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Replace values in a matrix

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KF
KF 2020-12-14
评论： Rik 2020-12-14
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.
  2 个评论
KALYAN ACHARJYA
KALYAN ACHARJYA 2020-12-14
"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
KF 2020-12-14
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.

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采纳的回答

Rik
Rik 2020-12-14
Ran in:
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
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  2 个评论
KF
KF 2020-12-14
Perfect, thank you!! Knew there would be a simplified way :)
Rik
Rik 2020-12-14
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).

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更多回答（1 个）

David Hill
David Hill 2020-12-14
a=A==1;
b=A==2;
A(A==3)=1;
A(A==4)=2;
A(A==5)=3;
A(b)=4;
A(a)=5;

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