to
>> assign rows
>> such that I can put them in A(1), A(2), etc., all with the dimension 1 x
>> 1300.
>>
>> Thanks!
>
> By definition A will be multidimensional since you have A(x) and each A(x)
> has a length of 1300.
>
> If you need 1 x 1300 vectors you could use a unique name, i.e. A1=...,
> A2=...

*snip*

NO. Do NOT use EVAL like this. See Q4.6 in the newsgroup FAQ for an
explanation why this is a Bad Idea.

> or maybe use a cell array where each cell contains a 1x1300 array
>
> c=cell(32,1);
> for n = 1:rows
> c{n}=S(n,:);
> end

To the OP: I would instead just use S(n, :) wherever in your code you would
have used A1, A2, ... or A(1), A(2), ... IMO it's no more difficult to read
than A1, A2, A(1), A(2), etc. and it avoids creating lots of smaller
variables in your workspace.

--
Steve Lord
slord@mathworks.com
comp.soft-sys.matlab (CSSM) FAQ: http://matlabwiki.mathworks.com/MATLAB_FAQ

...
> NO. Do NOT use EVAL like this. See Q4.6 in the newsgroup FAQ for an
> explanation why this is a Bad Idea.
>
...
> To the OP: I would instead just use S(n, :) wherever in your code you would
> have used A1, A2, ... or A(1), A(2), ... IMO it's no more difficult to read
> than A1, A2, A(1), A(2), etc. and it avoids creating lots of smaller
> variables in your workspace.
>
> --
> Steve Lord
> slord@mathworks.com
> comp.soft-sys.matlab (CSSM) FAQ: http://matlabwiki.mathworks.com/MATLAB_FAQ
>

From the FAQ mentioned, "Now, if you still really want to create variables with dynamically generated names, you need to use EVAL. With EVAL, you use MATLAB commands to generate the string that will perform the operation you intend."

I agree this is not the best way to go about things and I wouldn't do it, but the FAQ itself spells out how to do what the OP wants to do using EVAL.

My advice to the OP is use what works for you to solve your problem. When you get more experience with Matlab, follow Steve's advice above and use S(n, :) instead of creating more variables.

"Doug " <nospam@thx.com> wrote in message
> From the FAQ mentioned, "Now, if you still really want to create variables with dynamically generated names, you need to use EVAL. With EVAL, you use MATLAB commands to generate the string that will perform the operation you intend."
>
> I agree this is not the best way to go about things and I wouldn't do it, but the FAQ itself spells out how to do what the OP wants to do using EVAL.
>
> My advice to the OP is use what works for you to solve your problem. When you get more experience with Matlab, follow Steve's advice above and use S(n, :) instead of creating more variables.


I would say that this last bit is backwards. EVAL should be unknown to beginners and rarely used by advanced users. When a beginner starts using eval, he will come back to this NG with one (or more) of the following complaints:

1. My code is very slow, help me speed it up.
2. My code doesn't work and I cannot figure out why.
3. MATLAB closed by itself and I don't know why.

I have seen it again and again. EVAL has its place, but IMO, it is not in a beginner's hands.

"Matt Fig" <spamanon@yahoo.com> wrote in message
> >
> > My advice to the OP is use what works for you to solve your problem. When you get more experience with Matlab, follow Steve's advice above and use S(n, :) instead of creating more variables.
>
>
> I would say that this last bit is backwards. EVAL should be unknown to beginners and rarely used by advanced users. When a beginner starts using eval, he will come back to this NG ....

My advise was to "use what works for you to solve your problem". How could you possibly think that is backwards? If the OP gets results today he/she will be inclined to use Matlab again. If that person is scolded for not thinking like an experienced programmer they will avoid Matlab and this NG.

If that person runs into a problem later and comes back to this NG and gets more advice and learns more that way... why isn't that a good thing?

Doug wrote:
...
> My advise was to "use what works for you to solve your problem".
> How could you possibly think that is backwards? If the OP gets
> results today he/she will be inclined to use Matlab again. If that
> person is scolded for not thinking like an experienced programmer
> they will avoid Matlab and this NG.
>
> If that person runs into a problem later and comes back to this NG
> and gets more advice and learns more that way... why isn't that a
> good thing?

I'd say it's not a good thing to reinforce poor practice initially as
what one learns first is the hardest to unlearn.

--