I have two matrices, each with about ten rows. I am trying to stack the rows on top of each other consecutively. For example I have matrices A and B, and I have made a for loop to make it repeat the processes of stacking their rows. The problem now is that when I run the script, it only does it for the first row and doesnt continue on. The code is written below
A = [1 2 3 4 5 6 7 8 9 10];
B= [11 12 13 14 15 16 171 18 19 20];
for i2= 1:10
l= A(i2,1);
p= B(i2,1);
for i4= 1:10
i4 = [l;p] ;
end
end
% so the answer will be
% i4 = [1 11 2 12 3 13 4 14 5 15 6 16 7 17 8 18 9 19 10 20]
Please I need some help
I haven't got matlab to test this but:
new = []
for i= 1:10
l= A(i,:);
p= B(i,:);
for j= 1:10
new(end+1) =[l(j),p(j)] ;
end
end
There is surely a nicer way but this should work.
From:
https://stackoverflow.com/a/17766565/7252268
A = [1 2 3 4 5 6 7 8 9 10];
B= [11 12 13 14 15 16 171 18 19 20];
i4 = zeros(size(A,1), size(A,2)+size(B,2));
i4(:,1:2:end) = A;
i4(:,2:2:end) = B;
Alternatively to the solution of Ekin Inceleme you can use
i4 = reshape([A.' B.'].', 1, 2*size(A,2));
providing that A and B are equally sized. However, this one here is not so easy to read.
You don't need any loops!
just try this:
i4 = [A,B];
Related
I have a matrix A in Matlab of dimension Nx(N-1), e.g.
N=5;
A=[1 2 3 4;
5 6 7 8;
9 10 11 12;
13 14 15 16;
17 18 19 20];
I want to rearrange the elements of A in a certain way. Specifically I want to create a matrix B of dimension (N-1)xN such that:
for i=1,...,N,
B(:,i) collects
1) the first i-1 elements of the i-1th column of A and
2) the last N-i elements of the ith column of A.
Notice that for i=1 the i-1th column of A does not exist and therefore 1) is skipped; similarly, for i=N theith column of A does not exist and therefore 2) is skipped.
In the example above
B=[5 1 2 3 4
9 10 6 7 8
13 14 15 11 12
17 18 19 20 16];
This code does what I want. I am asking your help to vectorise it in an efficient way.
B=zeros(N-1,N);
for i=1:N
if i>1 && i<N
step1=A(1:i-1,i-1);
step2=A(i+1:N,i);
B(:,i)=[step1;step2];
elseif i==1
B(:,i)=A(i+1:N,i);
elseif i==N
B(:,i)=A(1:i-1,i-1);
end
end
Extract the lower and upper triangular matrices of A. Then reassemble them with a "diagonal shift":
u = triu(A);
l = tril(A,-1);
B = padarray(u(1:end-1,:),[0 1],'pre') + padarray(l(2:end,:),[0 1],'post');
Another valid approach using logical indexing combined with tril and triu:
B = zeros(size(A'));
B(tril(true(size(B)))) = A(tril(true(size(A)), -1));
B(triu(true(size(B)), 1)) = A(triu(true(size(A))));
Result:
>> B
B =
5 1 2 3 4
9 10 6 7 8
13 14 15 11 12
17 18 19 20 16
I have two matrices in Matlab A and B respectively of dimension MxN and GxN.
M can be greater or smaller than G.
A and B do not contain identical rows.
I want to construct a matrix C of dimension Hx(N+2) in the following way
C=[];
for i=1:size(A,1)
%if A(i,1:end-1) coincides with a row in B(:,1:end-1) (it can coincide with only one row at most)
%then C=[C;A(i,1:end) B(j,end)]; %where j is the index of the identical row in B
%otherwise impose C=[C;A(i,1:end) 0]
end
for i=1:size(B,1)
%if B(i,1:end-1) does not coincide with any row in A(:,1:end-1)
%then impose C=[C; B(i,1:end-1) 0 B(i,end)];
end
For example:
A=[1 2 3 4 5 100; 6 7 8 9 10 101; 11 12 13 14 15 102];
B=[6 7 8 9 10 103; 15 16 17 18 19 104]
C=[1 2 3 4 5 100 0; 6 7 8 9 10 101 103; 11 12 13 14 16 102 0; 15 16 17 18 19 0 104]
As M and G can be very high, I am looking for the fastest way to perform this.
