Thread Subject:

Indexing matrix with dependencies

To the MATLAB community,

I'm a novice to the MATLAB language and its many functions and so on, and have had difficulty finding much relevant information on matrix indexing with 'dependencies'.

I was wondering if anyone knew of a function call(s) that would effectively solve the following.

I have a matrix of values A (for example, 3 columns by 6 rows):

A =

1 1 3
1 0 4
1 1 1
1 1 2
2 0 1
2 1 4

I need to print to B a list (of counts or of values) of the third column dependent on the first or the second column, or both. For example, could I get printed, for values of 1 in column 1, the values of the third column dependent on, say, zero in the second?

So B looks like:

3 1 2 4

There is probably an easy solution with logical indexing, but I'm more than a little lost in the logic (so to speak).

Thanks!

Matt

On 7/19/2012 6:11 AM, Matthew Hyett wrote:
> To the MATLAB community,
>
> I'm a novice to the MATLAB language and its many functions and so on,
> and have had difficulty finding much relevant information on matrix
> indexing with 'dependencies'.
>
> I was wondering if anyone knew of a function call(s) that would
> effectively solve the following.
>
> I have a matrix of values A (for example, 3 columns by 6 rows):
>
> A =
>
> 1 1 3
> 1 0 4
> 1 1 1
> 1 1 2
> 2 0 1
> 2 1 4
>
> I need to print to B a list (of counts or of values) of the third column
> dependent on the first or the second column, or both. For example, could
> I get printed, for values of 1 in column 1, the values of the third
> column dependent on, say, zero in the second?
>
> So B looks like:
>
> 3 1 2 4
...

Well, I don't see how B follows from your description at all--the only
value where A(:,1)==1 & A(:,2)==0 --> A(:,3)==4

For your result vector A(:,[1:2]) looks like

A =
      1 1
      1 0
      1 1
      2 1
 >>

The problem as described --

 >> A=[1 1 3;1 0 4; 1 1 1; 1 1 2;2 0 1;2 1 4];
 >> B=A(ismember(A(:,[1:2]),[1 0],'rows'),3)
B =
      4
 >> inSet = sum(ismember(A(:,[1:2]),[1 0],'rows'))
ans =
      1
 >>

Salt conditions to suit...

Look at the doc for

doc ismember % and friends for such comparisons/selection processes

--

"Matthew Hyett" <m.hyett@unsw.edu.au> wrote in message
news:ju8q0d$nnn$1@newscl01ah.mathworks.com...
> To the MATLAB community,
>
> I'm a novice to the MATLAB language and its many functions and so on, and
> have had difficulty finding much relevant information on matrix indexing
> with 'dependencies'.
>
> I was wondering if anyone knew of a function call(s) that would
> effectively solve the following.
>
> I have a matrix of values A (for example, 3 columns by 6 rows):
>
> A =
>
> 1 1 3
> 1 0 4
> 1 1 1
> 1 1 2
> 2 0 1
> 2 1 4
>
> I need to print to B a list (of counts or of values) of the third column
> dependent on the first or the second column, or both. For example, could
> I get printed, for values of 1 in column 1, the values of the third column
> dependent on, say, zero in the second?

I think you mean "one in the second", right?

Let's build the indexing expression. You want to grab all elements from
column 3 of A in rows that meet a certain criterion.

B = A(something, 3)

That criterion is that the corresponding elements in column 2 of A are equal
to 1, if I understand what you're asking.

something = A(:, 2) == 1

Read A(:, 2) as the intersection of all rows of A and column 2 of A.
Similarly something like A([2 4], [2 3]) would be the elements at the
intersections of rows 2 and 4 of A and columns 2 and 3 of A, or in this case
[0 4; 1 2].

So your indexing expression is:

B = A(A(:, 2) == 1, 3)

> So B looks like:
>
> 3 1 2 4

Since you're extracting from a single column in the indexing expression
above, B comes out as a column vector; if you need it to be a row vector,
transpose it.

--
Steve Lord
slord@mathworks.com
To contact Technical Support use the Contact Us link on
http://www.mathworks.com

"Matthew Hyett" <m.hyett@unsw.edu.au> wrote in message <ju8q0d$nnn$1@newscl01ah.mathworks.com>...
> To the MATLAB community,
>

Thanks to you both. Yes, Steven, I meant "one in the second." - apologies (sleepless).
I will have a try at these suggestions.

Best,
Matt