How do I add a value to a field in each element of a struct array?

Question

Brandon Kuczenski 2014-4-29

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/127661-how-do-i-add-a-value-to-a-field-in-each-element-of-a-struct-array

评论： Image Analyst 2014-4-29

在 MATLAB Online 中打开

Say I have a struct array with a numeric field:

>> a=struct('bar',{47 52})

I want to increment that field across each element of the array, something like the following:

>> [a.bar] = [a.bar] + 1; % this fails

The nearest I can figure out is the following, which is cumbersome:

>> inc = num2cell([a.bar]+1);
>> [a.bar] = inc{:};

Is there a way to do this without creating an intermediate variable? Thanks in advance.

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Image Analyst 2014-4-29

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/127661-how-do-i-add-a-value-to-a-field-in-each-element-of-a-struct-array#answer_135086

在 MATLAB Online 中打开

Well why are you messing around with cell arrays? Why make it way more complicated than it needs to be??? I don't see any reason for a cell array. I think you need to read the FAQ: http://matlab.wikia.com/wiki/FAQ#What_is_a_cell_array.3F

Try using just regular numerical arrays:

% Create structure.
a=struct('bar',[47 52])
a.bar % Report to command window.
% Add 1
a.bar = a.bar+1
a.bar % Report to command window.

In the command window:

a = 
    bar: [47 52]
ans =
    47    52
a = 
    bar: [48 53]
ans =
    48    53

2 个评论
显示无隐藏无

Brandon Kuczenski 2014-4-29

Your 'a' is not a struct array- it is a struct with one field whose contents are an array. In my version, a(1).bar = 47, a(2).bar=52, and a.bar(2) gives an error.

Image Analyst 2014-4-29

在 MATLAB Online 中打开

Sorry, I misunderstood. A fast, straightforward and intuitive solution is to simply use a for loop:

for k = 1:length(a)
  a(k).bar = a(k).bar+1;
end

Don't believe the hype about a for loop being slow. It's not. Assuming you have less than several million structures in the array, it should be very fast.

请先登录，再进行评论。