Delete some specific numbers from an array

1 visualización (últimos 30 días)
Rongyu Chu
Rongyu Chu el 2 de Oct. de 2019
Editada: Andrei Bobrov el 3 de Oct. de 2019
So, the problem is like this. I have an array of random numbers. For example [10 9 5 5 7 3 5 8 3 5 6 3 7 10 ]. First, I have to delete some of the numbers to make any point's value can not between its adjecent numbers. For example 9 is between 10 and 5, so I will delete 9. I tried to use a for loop but that does not work.
  2 comentarios
Walter Roberson
Walter Roberson el 2 de Oct. de 2019
Hint: sign(diff(TheVector)) will be the same for two entries in a row if the middle value is between the two other values (or if the three values are all equal)
Andrei Bobrov
Andrei Bobrov el 3 de Oct. de 2019
We delete 7, 7 is between 3 ans 10?

Iniciar sesión para comentar.

Respuestas (2)

Andrei Bobrov
Andrei Bobrov el 2 de Oct. de 2019
Editada: Andrei Bobrov el 3 de Oct. de 2019
A = [10 9 5 5 7 3 5 8 3 5 6 3 7 10 ];
B = sign(diff(A(hankel(1:3,3:numel(A)))',1,2));
out = A( [true; prod(B,2) ~= 1 ; true] );
  2 comentarios
Rongyu Chu
Rongyu Chu el 3 de Oct. de 2019
My final purpose is not just delete 9 among 10 and 5. This is just an example. I will have a dat file which contians over 10,000 numbers and I need it to process automatically. Thank you for your help.
Andrei Bobrov
Andrei Bobrov el 3 de Oct. de 2019
I'm fix.

Iniciar sesión para comentar.


Rongyu Chu
Rongyu Chu el 3 de Oct. de 2019
So, here is the cide I wrote.
N = load('exp.dat');
len = length(N);
for i = 2:(len-3)
if N(i+1)>N(i)>N(i-1)
N(i) = [];
len = len - 1;
%i = i - 1;
elseif N(i+1)<N(i)<N(i-1)
N(i) = [];
len = len - 1;
%i = i - 1;
end
end
I tired to use if sentence to compare the value and delete those numers whose value is in the middle of its adjcent numbers. However, this code does not work well. I am wondering why.
  3 comentarios
Rongyu Chu
Rongyu Chu el 3 de Oct. de 2019
thank you. I realized that problem. But now I found a new problem. After I delete some numbers from an array, the arrat size gonna change. So it's not always len-1. But, how should I update that information ?
Walter Roberson
Walter Roberson el 3 de Oct. de 2019
Index from highest to lowest so that when you delete, you do not need to examine the entries that "fell down".

Iniciar sesión para comentar.

Categorías

Más información sobre Matrices and Arrays en Help Center y File Exchange.

Etiquetas

Community Treasure Hunt

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

Start Hunting!

Translated by