Access value in cell arrays

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TN
TN on 29 Jun 2013
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
  2 Comments
per isakson
per isakson on 29 Jun 2013
Is A supposed to be a cell arrays of structures?
A field named "str" holding a numerical value isn't that confusing?
TN
TN on 29 Jun 2013
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.

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Accepted Answer

per isakson
per isakson on 29 Jun 2013
Edited: per isakson on 29 Jun 2013
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
  4 Comments
TN
TN on 29 Jun 2013
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
Matt J on 29 Jun 2013
You can Accept-click per's Answer, then, since that was what you were looking for.

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More Answers (2)

Matt J
Matt J on 29 Jun 2013
Edited: Matt J on 29 Jun 2013
No. You can do this instead
A(1).str = 1;
A(2).str = 2;
...
A(10).str = 10;
B=[A(:).str]
  3 Comments
Matt J
Matt J on 29 Jun 2013
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
per isakson on 29 Jun 2013
Edited: per isakson on 29 Jun 2013
I agree.
However, for some reason the cell array may contain structures with only some fields in common.

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James Tursa
James Tursa on 29 Jun 2013
Another variation:
x = [A{:,1}];
B = [x.str];
  2 Comments
TN
TN on 29 Jun 2013
Yes... thanks very much, James. Your solution works great for me too!
Matt J
Matt J on 30 Jun 2013
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.

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