Replacing values in a Matrix
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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
Antworten (4)
dpb
am 15 Aug. 2016
A(A==yourvalue)=NewValue;
5 Kommentare
dpb
am 4 Feb. 2022
Bearbeitet: dpb
am 4 Feb. 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.
Thorsten
am 15 Aug. 2016
Bearbeitet: Thorsten
am 15 Aug. 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 Kommentar
BJ Anderson
am 12 Mär. 2019
Bearbeitet: BJ Anderson
am 12 Mär. 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 Kommentar
BJ Anderson
am 12 Mär. 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.
Stephen23
am 5 Nov. 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]
old = [12,25,40];
new = [99,23,42];
B = interp1(old,new,A)
0 Kommentare
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