CODEWITHSUNDEEP
matlabmatrixmatrix-indexing
jrenzhile
jrenzhile

Reputation: 121

replace indices in matrix with other values in matlab

Suppose now I have a matrix

S = [1 1 1 2 2 2;
     1 1 1 2 2 2;
     2 2 2 2 1 1;
     2 2 2 2 1 1;
     2 2 2 2 1 1]

And another matrix

A = [1 2;
     2 4]

The first row in A is the unique indices of S, and the second row contains the values that the values in the first row will be replaced. That is, all "1"s in S will be replaced by 2, and all "2"s will be replaced by 4. Finally I'll get a matrix

SS = [2 2 2 4 4 4;
      2 2 2 4 4 4;
      4 4 4 4 2 2;
      4 4 4 4 2 2;
      4 4 4 4 2 2]

Right now what I'm doing is:

SS = zeros(size(S));
for i = 1:size(A,2)
    SS(S==index(A(1, i)) = A(2,i);
end

Now, I have a pretty big matrix, and using a for loop is a little bit slow. Is there a faster way to do that?

Upvotes: 3

Views: 2996

Answers (3)

tmpearce
tmpearce

Reputation: 12693

Use the second output of ismember to give you indices of the values in row 1 of A. Use these indices to directly create matrix SS.

Example (changed initial values for clarity):

S = [5 5 5 3 3 3; 5 5 5 3 3 3; 3 3 3 3 5 5; 3 3 3 3 5 5; 3 3 3 3 5 5]; A = [5 3; 2 4];

>> [~, Locb] = ismember(S,A(1,:))
Locb =

     1     1     1     2     2     2
     1     1     1     2     2     2
     2     2     2     2     1     1
     2     2     2     2     1     1
     2     2     2     2     1     1

>> SS = reshape(A(2,Locb),size(S))
SS =

     2     2     2     4     4     4
     2     2     2     4     4     4
     4     4     4     4     2     2
     4     4     4     4     2     2
     4     4     4     4     2     2

Upvotes: 3

Eitan T
Eitan T

Reputation: 32930

You could go about this with an arrayfun one-liner, like this:

SS = arrayfun(@(x)A(2, (A(1, :) == x),  S)

Upvotes: 1

Indranil Sinharoy
Indranil Sinharoy

Reputation: 523

If I have understood your question correctly, I would use numpy array instead of standard python arrays or lists. Then the code becomes very simple as shown below:

# Import numpy
from numpy import array, zeros, shape
# Create the array S
S = array([[1,1,1,2,2,2],[1,1,1,2,2,2],[2,2,2,2,1,1],[2,2,2,2,1,1],[2,2,2,2,1,1]])
# Create the array A
A = array([[1,2],[2,4]])
# Create the empty array SS
SS = zeros((shape(S)))
# Actual operation needed 
SS[S==A[0,0]]=A[1,0]
SS[S==A[0,1]]=A[1,1]

Now if you see the array SS, it will look as follows:

SS
array([[ 2.,  2.,  2.,  4.,  4.,  4.],
       [ 2.,  2.,  2.,  4.,  4.,  4.],
       [ 4.,  4.,  4.,  4.,  2.,  2.],
       [ 4.,  4.,  4.,  4.,  2.,  2.],
       [ 4.,  4.,  4.,  4.,  2.,  2.]])

Sorry for the confusion earlier. I had (for some reason) assumed that this question was for Python (my bad!). Anyways, the answer for MATLAB is very similar:

SS = zeros(size(S))
SS(S==A(1,1))=A(2,1)
SS(S==A(1,2))=A(2,2)

Upvotes: 1

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