Creating a submatrix from a matrix

I have a matrix for example; A = {1 3 7; 10 1 2; 11 5 9; 12 2 5] (however, mine is much bigger) And I would like to keep the rows where all the elements in the row are less than or equal to 7 to put into a new submatrix called B. How would I go about this?
Many thanks for any help you can give :)

 采纳的回答

A = [1 3 7; 10 1 2; 11 5 9; 12 2 5]
your_mat = A(all(A<=7,2),:);
I looked at your question history. Please accept an answer if it helped you. It is the only "payment" contributors in this forum receive.

5 个评论

Hi there,
I'm so sorry Jose... I didn't realise that that is what I should do! Please accept my sincerest apologies! Many thanks for flagging this up to me! Serves my right for not reading the rules!
Happy New year! :) Bran
No worries, and no apologies needed. It's not really a rule, but something nice to do for the contributors. Anyway...
Cheers!
Hi there,
Just a quick question. I have applied this code to another matrix modifying it slightly to read;
your_mat = B2(all((-8/3)<=B2<=(8/3),2),:)because I want to keep all rows where the all the column numbers are greater than -8/3 or less than 8/3 however, the matrix I get is the exact same size as the original B2 and I know that this is not the case. Any ideas what could be going wrong??
Many thanks in advance :) Bran
(-8/3)<=B2<=(8/3)
is not valid Matlab syntax. You could try instead:
abs(B2) <= 8/3
Many thanks, that worked for me :)

请先登录,再进行评论。

更多回答(3 个)

Thomas
Thomas 2013-1-8
编辑:Thomas 2013-1-8
A = [1 3 7; 10 1 2; 11 5 9; 12 2 5]
out=A(find(sum(A<=7,2)==size(A,2)),:) % rows with elements <=7
A = {1 3 7; 10 1 2; 11 5 9; 12 2 5};
B=cell2mat(A);
out1=B(B<7);
n=numel(out1);
n1=floor(sqrt(n));
m1=ceil(n/n1);
out=cell(1,n1*m1);
out(1:n)=num2cell(out1)';
B=reshape(out,n1,m1)
nabin
nabin 2014-5-8
I have a matrix A=[1 2 3; 1 2 9; 2 3 4]. I want a matrix B whose column 1 is equal to 1. How can I do this? B=[1 2 3; 1 2 9]

类别

帮助中心File Exchange 中查找有关 Creating and Concatenating Matrices 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by