Array of struct manipulating

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Majed
Majed 2011-11-20
I have array of struct
x.a
x.b
x.c
I want to do the following two operations
  1. x(:).a = x(:).a +10
  2. x(:).c = x(:).a / x(:).b
for all element without using for loop
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Majed
Majed 2011-11-21
a is a struct member of type integer (I was mistaken )
and x(:) is the whole array of struct
Jan
Jan 2011-11-22
@Majed: Why do you want to avoid the FOR loop?! It would be much nicer and most likely more efficient than creating intermediate vectors, split them to a cell and distribute it over different fields again. The elements are independent from eachotehr, the calculations are independent, so a FOR loop is the straightest method.

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回答（4 个）

David Young
David Young 2011-11-20
It depends whether your data is a structure of arrays or an array of structures. See this video for the difference (and google matlab structures and arrays for more information).
If you choose to use a structure of arrays, you can just do
x.a = x.a + 10;
etc. If you use an array of structures, it's more fiddly.
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Majed
Majed 2011-11-21
I mentioned in the question that I have array of struct,
I have no solution for that !!

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Jan
Jan 2011-11-21
Unfortunately you did not answer my question about the size and type of the fields and the dimensions of the struct. But let me guess at least - perhaps this helps:
x(1).a = 2;
x(2).a = 3;
x(1).b = 17;
x(2).b = 23;
result = [x.a] ./ [x.b]
  1 个评论
Majed
Majed 2011-11-21
That did not work for my case,
I need to assign this to an existing element in the struct

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Fangjun Jiang
Fangjun Jiang 2011-11-21
%%construct a structure array
M=3;
s=struct('a',repmat({1},M,1),'b',repmat({2},M,1),'c',repmat({3},M,1));
%process
N=length(s);
temp=[s.a]+10;
temp=mat2cell(temp,1, ones(N,1));
[s.a]=temp{:};
temp=[s.a]./[s.b];
temp=mat2cell(temp,1, ones(N,1));
[s.c]=temp{:};
Jan
Jan 2011-11-22
Fangjun's approach is nice and works without loops. But creating the temporary arrays wastes a lot of time. A simple FOR loop would be more efficient:
M = 3;
s0 = struct('a', repmat({1},M,1), 'b',repmat({2},M,1), 'c',repmat({3},M,1));
tic
for k = 1:1000
s = s0;
N = length(s);
temp = [s.a]+10;
temp = mat2cell(temp,1, ones(N,1));
[s.a] = temp{:};
temp = [s.a]./[s.b];
temp = mat2cell(temp,1, ones(N,1));
[s.c]= temp{:};
end
toc
% Elapsed time is 0.237474 seconds.
tic
for k = 1:1000
s = s0;
for i = 1:numel(s)
s(i).a = s(i).a + 10;
s(i).c = s(i).a / s(i).b;
end
end
toc
% Elapsed time is 0.024864 seconds.
For this tiny data set the FOR loop is 10 times faster. For larger data with N=300, I get at least a speedup factor of 3.
  1 个评论
Fangjun Jiang
Fangjun Jiang 2011-11-22
Good point, Jan. However, when I changed M to be 300. The result is opposite.
Elapsed time is 3.575805 seconds.
Elapsed time is 5.151734 seconds.
Elapsed time is 3.550913 seconds.
Elapsed time is 5.179092 seconds.
Elapsed time is 3.552488 seconds.
Elapsed time is 5.137140 seconds.

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