Setting the properties of an Object Array

2012 10 1
2 个回答
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1 个投票

The documentation and experimentation make it clear that I can do this to get the property values of an array of objects:
Array_of_property_values=[Object_Array(:).Property];
The question is how can I do the reverse without major use of subsasgn. For example I want to be able to do something like the following
[Object_Array(:).Property]=Array_of_property_values_to_assign;
I just can't seem to make that work. I have tried many different permutations but nothing seems to work. Does any on have any thoughts on how to make that work.
I have tried the following with no success:
[Object_Array(1:2).Property]=[10,20];
% suggestion from Stackoverflow
[Object_Array(:).Property]=deal(Array_of_property_values_to_assign);

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 采纳的回答

Matt J
Matt J 2012-10-1

1 个投票

args=num2cell(Array_of_property_values_to_assign);
[Object_Array(:).Property]=deal(args{:});

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Nicholas Dinsmore
Nicholas Dinsmore 2012-10-1
Matt, That is great thank you. Is there a way to do that in a single line of code?
Matt J
Matt J 2012-10-1
编辑：Matt J 2012-10-1
If Array_of_property_values_to_assign is already a cell array, then you can just do
[Object_Array(:).Property]=deal(Array_of_property_values_to_assign{:});
Otherwise, no, there is no way to do it in 1 line, except for fancy subsasgn methods.
per isakson
per isakson 2012-10-3
- or write an appropriate version of deal.

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更多回答（1 个）

Grzegorz Lippe
Grzegorz Lippe 2013-6-11

0 个投票

Hello,
this isn't working for me and I can't figure out what I'm doing wrong:
>> a = num2cell(1:3)
a =
[1] [2] [3]
>>[p(:).h_fP_star] = deal(a{:});
>>[p.h_fP_star]
ans =
3 3 3
Could anybody help?

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