Decimal increments in a for loop

8 vues (au cours des 30 derniers jours)
nth24601
nth24601 le 21 Oct 2018

Hello, I am trying to make a for loop in which i goes up by decimal increments and for each value of i (1, 1.1, 1.2, 1.3) a different value for a function is calculated and stored. I am aware that there are such threads already existing but none of them actually seem to solve the problem. My code so far looks like that:

z = 1:0.1:100;
for i=1:length(z)
  if i<10
      y(:,i) = function of i
  elseif (10<=i)&&(i<50)
      y(:,i) = another function of i
  elseif (50<=i)&&(i<=100)
      y(:,i) = another function of i
  end
end

The problem with that is that i does not actually go up by 0.1, it goes up by 1, and the function calculates values for i equal to 1,2,3,... etc I want it to take i=1, calculate the value of y for that, then take i=1.1, calculate the value for that etc.

Réponse acceptée

Walter Roberson
Walter Roberson le 22 Oct 2018
Modifié(e) : Walter Roberson le 24 Oct 2018
  1 commentaire
nth24601
nth24601 le 24 Oct 2018
Thanks, the answers in this link helped me solve this problem.

Connectez-vous pour commenter.

Plus de réponses (1)

madhan ravi
madhan ravi le 21 Oct 2018
z = 1:0.1:100;
for i=1:length(z)
if z(i)<10
y(:,i) = function of i
elseif (10<=z(i))&&(z(i)<50)
y(:,i) = another function of i
elseif (50<=z(i))&&(z(i)<=100)
y(:,i) = another function of i
end
end
  3 commentaires
nth24601
nth24601 le 22 Oct 2018
This gets close to the answer but it doesnt make i go by increments of 0.1, it only creates a 1000 entries for i, but i keeps going up by 1. I need the final value of i to be 100. Thank you for the answer though.
madhan ravi
madhan ravi le 22 Oct 2018
You can’t index 0.1 as an index , See indexing in MathWorks Page you will understand only an integer can be used as an index not a float

Connectez-vous pour commenter.

Catégories

En savoir plus sur Matrix Indexing dans Help Center et File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by