Access value in cell arrays

A{1,1}.str = 1;
A{2,1}.str = 2;
... (so on)
A{10,1}.str = 10;

Can I say:

B = A{:,1}.str;

so that:

B=[1 2 3 4 5 6 7 8 9 10];

Thanks very much

[per isakson]

Is A supposed to be a cell arrays of structures?

A field named "str" holding a numerical value isn't that confusing?

[TN]

Yes... A is a cell arrays. Each cell array is a structure.

'str' is just a name. It does not mean to be a string. Sorry for the confusion.

Answer 1

[per isakson]

What you look for can be achieved with cellfun, see below.

Indexing like C{:}.field is not supported (AFAIK).

The script

    S1.field=1;
    S2.field=2;
    S3.field=3;
    C = { S1, S2, S3 };
    C{1}.field
    C{2}.field
    C{3}.field
    C{:}.field

returns

    ans =
         1
    ans =
         2
    ans =
         3
    Bad cell reference operation.

And

    vec = cellfun( @(S) S.field, C, 'uni', true )

returns

    ans =
         1     2     3

.

EDIT

The single command with cellfun is justified in case the cells of the cell array contain structures with only some fields in common.

Example:

    S1.field=1;
    S2.field=2;
    S3.field=3;
    S1.field1=1;
    S2.field2=2;
    S3.field3=3;
    C = { S1, S2, S3 };
    vec = cellfun( @(S) S.field, C, 'uni', true )

returns

    vec =
         1     2     3

[TN]

Thanks for your response. I am looking for a single command like the way Matlab handling the normal matrix array instead of using the "for" loop.

[Matt J]

per gave you a single command solution

 vec = cellfun( @(S) S.field, C, 'uni', true )

[TN]

Thanks very much, Per and Matt. That's it. Sorry... somehow i didn't see the few last lines from "Per". You are all genious!

[Matt J]

You can Accept-click per's Answer, then, since that was what you were looking for.

Answer 2

[Matt J]

No. You can do this instead

   A(1).str = 1;
   A(2).str = 2;
   ... 
   A(10).str = 10;

   B=[A(:).str]

[TN]

Hi Matt, Thanks for your response... but my problem is when A is a cell "{}" array (not a matrix "()" array).

[Matt J]

It would not make sense to hold structures having the same fields inside cells. It just makes them harder to get to (as you've discovered).

[per isakson]

I agree.

However, for some reason the cell array may contain structures with only some fields in common.

Answer 3

[James Tursa]

Another variation:

x = [A{:,1}];
B = [x.str];

[TN]

Yes... thanks very much, James. Your solution works great for me too!

[Matt J]

TN, if James solution works for you, there is really no reason to be carrying around A. You may as well just use x. As per said, it might make sense if A{i} were structs with different fields, but James' approach will not work if that is the case.