How to bold in a sprintf function?

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Elena
Elena on 29 Mar 2022
Commented: Jan on 29 Mar 2022
For below, m and b are going to be numbers, how can i make only those parts of the statement bold? I know its \bf somewhere but i couldnt get it to work.
sprintf('Y = %.2f X + %.2f',m,b)
As a follow up, I also have this where im trying to make every 0 or multiple of 10 bold on the x-axis, how can i incoorporate that? This is what i have, once again could get \bf to work anywhere
endAt = length(myAx.XAxis.TickLabels)
for i=0:10:endAt
myAx.XAxis.TickLabels{0} = i;
end

Accepted Answer

Star Strider
Star Strider on 29 Mar 2022
You can do that in a a text call (with any text objects, such as title, xlabel, etc.), nowhere else.
m = pi;
b = exp(1);
text(0.3, 0.7, sprintf('Y = \\bf%.2f\\rm X + \\bf%.2f\\rm',m,b))
.

More Answers (2)

Voss
Voss on 29 Mar 2022
m = 2;
b = 1;
% set up normal and bold strings, for comparison:
str_normal = sprintf('Y = %.2f X + %.2f',m,b)
str_normal = 'Y = 2.00 X + 1.00'
% use \\ to "escape" the \, i.e., allow a backslash to "pass-through" sprintf()
% without interpretation:
str_bold = sprintf('Y = {\\bf%.2f} X + {\\bf%.2f}',m,b)
str_bold = 'Y = {\bf2.00} X + {\bf1.00}'
% make some lines to put in a legend, using str_normal and str_bold as
% their names:
x = 0:30;
h_normal = plot(x,m*x+b);
hold on
h_bold = plot(x,m*x+b);
legend([h_normal h_bold],{str_normal str_bold});
% set up the XTickLabels:
xtick_label = cell(size(x));
for ii = 1:numel(x)
if mod(x(ii),10) == 0 % x(ii) is a multiple of 10 (0 is a multiple of 10 too)
xtick_label{ii} = sprintf('{\\bf%d}',x(ii));
else
xtick_label{ii} = sprintf('%d',x(ii));
end
end
set(gca(),'XTick',x,'XTickLabel',xtick_label)

Jan
Jan on 29 Mar 2022
Edited: Jan on 29 Mar 2022
Ticks = linspace(0, 30, 7);
ax = axes('XLim', [0, 30], 'XTick', Ticks, ...
'TickLabelInterpreter', 'latex');
for k = 1:numel(Ticks)
if rem(Ticks(k), 10) == 0
ax.XAxis.TickLabels{k} = ['\bf' ax.XAxis.TickLabels{k}];
end
end
% Or without a loop:
m = (rem(Ticks, 10) == 0);
ax.XAxis.TickLabels(m) = strcat('\bf', ax.XAxis.TickLabels(m)):
  2 Comments
Elena
Elena on 29 Mar 2022
thank you! i just realised i can only accept one answer but i really appreciate it
Jan
Jan on 29 Mar 2022
@Elena: You are welcome.
Accepting one answer is fine. The main point of accepting is to indicate, that this question does not need further attention. I have more points than I can eat, so I'm glad, if I can help to solve the problems. :-)

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