How to store output of an array?

Question

yashvin 2015-7-13

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/229597-how-to-store-output-of-an-array

评论： Guillaume 2015-7-13

The code works perfectly.

 a=[0 200 900 1000 1200 1798 1799 1801 2000 2700 3601]
N = round(max(a)/900); 
for k = 1:N
 [~, ind(k)] = min(abs(a - 900*k)); 
 end

For more arrays of a, i want to store the corresponding ind value. How can i do it?

example,

a=[0 200 900 1000 1200 1798 1799 1801 2000 2700 3601]
   a=[0 1800  2000 2700 3600]
    N = round(max(a)/900); 
   for k = 1:N
   [~, ind(k)] = min(abs(a - 900*k)); 
   end

For each array of a, how do i store the corresponding ind values?

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Answer 1

Guillaume 2015-7-13

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https://ww2.mathworks.cn/matlabcentral/answers/229597-how-to-store-output-of-an-array#answer_185854

在 MATLAB Online 中打开

Here is a non-loopy version of your original code (I have no idea if it's faster than your code):

N = (1:round(max(a)/900))*900;
[~, ind] = min(abs(bsxfun(@minus, a', N)));

To do the same for several a arrays, first stores all these arrays into a cell array:

allAs = {
     [0 200 900 1000 1200 1798 1799 1801 2000 2700 3601];
     [0 1800  2000 2700 3600]
        };

Then use a loop or cellfun to repeat the above or your code over the cell array:

[~, allInds] = cellfun(@(a) min(abs(bsxfun(@minus, a', (1:round(max(a)/900))*900))), ...
                       allAs, 'UniformOutput', false)

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yashvin 2015-7-13

Ah sorry! My bad you are right. Being a beginner, I meant very common apporach, as I have no in depth knowledge of cellfun. I have a feeling that the solution is very simple. So any solution that is less complex than cellfun.

The problem is that ind which is the output of min cannot be specified in terms of cell . Else for each array, i could have stored it as a cell.

Sorry for the misinterpretation

Guillaume 2015-7-13

在 MATLAB Online 中打开

cellfun is just a simple replacement for a loop that avoids you having to keep track of the numbers of elements and having to predeclare the result array. The equivalent code would be:

allInds = cell(size(allAs));
for iter = 1:numel(allAs)
   a = allAs{iter};
   [~, allInds{iter}] = min(abs(bsxfun(@minus, a', (1:round(max(a)/900))*900)));
end

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