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April 22nd, 2008

MATLAB Basics: Array of structures vs Structures of arrays

This short video covers the difference between a “structure of arrays” and an “array of structures”.

By Doug Hull

12:28 UTC | Posted in MATLAB Basics, PodCast, Video | Permalink | 20 Comments »

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20 Responses to “MATLAB Basics: Array of structures vs Structures of arrays”

Emil replied on April 22nd, 2008 at 13:54 UTC :

It’s easier to create vector s by:
s = [people.SimmNum];
Tom replied on April 22nd, 2008 at 16:29 UTC :

This and Emil’s comment were quite helpful to me.

To take this one step further, one advantage of Arrays of Structures would be the ability to have vectors of different lengths for each numerical part, and a cell array of names of different lengths for each name part.

For example:

people(1).name=’Erin’
people(2).name={‘Joan’ ‘Carla’}
people(1).sim=44
people(2).sim=[561 49]
Doug replied on April 23rd, 2008 at 08:01 UTC :

Emil,

Yes, you are correct. I can not believe I over looked that. There are so many ways to do certain things in MATLAB.

Doug
Ryan Gray replied on April 24th, 2008 at 14:31 UTC :

If you have an array that you want to assign to a field across a structure array, you can do this:

[structArray(:).fieldname] = deal(num2cell(array))

Of course, the length of array should be the same as the length of structArray, but if structArray doesn’t exist yet, then you can do:

[structArray(1:length(array)).fieldname] = deal(num2cell(array))

If array is already a cell array, you can avoid using both deal and num2cell (although they will still work) and use:

[structArray(:).fieldname] = array{:}
or
[structArray(1:length(array)).fieldname] = array{:}

An example, modifying the struct array ‘people’ from the video:

names = {‘Larry’,'Moe’,'Curly’};
[people(:).name] = names{:};
liuzhi replied on May 14th, 2009 at 13:36 UTC :

In one of my program, I’d like to use an array of array of structures, such like
```
s0 = struct('field1', [v1 v2], 'field2', [v3 v4]);
s1 = struct('field1', [v1' v2'], 'field2', [v3' v4']);
```
I want another data structure s = [s0, s1] so that I can access s0(1) like s(0,1). How can I do that?
Doug replied on May 14th, 2009 at 19:19 UTC :

You want to use a cell array here:
```
s{1} = s0
s{2} = s1
s{1}.field1
s{1}.field1(1,3)
```
Doug
Dan replied on August 4th, 2009 at 19:55 UTC :

I may be wrong, but one of the examples in Post 4 (Ryan Gray) doesn’t work as advertised. Specifically,
```
[structArray(1:length(array)).fieldname] = deal(num2cell(array))
```
ends up assigning the whole array to every instance of fieldname, instead of distributing the array across those instances. What does seem to work is:
```
cell_array = num2cell(array);
[structArray(1:length(array)).fieldname] = cell_array{:};
```
I leave it as an exercise for a MATLAB guru to splain why this is the case.
Sandy replied on December 11th, 2009 at 20:39 UTC :

Hi

I’m new at this –

How do I save an array of structures to a .mat file so that I can use it later? I’m able to “save” it and then “load” it into a different script, but not able to unpack the components.

Thanks,

Sandy
Doug replied on December 14th, 2009 at 16:11 UTC :

@Sandy,

What do you mean by “unpack” the components?

Doug
Halfdan replied on March 2nd, 2010 at 03:19 UTC :

I would also like to know how to save an array of structures, rather than just a single structure. I know that there’s a memory waste connected with array of structs, but I really find array of struct to be a very easy way of organizing certain kinds of data.
Doug replied on March 2nd, 2010 at 18:03 UTC :

@hakdan,

I do not understand your question. What have you tried?

Doug
Emil replied on May 13th, 2010 at 12:00 UTC :

Hei,

I ran into the same problem described by the previous two posts.
```
names = {'Peter','Paul','Mary'};
ages = {25,29,26};
[people(1:3).name] = names{:};
[people(1:3).age] = ages{:};
```
Now
```
save('test','-struct','people')
```
yields:
??? Error using ==> save
The argument to -STRUCT must be the name of a scalar structure variable.

whereas
```
person = people(1);
save('test','-struct','person')
```
is no problem.

