Question of mean
3 visualizaciones (últimos 30 días)
Mostrar comentarios más antiguos
When I write a program about consumption (C) with 10^5 trajectories, I want to calculate the expectation of them with the command mean(C(:,j)). However, at j=1, all the C are equal to 0.86, but the mean(C(:,1))=0.859999999999891. Would anyone tell me why it happens and how to correct it please? Thank you.
0 comentarios
Respuesta aceptada
Tom Lane
el 21 de Mayo de 2012
As Oleg and Titus imply, one idea is to change the test so that it allows for things that differ by a small amount to be treated as roughly equal. Another thing that sometimes works is to refine your answer by another application of mean. This works in some cases but doesn't change the fundamental fact that two theoretically identical things may be slightly different when represented by double precision on a computer. Anyway, here's what I had in mind:
>> x = repmat(0.86,1e5,1);
>> fprintf('%25.18g\n',x(1))
0.85999999999999999
>> m = mean(x); fprintf('%25.18g\n',m)
0.86000000000030574
>> m = m + mean(x-m); fprintf('%25.18g\n',m)
0.85999999999999999
0 comentarios
Más respuestas (1)
Titus Edelhofer
el 21 de Mayo de 2012
Hi,
0.86 is not exactly representable in computers double precision representation:
fprintf('%.20f\n', 0.86)
0.85999999999999999000
Titus
3 comentarios
Ver también
Categorías
Más información sobre Loops and Conditional Statements en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!