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Why is format long providing a strange result?

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Connor Taylor
Connor Taylor on 5 Dec 2019
Edited: Adam Danz on 6 Dec 2019
I am having trouble with the output from a code I am running. The answers are correct, but the Matlab command window is outputting them in a funny way.
Instead of printing 19.9998.... it gives this as 0.199998... with 1.0e+02. This is technically correct, but just looks a bit shabby.
I have already played around with the format settings in Customise toolbar > Command window > Text display. This still hasn't fixed the issue.
Format long seems to work well for the code below where dt=2. However changing dt to dt=0.2 is what is causing the problem.
Many thanks in advance.
clc
format long
r=0.5;
k=20;
dt=2;
t_run=30.0;
n_max=t_run/dt +1;
xtrue=zeros(n_max,1);
n=1
x0=0.5;
time=0:dt:t_run;
for n=1:n_max
xtrue(n)=k/(1+((k-x0)*exp(-r*2*(n-1)))/x0);
disp([n xtrue(n)])
end
plot(time,xtrue)
xlim([0 30])
ylim([0 30])
xlabel('Time')
ylabel('Population')
title('True Solution')
hold on
%% True plot with dt = 0.2
dt=0.2;
n_max=t_run/dt +1;
xtrue=zeros(n_max,1);
n=1
time=0:dt:t_run;
for n=1:n_max
xtrue(n)=k/(1+((k-x0)*exp(-r*0.2*(n-1)))/x0);
disp([n xtrue(n)])
end
plot(time,xtrue)
legend('dt=2','dt=0.2')
hold off

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Answers (2)

Steven Lord
Steven Lord on 5 Dec 2019
I recommend using format longg instead of format long for this case.

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Adam Danz
Adam Danz on 5 Dec 2019
Edited: Adam Danz on 6 Dec 2019
It's because of the way you're combing two variables in the output display.
Look at the last iteration of your 2nd for-loop.
% Your 2nd for-loop
for n=1:n_max
xtrue(n)=k/(1+((k-x0)*exp(-r*0.2*(n-1)))/x0);
disp([n xtrue(n)])
end
  • n is equal to 151
  • xtrue(n) is equal to 19.999761399036565 (long format)
  • when you combine them using [n xtrue(n)], the long format reformats the output.
format long
[n xtrue(n)]
% ans =
% 1.0e+02 *
% 1.510000000000000 0.199997613990366
format short
[n xtrue(n)]
% ans =
% 151.0000 19.9998
If you'd like to specify the precision, use fprintf() instead of disp().
format long % or whatever format you want - you'll get the same results
fprintf('%.4f, %.4f\n',n, xtrue(n))
% Result:
% 151.0000, 19.9998
% Or maybe you want
fprintf('%d, %.4f\n',n, xtrue(n))
% Result:
% 151, 19.9998

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