Hi,
I've got an issue splitting a matrix into two seperate matrices using a For loop for rows that have the number 11 in it and one that doesnt.
The matrix X = [1 2 3 11 5; 77 1 2 11 3; 1 2 3 4 5; 1 2 3 4 5; 1 11 3 4 5; 1 2 3 4 5]
I've tried the following..
for i=1:length(J)
if J(i,:)==11
A(i,:)=J
else J(i,:)~=11
B(i,:)=J
end
end
end

回答 (1 件)

Star Strider
Star Strider 2021 年 10 月 4 日

0 投票

No loops necessary —
X = [1 2 3 11 5; 77 1 2 11 3; 1 2 3 4 5; 1 2 3 4 5; 1 11 3 4 5; 1 2 3 4 5]
X = 6×5
1 2 3 11 5 77 1 2 11 3 1 2 3 4 5 1 2 3 4 5 1 11 3 4 5 1 2 3 4 5
X11 = X(any(X==11,2),:)
X11 = 3×5
1 2 3 11 5 77 1 2 11 3 1 11 3 4 5
Xnot11 = X(~any(X==11,2),:)
Xnot11 = 3×5
1 2 3 4 5 1 2 3 4 5 1 2 3 4 5
This approach simply uses logical indexing. See the documentation for the any function to udnerstand how it works.
.

2 件のコメント

John Smith
John Smith 2021 年 10 月 4 日
編集済み: John Smith 2021 年 10 月 4 日
Without loop I was also able to do it with
K = X == 11;
A = X(sum(K,2)>0,:)
B = X(~sum(K,2)>0,:)
For my assignment however I have to do it with a loop unfortunately
Star Strider
Star Strider 2021 年 10 月 4 日
O.K.
Frist, don’t use the length function here. It returns the greatest dimension, and that may not always be the number of rows. Use the size function instead, and specify the required dimension.
Second, it is still possible to use the any function in the loop. The alternative is to use nested loops, and scan the elements of each row to see if any of them equal 11, then store the row appropriately if they do.
.

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