How do I store data that meet conditions of an if statement

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NA 2021-2-1

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https://ww2.mathworks.cn/matlabcentral/answers/733083-how-do-i-store-data-that-meet-conditions-of-an-if-statement

评论： NA 2021-2-3

采纳的回答： Jan

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Hi All,

I'm slightly struggling trying to get my code to work, and I hope someone can point me in the right direction.

Very briedly, I have two vectors, 'xx' and 'yy'. I created a matrix using these two vectors, 'xy'. If xy(:,1) is ==1, I would like to store the corresponding cell in xy(:,2) in a seperate variable, 'data'. Otherwise, if the logical expression = 0, I'd like that cell to be a NaN

xx=[-3; -3; -3; -3; -3; -3; 1; -2; -2; -3; -3; -2; 1; 1; 1; -2; 1; 1; ];
yy=[26.28; 36.96; 90.00; 90.00; 90.00; 90.00; 27.85; 29.97; 47.03; 90.00; 67.22; 78.87; 19.60; 9.00; 2.00; 3.41; 1.88; 2.50];
xy=[xx yy];
for i=1:size(xy(:,1),1)
    if xy(:,1)==1
       data=xy(:,2);
     else
       data=NaN;
    end
end

Thanks in advance.

Cheers

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Answer 1

Jan 2021-2-1

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https://ww2.mathworks.cn/matlabcentral/answers/733083-how-do-i-store-data-that-meet-conditions-of-an-if-statement#answer_611873

编辑：Jan 2021-2-1

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xx = [-3; -3; -3; -3; -3; -3; 1; -2; -2; -3; -3; -2; 1; 1; 1; -2; 1; 1; ];
yy = [26.28; 36.96; 90.00; 90.00; 90.00; 90.00; 27.85; 29.97; ...
      47.03; 90.00; 67.22; 78.87; 19.60; 9.00; 2.00; 3.41; 1.88; 2.50];
xy = [xx yy];
data  = nan(size(xx));
match = (xx == 1);
data(match) = xy(match, 2);

This "vectorized" method is faster and nicer than a loop. But with a loop:

data = nan(size(xy, 1), 1);
for i = 1:size(xy, 1)    % better than: size(xy(:,1),1)
    if xy(i, 1) == 1
       data = xy(i, 2);
    end
end

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dpb 2021-2-2

I posted a demonstration of what happens with dynamic allocation just a day or so ago at<Comment_1298378>

NA 2021-2-3

Thanks dpb, thats very useful information. Cheers

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Answer 2

dpb 2021-2-1

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Don't need any loops nor the explicity xy array that is duplicate of existing data.

data=nan(size(x,1),2);              % initialize to NaN
ix=(xx==1);                         % logical addressing vector of wanted locations
data(ix,:)=[xx(ix) yy(ix)];         % set data values

Alternatively,

data=[xx yy];                       % initialize to data
data(data(:,1),:)=nan;              % set unwanted values to NaN

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NA 2021-2-1

I wrote it usuing a for loop because it's a subset of a larger code (and needs to be incorporated within a loop). But thank you for your feedback :) Cheers

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