For loop: Odd numbers turn 0 when checked if even

Hello,
Sorry if the title is misleading. In short, when code checks if number is even; it assigns it to evens. However, if number is odd; it turns it into 0 and assigns it to even.
Here is the output...
48 55 6 66 89 11 44 29 99 61 26 14 55 83 84 84 21 55 88 13
48 0 6 66 0 0 44 0 0 0 26 14 0 0 84 84 0 0 88
48 is even, 55 is not so it turns into 0, 6 is even, and so on....
integers = randi([1 100], 1, 20);
evens = ones(1);
odds = ones(1);
for k = 1:length(integers)
if mod(integers(k), 2) == 0 % check if number is even
evens(k) = integers(k);
end
odds(k) = integers(k); % else if number is odd
end

 Respuesta aceptada

KSSV
KSSV el 17 de Nov. de 2018
Editada: KSSV el 17 de Nov. de 2018
integers = randi([1 100], 1, 20);
evens = zeros([],1) ;
odds = zeros([],1) ;
c1 = 0 ;c2 = 0 ;
for k = 1:length(integers)
if mod(integers(k), 2) == 0 % check if number is even
c1 = c1+1 ;
evens(c1) = integers(k);
else
c2 = c2+1 ;
odds(c2) = integers(k); % else if number is odd
end
end
YOu need not to use loop even.....you can straight away get what you want.
integers = randi([1 100], 1, 20);
evens = integers(mod(integers,2)==0) ;
odds = integers(mod(integers,2)~=0) ;

3 comentarios

Obadah M.
Obadah M. el 17 de Nov. de 2018
Thanks it worked.
What is the explaination behind this?
KSSV
KSSV el 17 de Nov. de 2018
Editada: KSSV el 17 de Nov. de 2018
YOu have initiliazed your evens/odds to zeros ...and when conditon is met..the zero will be repalced..and where condition is not met..zero remains same. In the given code..I am writitng value only when condition is met.
Obadah M.
Obadah M. el 17 de Nov. de 2018
Alright thanks!

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre Loops and Conditional Statements en Centro de ayuda y File Exchange.

Productos

Versión

R2017b

Etiquetas

Preguntada:

el 17 de Nov. de 2018

Editada:

el 17 de Nov. de 2018

Community Treasure Hunt

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

Start Hunting!

Translated by