Convert list of points into string and group it in tens
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Hello,
I have a list of points which I want to convert to string and format it in a certain way.
I have manged to do this, but the code looks a bit complicated and wonder how this could be rewrite in a more efficient way.
points = randi(10,36,2);
points_len = size(points,1);
sPoints = '';
for kk = 1:points_len
if rem(kk+9, 10) == 0
if kk > points_len - rem(points_len,10)
endPoint = points_len ;
else
endPoint = kk+9;
end
sPoints = [sPoints, sprintf('Points %0.0f to %0.0f\n', kk, endPoint)];
end
sPoints = [sPoints, num2str(points(kk,:)), newline];
end
sPoints
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Rik
am 21 Okt. 2022
I would suggest putting each section in a cell, so you can easilly add the headers:
points = randi(10,36,2);
PointsPerSection = 10;
NumRows = ceil(size(points,1)/PointsPerSection);
rows = zeros(NumRows,1);
rows(1:(end-1)) = PointsPerSection;
rows(end) = size(points,1)-sum(rows); % Determine last element dynamically
data = mat2cell(points,rows,2)
for n=1:numel(data)
tmp = data{n};
HeaderPart = sprintf('Points %d to %d\n' ,...
(n-1)*PointsPerSection + 1 ,...
(n-1)*PointsPerSection + size(tmp,1) );
% The sprintf function allows providing arrays, but will process them
% by column, while humans will read the text by row. That is why we
% need to use .' to transpose the data.
DataPart = sprintf('%.0f %.0f\n',tmp.');
data{n} = [HeaderPart DataPart];
end
sPoints = horzcat(data{:});
disp(sPoints)
3 Kommentare
Rik
am 21 Okt. 2022
You're welcome.
As to your question: habit. A single quote is not just the transpose, but the conjugate transpose. For real numbers there is no difference, but if you forget when working with complex number you will have a very difficult problem to debug. I try to learn myself good habits to avoid such bugs from ever occuring. For similar reasons I no longer use length when I actually mean numel (or I use size with an input argument).
a = 1+i;
a' , a.'
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