Replacing values in a Matrix

1.113 visualizaciones (últimos 30 días)
gooniyath
gooniyath el 15 de Ag. de 2016
Respondida: Stephen23 el 5 de Nov. de 2023
Hi,
I have a matrix similar to this:
A =[ ]
25 40 40 40 25 25 25 25 25 25
25 40 40 40 40 25 25 25 25 25
25 25 40 40 40 40 25 25 25 25
25 25 40 40 40 40 40 25 25 25
25 25 25 40 40 40 40 40 25 25
25 25 25 40 40 40 40 40 40 25
25 25 12 12 40 40 40 40 40 40
25 12 12 12 12 40 40 40 40 40
12 12 12 12 12 12 40 40 40 40
12 12 12 12 12 12 25 40 40 40
12 12 12 12 12 25 25 40 40 40
How do I write a script to replace all the 25's, with a certain value, and the 40's with another value and 12's with another value?
Thanks
  2 comentarios
Nilesh
Nilesh el 5 de Nov. de 2023
give a single matlab commandthat will overwritethe 3rd entryin vectorB with 12
DGM
DGM el 5 de Nov. de 2023
B(3) = 12;
If you need to resort to copy-pasting such a simple question, then good luck writing anything.

Iniciar sesión para comentar.

Respuestas (4)

dpb
dpb el 15 de Ag. de 2016
A(A==yourvalue)=NewValue;
  5 comentarios
Yogya Chawla
Yogya Chawla el 4 de Feb. de 2022
can you explain this syntax
dpb
dpb el 4 de Feb. de 2022
Editada: dpb el 4 de Feb. de 2022
A use of "logical indexing", one of the most powerful of MATLAB features, illustrated at <MatrixIndexingByLogicalExpression>.
A==yourvalue is a logical vector true where the values of A match yourvalue, false elsewhere. MATLAB then assigns the RHS to the true locations; ignoring the false positions.
It is the one way one can address an array with 0/1, but the values must be of class logical, not numeric.

Iniciar sesión para comentar.


Thorsten
Thorsten el 15 de Ag. de 2016
Editada: Thorsten el 15 de Ag. de 2016
[Aval, ~, indAval] = unique(A);
Define the new values. Values are ordered from the smallest value to replace with to the largest, i.e., to replace 12 with 41, 25 with 26 and 40 with 13 defise Avalnew as
Avalnew = [41; 26; 13];
Anew = Avalnew(indAval);
Anew = reshape(Anew, size(A));
  1 comentario
Javier Cabello
Javier Cabello el 17 de Abr. de 2020
Much appreciated. Very flexible and smart solution!

Iniciar sesión para comentar.


BJ Anderson
BJ Anderson el 12 de Mzo. de 2019
Editada: BJ Anderson el 12 de Mzo. de 2019
The real answer you're looking for is changem:
The syntax looks like this:
B = changem(A,[0 0],[9 8])
where the latter two arguments are vectors, wherein the all elements in the last vector are replaced with their counterparts in the first vector, within data array A.
Syntax
mapout = changem(Z,newcode,oldcode)
Description
mapout = changem(Z,newcode,oldcode) returns a data grid mapout identical to the input data grid, except that each element of Z with a value contained in the vector oldcode is replaced by the corresponding element of the vector newcode.
oldcode is 0 (scalar) by default, in which case newcode must be scalar. Otherwise, newcode and oldcode must be the same size.
Examples
Invent a map:
A = magic(3)
A =
8 1 6
3 5 7
4 9 2
Replace instances of 8 or 9 with 0s:
B = changem(A,[0 0],[9 8])
B =
0 1 6
3 5 7
4 0 2
  1 comentario
BJ Anderson
BJ Anderson el 12 de Mzo. de 2019
A quick update on changem:
Sadly, if one inspects the actual code within changem, it functions as a loop. While it is a handy one-liner, it does not have the time-savings of moving from a looped function to an matrix-operation function.

Iniciar sesión para comentar.


Stephen23
Stephen23 el 5 de Nov. de 2023
A = [25,40,40,40,25,25,25,25,25,25; 25,40,40,40,40,25,25,25,25,25; 25,25,40,40,40,40,25,25,25,25; 25,25,40,40,40,40,40,25,25,25; 25,25,25,40,40,40,40,40,25,25; 25,25,25,40,40,40,40,40,40,25; 25,25,12,12,40,40,40,40,40,40; 25,12,12,12,12,40,40,40,40,40; 12,12,12,12,12,12,40,40,40,40; 12,12,12,12,12,12,25,40,40,40; 12,12,12,12,12,25,25,40,40,40]
A = 11×10
25 40 40 40 25 25 25 25 25 25 25 40 40 40 40 25 25 25 25 25 25 25 40 40 40 40 25 25 25 25 25 25 40 40 40 40 40 25 25 25 25 25 25 40 40 40 40 40 25 25 25 25 25 40 40 40 40 40 40 25 25 25 12 12 40 40 40 40 40 40 25 12 12 12 12 40 40 40 40 40 12 12 12 12 12 12 40 40 40 40 12 12 12 12 12 12 25 40 40 40
old = [12,25,40];
new = [99,23,42];
B = interp1(old,new,A)
B = 11×10
23 42 42 42 23 23 23 23 23 23 23 42 42 42 42 23 23 23 23 23 23 23 42 42 42 42 23 23 23 23 23 23 42 42 42 42 42 23 23 23 23 23 23 42 42 42 42 42 23 23 23 23 23 42 42 42 42 42 42 23 23 23 99 99 42 42 42 42 42 42 23 99 99 99 99 42 42 42 42 42 99 99 99 99 99 99 42 42 42 42 99 99 99 99 99 99 23 42 42 42

Categorías

Más información sobre Matrix Indexing 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!

Translated by