Create a vector with for loop with mathematical operations

2 visualizaciones (últimos 30 días)
w=[5 4 3 2]
Use this vector in a mathematical expression to create the following vectors:
My code snippet is below, code for that question. When I run it, I can't get expected result. I just want to create a vector as noted above.How can I solve that question?
w=[5 4 3 2]
for i= w
a3=[1/(i+i)]
b3 =[i^i]
c3 =[i/sqrt(i)]
d3 =[(i^2)/(i^i)]
end

Respuesta aceptada

KSSV
KSSV el 10 de Mayo de 2020
Read about ., element by element operations.
w = [5 4 3 2] ;
a = 1./(w+w) ;
b = w.^w ;
c = w./sqrt(w) ;
d = w.^2./w.^w ;

Más respuestas (1)

Toder
Toder el 10 de Mayo de 2020
To fix your current code, you need to use indices in the for loop.
w=[5 4 3 2]
for i = w
a3(6-i)=[1/(i+i)]
b3(6-i) =[i^i]
c3(6-i) =[i/sqrt(i)]
d3(6-i) =[(i^2)/(i^i)]
end
This assigns a3(1) in the first loop, a3(2) in the second, etc. Your code reassigns the scalar a3 each loop. The indices I use are somewhat strange because we're looping over w, which is fine, it just makes indices harder to keep track of. You could do it differently by using a counter:
w=[5 4 3 2]
j = 1
for i = w
a3(j)=[1/(i+i)]
b3(j) =[i^i]
c3(j) =[i/sqrt(i)]
d3(j) =[(i^2)/(i^i)]
j = j+1
end
Or you could have your for loop count from 1 to 4:
w=[5 4 3 2]
for i = 1:4
a3(i)=[1/(w(i)+w(i))]
b3(i) =[w(i)^w(i)]
c3(i) =[w(i)/sqrt(w(i))]
d3(i) =[(w(i)^2)/(w(i)^w(i))]
end
All this being said, I recommend "vectorizing" your code where possible. Dot operators perform arithmetic element-by-element, see the documentation for dot multiplication here. For instance, the whole a3 vector can be defined this way:
a3 = 1./(2*w)
  5 comentarios
David Banahene
David Banahene el 23 de En. de 2022
I have a vector a=[1:5] and would want to know how to plot (a-3)^3.

Iniciar sesión para comentar.

Categorías

Más información sobre Loops and Conditional Statements en Help Center y File Exchange.

Productos

Community Treasure Hunt

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

Start Hunting!

Translated by