How to pass a struct array to another struct array?

Question

I am trying to pass a struct array to another without success.

s(1).f = (1:3);
s(2).f = (4:6);
s(3).f = (7:9);

q(1).n = 'nameA';
q(2).n = 'nameB';
q(3).n = 'nameC';
q(3).f = [];

q.f = s.f

The field n shouldn't be modified.

Do I miss something?

Answer 1

How about arrayfun

q = arrayfun( @(x,y) setfield( x, 'f', y.f ), q, s );

Apparently setfield is only for setting one struct element in struct array -- thus the arrayfun .

EDIT:
a much better answer given by Dan .

Answer 2

You can assign each field of an array of structs directly to a cell array and then you can use deal to convert a struct to a cell array:

s(1).f = (1:3);
s(2).f = (4:6);
s(3).f = (7:9);

q(1).n = 'nameA';
q(2).n = 'nameB';
q(3).n = 'nameC';

c = cell(3,1);
[c{:}] = deal(s.f);
[q.f] = c{:};

Here is a good article on this sort of thing

Edit: Or as Shai points out you can just go

[q.f] = s.f

Answer 3

我想让Shai的建议更加明显，因为它更容易阅读。

[qf] = sf

Answer 4

Though I'd say @Dan's answer is pretty canon for this question, I'd like to present an alternative:

s(1).f = (1:3);
s(2).f = (4:6);
s(3).f = (7:9);

[q(1:length(s)).f] = s.f;

Though slightly more verbose than the [qf] = sf syntax, it has the advantage of functioning as expected even if q has not been preallocated to the correct size to be a copy of s .

For example:

s(1).f = (1:3);
s(2).f = (4:6);
s(3).f = (7:9);

[q.f] = s.f;

Returns qf as a 1x1 struct equal to s(1).f