How can I extract matrix without NaN values

I have two given matrices A and B in same size 5 x 5
A=[1 3 0 3 5; 2 3 4 1 0; 0 1 3 0 1; 2 4 5 4 4; 5 5 2 0 1]
B=[NaN NaN NaN NaN NaN; 2 3 4 1 NaN; 1 2 2 1 NaN; NaN 4 5 4 4; NaN 4 2 0 0]
How I could extract two sub-matrices of them - let's call them A1 and B1; by that B1 contains only numeric values of B, and A1 contains the corresponded values of A with same allocation index with numeric values of B. They should be like this
A1=[2 3 4 1; 0 1 3 0; 4 5 4 4; 5 2 0 1]
B1=[2 3 4 1; 1 2 2 1; 4 5 4 4; 4 2 0 0]
I tried for loops for a simple solution but it messed up. Thank you for your help~

 Akzeptierte Antwort

KSSV
KSSV am 28 Apr. 2020

1 Stimme

There should be more elegant method wituhut using loops.
clc; clear all ;
A=[1 3 0 3 5; 2 3 4 1 0; 0 1 3 0 1; 2 4 5 4 4; 5 5 2 0 1]
B=[NaN NaN NaN NaN NaN; 2 3 4 1 NaN; 1 2 2 1 NaN; NaN 4 5 4 4; NaN 4 2 0 0]
A1=[2 3 4 1; 0 1 3 0; 4 5 4 4; 5 2 0 1]
B1=[2 3 4 1; 1 2 2 1; 4 5 4 4; 4 2 0 0]
[m,n] = size(B) ;
A1 = cell([],1) ;
B1 = cell([],1) ;
count = 0 ;
for i = 1:m
idx = ~isnan(B(i,:)) ;
if any(idx)
count = count+1 ;
A1{count,1} = A(i,idx) ;
B1{count,1} = B(i,idx) ;
end
end
% The below works only when each cell has same number of elements
A1 = cell2mat(A1) ;
B1 = cell2mat(B1) ;

3 Kommentare

Kien Pham
Kien Pham am 28 Apr. 2020
Thank you for spending time with this question. The code you posted returns A1 and B1 as cell instead of matrices. It that possible to attain the matrices but not cell?
KSSV
KSSV am 28 Apr. 2020
You can use cell2mat. I have given in the code right.
Kien Pham
Kien Pham am 28 Apr. 2020
ahhh my fault :) I gazed on the loop part... Have a nice day :)

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Weitere Antworten (1)

dpb
dpb am 28 Apr. 2020
Bearbeitet: dpb am 28 Apr. 2020

1 Stimme

A1=A.'; B1=B.'; % so can operate by row instead column...
isf=isfinite(B1);
>> B1=reshape(B1(isfinite(B1)),4,[]).'
B1 =
2 3 4 1
1 2 2 1
4 5 4 4
4 2 0 0
>> A1=reshape(A1(isf),4,[]).'
A1 =
2 3 4 1
0 1 3 0
4 5 4 4
5 2 0 1
>>

1 Kommentar

Kien Pham
Kien Pham am 28 Apr. 2020
Bearbeitet: Kien Pham am 28 Apr. 2020
Thank you for the response, it works well.

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am 28 Apr. 2020

Bearbeitet:

am 28 Apr. 2020

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