How to plot a cell array containing characters?
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Hi everyone.
I have a cell array (1x1000) that looks like this:
comment=[artefact artefact 0 0 0 0 ...] ;
How can I plot the strings of this array against other values already plotted as lines?
3 个评论
Jan
2022-4-15
comment=[artefact artefact 0 0 0 0 ...] ;
This is not a cell array. Please post some code, which really procudes the variables you are talking of.
I have no idea, how you want to "plot" a string or CHAR vector.
Mariagrazia Ambrosino
2022-4-15
编辑：Mariagrazia Ambrosino
2022-4-15
I got this result as an export from c ++. it is stored as a 1x1000 cell and inside there are two strings. I should add these 2 strings to an already existing plot.
it is a very difficult subject for me, I hope I have explained myself.
I have to add the 'artefact' strings on the red lines:
Image Analyst
2022-4-19
Again, attach your cell array with the paperclip icon
save('Answers.mat', 'comment');
after you read this link:
回答（1 个）
% you have a 1x1000 cell array called 'comment':
comment = repmat({0},1,1000);
comment([1 2]) = {"artefact"};
disp(comment);
% and you have already plotted the lines:
figure
xlim([0 5]*1e6);
ylim([-4000 4000]);
hold on
plot([0 0 NaN 1.3 1.3 NaN 4.52 4.52 NaN]*1e6,[-4000 4000 NaN -4000 4000 NaN -4000 4000 NaN],'--r');
% here's how you can make the texts:
text(0,0,comment{1});
text(1.3e6,0,comment{2},'Rotation',90)
text(4.52e6,0,comment{2},'Rotation',-45);
4 个评论
Mariagrazia Ambrosino
2022-4-19
You can use isstring or ischar to determine whether something is a string or a character array, respectively.
% some comments:
words = split('now is the winter of our discontent');
% put them into a cell array at random indices:
idx = cumsum(randi(5,1,numel(words)));
comment = repmat({0},1,idx(end));
comment(idx) = words
% make text objects showing the comments (use ischar to determine whether
% a particular cell of cell array 'comment' contains a comment)
for ii = 1:numel(comment)
if ischar(comment{ii})
text(ii,ii,comment{ii});
end
end
grid on
xlim([0 ii+1]);
ylim([0 ii+1]);
Mariagrazia Ambrosino
2022-4-20
Voss
2022-4-20
You're welcome! If you have any other questions, let me know. Otherwise, if my answer satisfactorily addresses what you're trying to do, please click 'Accept this Answer'. Thanks!
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2022-4-15
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