Replace values in a matrix

Hi,

I have a 250x250 matrix, each cell having either 1,2,3,4,5 or NaN, derived from kmeans index. I'm looking to reorder my kmeans values and have found a new order for these values based on populatation e.g. 3 4 5 2 1, 3 being most populated, 1 being least. I'm looking for a way to tell the system to make 3 the new 1, 4 the new 2 etc.. i.e. replace all 3s with 1, all 4s with 2, all 5s with 3, all 2s with 4, all 1s with 5. However, when I use a loop, I end up losing data as it thinks the converted numbers were my original ones and replaces them to a new value.

[KALYAN ACHARJYA]

"I end up losing data as it thinks the converted numbers were my original ones and replaces them to a new value"

Copy the data with temp variable. The initial part of the question is not clear to me. Please make it easy to understand with simple examples.

[KF]

So, let's say for simplicity I have a 10x10 matrix,

A = [ ]

NaN  NaN  NaN   3    3    3    2   NaN  NaN  NaN
NaN   4    4    3    3    3    3    3   NaN  NaN
NaN  NaN   4    3    3    3    5    2   NaN  NaN
NaN   3    3    3    3    1    3    3    3   NaN
4     1    4    2    5    5    5    3   NaN  NaN
4     2    4    3    5    5    5    3    3   NaN
NaN   2    4    3    3    3    3    3   NaN  NaN
NaN   3    3    3    2    1    5    4   NaN  NaN
NaN  NaN  NaN   3    1    4    1    3   NaN  NaN
NaN  NaN  NaN   3    3    3    3    3   NaN  NaN

How would I replace all the 3s with 1, all of the 4s with 2, all of the 5s with 3, all of the 2s with 4 and all of the 1s with 5.

The NaNs should stay as they are.

Answer 1

[Rik]

Because you are only using integers, you can even use indexing to do the replacement:

A=[NaN  NaN  NaN   3    3    3    2   NaN  NaN  NaN
NaN   4    4    3    3    3    3    3   NaN  NaN
NaN  NaN   4    3    3    3    5    2   NaN  NaN
NaN   3    3    3    3    1    3    3    3   NaN
4     1    4    2    5    5    5    3   NaN  NaN
4     2    4    3    5    5    5    3    3   NaN
NaN   2    4    3    3    3    3    3   NaN  NaN
NaN   3    3    3    2    1    5    4   NaN  NaN
NaN  NaN  NaN   3    1    4    1    3   NaN  NaN
NaN  NaN  NaN   3    3    3    3    3   NaN  NaN];
dict=[5 4 1 2 3];
L=~isnan(A);
A(L)=dict(A(L))

A = 10×10

NaN NaN NaN 1 1 1 4 NaN NaN NaN NaN 2 2 1 1 1 1 1 NaN NaN NaN NaN 2 1 1 1 3 4 NaN NaN NaN 1 1 1 1 5 1 1 1 NaN 2 5 2 4 3 3 3 1 NaN NaN 2 4 2 1 3 3 3 1 1 NaN NaN 4 2 1 1 1 1 1 NaN NaN NaN 1 1 1 4 5 3 2 NaN NaN NaN NaN NaN 1 5 2 5 1 NaN NaN NaN NaN NaN 1 1 1 1 1 NaN NaN

●

[KF]

Perfect, thank you!! Knew there would be a simplified way :)

[Rik]

You're welcome. If you happen to have an equivalent problem with non-integers; you can use unique to generate a list of indices (use the first and third output).

Answer 2

[David Hill]

a=A==1;
b=A==2;
A(A==3)=1;
A(A==4)=2;
A(A==5)=3;
A(b)=4;
A(a)=5;