The following would work
```
save('test','people')
```
but is very unhandy as
```
x = load('test');
```
makes x a 1×1 struct with field people, and would require additionally
```
x = x.people;
```
So the question is:
What is the best way to save a struct array?

Emil

by the way, if ages was a 1×3 double [25,29,26] and not a 1×3 cell:
How would I deal it across the struct array people? This is the same problem as adressed by Dan on Aug 4th
Doug replied on May 13th, 2010 at 18:20 UTC :

@Emil,

Shy not just save it? Why are you trying to pull them out of the structure?

Doug
Huw replied on May 14th, 2010 at 09:44 UTC :

I also found the same problem with post 4 as described in post 7 and would be grateful if someone could explain why the solution needs to use 2 lines of code. e.g.
```
weather = repmat(struct('temp', 72, 'rainfall', 0.0),1,3);
nums = [23 10 12];
cellArray = num2cell(nums - [weather(:).temp])
[weather(:).rainfall] = cellArray{:}
```
Is there no way to assign each element of the result of num2cell to weather.rainfall on a single line.
annon replied on August 12th, 2011 at 00:39 UTC :

Hi
I’ve come across your blog a few times and have found it very helpful. So let me begin by thanking you.

I have a small question that I’ve not been able to find the solution to online. I’m hoping you may be able to shed some light on the matter.

The following structure has 2 elements.
```
people = struct(...
'name',{'bob', 'john'},...
'numKids',{0, 2}, ...
'kidsage',{[],[12,9]});
```
each element people(1), people(2) has three frields (name, numKids, kidsage)

Instead of declaring the structure in one line as above, I would like to define it one field at a time.
The closest working example I’ve been able to come up with is
```
people = struct

people.('name') = {'bob, 'john'}
people.('numkids') = {0, 2}
people.('kidsage') = {[], [12, 9]}
```
But defined in this way, the people structure only has one element.
ie.
```
people(1) =

name: {'bob' 'john'}
numkids: {[0] [2]}
kidsage: {[] [12, 9]}
```
Any ideas?
Doug replied on August 12th, 2011 at 19:09 UTC :

@Annon,

I do not know of a way to do that. You seem to be able to get the data in in other ways, why is it important to do it in this particular manner?

Doug

Steve L replied on August 19th, 2011 at 13:16 UTC :

Annon,

To do this you can use DEAL, but I think Doug’s question is a good one: why do you need/want to do it this way?

% Make sure people2 doesn't exist yet
% This way you can tell that the command performed
% the initial creation correctly
clear people2

% Define the cell array of names
S = {'bob', 'john'};

% DEAL the elements of the cell into the struct
[people2(1:2).name] = deal(S{:})

If you’re using release R14 or later, you can omit the DEAL as described in the Release Notes.

Omar replied on September 27th, 2011 at 21:36 UTC :

I want to have a .wav file into a vector so I can create de Spectra with the SPtool.

When I use the wavread() function and import the data into sptool it say thats an [array] so i couldnt create the spectrum, this tool only work with vectors. So how can i turn this array into vector?

a little help here please
nick replied on October 13th, 2011 at 15:43 UTC :

hi
i am new in MATLAB, can u plz tell whats d problem with the following statements (i need to assign an array of integers to a member of an array of structures )
n(1).loc=[x-1 y+1];
n(2).loc=[x y+1];
n(3).loc=[x+1 y+1];
n(4).loc=[x+1 y];
n(5).loc=[x+1 y-1];
why are these assignments wrong?? plz reply ASAP… thnx
Paul Torek replied on December 19th, 2011 at 15:45 UTC :

Thankyou thankyou thankyou!

You gave me the direction I needed to extract relevant info from the matlab dir filename command. Such as:

function Subdirs = OneLevelSubDirs(directory,match)
% xxx Need Help xxx
% Subdirs = OneLevelSubDirs(directory,match)

if nargin < 2 || isempty(match)
match = '*';
end

X = dir(directory);

Xdir = {X(:).name};
Xdir = Xdir(find(cell2mat({X(:).isdir})));
Subdirs = Xdir(~cellfun(@(x)ismember(x,{'.';'..'}),Xdir));

Did I mention thank you?