You can use ismember + indexing to do your task:
[idx1,idx2] = ismember(A(:,1:end-1), B(:,1:end-1), 'rows');
idx3 = ~ismember(B(:,1:end-1), A(:,1:end-1), 'rows');
C(idx1,:) = [A(idx1,:) B(idx2(idx1),end)];
C(~idx1,:) = [A(~idx1,:) zeros(sum(~idx1),1)];
C=[C;B(idx3,1:end-1) zeros(sum(idx3),1) B(idx3,end)];
You could also use intersect with a bit of preallocation to speed up the assignment (if M or G gets really large).
A=[1 2 3 4 5 100; 6 7 8 9 10 101; 11 12 13 14 15 102];
B=[6 7 8 9 10 103; 15 16 17 18 19 104];
C=[1 2 3 4 5 100 0; 6 7 8 9 10 101 103; 11 12 13 14 16 102 0; 15 16 17 18 19 0 104];
[M,N] = size(A);
G = size(B,1);
[tmp, idxA, idxB] = intersect(A(:,1:end-1),B(:,1:end-1),'rows')
idxBnotA = setdiff([1:G],idxB);
H = M + G - length(idxA);
C1 = zeros(H,N+1);
C1(1:M,1:N) = A;
C1(idxA,end) = B(idxB,end);
C1(M+1:end,1:end-2) = B(idxBnotA,1:end-1);
C1(M+1:end,end) = B(idxBnotA,end)
The input is an N-by-1 matrix. I need to reshape it to L-by-M matrix. The following is an example.
Input:
b =
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Set length = 18, Output:
X =
1 2 3
2 3 4
3 4 5
4 5 6
5 6 7
6 7 8
7 8 9
8 9 10
9 10 11
10 11 12
11 12 13
12 13 14
13 14 15
14 15 16
15 16 17
16 17 18
17 18 19
18 19 20
Because I have a very big matrix, using a loop to reshape is very inefficient. How can I improve the reshape speed?
Your example output matrix X is the perfect matrix to index a vector of length N to get what you want. It's also very easy to create using bsxfun:
N = 20;
b = rand(N,1);
M = 3; %// number of columns
L = N-M; %// Note that N-M is an upper limit for L!
idx = bsxfun(#plus, (0:L)', 1:M)
X = b(idx)
That's exactly what im2col (from the Image Processing Toolbox) does:
b = (1:20).'; %'// example data
L = 18; % // desired length of sliding blocks
x = im2col(b, [L 1]); % // result
I'd use horzcat. For example:
function X = reshaper(b,len)
diff = length(b) - len + 1;
X = b(1:len);
for i=2:diff
X = horzcat(X,b(i:len+(i-1)));
end
You could probably remove the for loop with some further thought.
I have a matrix
A= [1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16; 17 18 19 20]
I want to do some calculation on this matrix. But actually I do not need all the rows. So I have to discard some of the rows from the above matrix before doing a calculation. After discarding 3 rows, we will have a new matrix.
B= [1 2 3 4; 9 10 11 12; 17 18 19 20];
Now I have to use B to make some other calculations. So how can I discard some of the unwanted rows from a matrix in matlab? Any suggestion will be helpful. Thanks.
Try this: (Use when no. of rows to keep is lesser)
%// Input A
A = [1 2 3 4; 5 6 7 8; 9 10 11 12; 13 14 15 16; 17 18 19 20];
%// Rows (1-3,5) you wanted to keep
B = A([1:3, 5],:)
Output:
B =
1 2 3 4
5 6 7 8
9 10 11 12
17 18 19 20
Alternative: (Use when no. of rows to discard is lesser)
%// rows 2 and 3 discarded
A([2,3],:) = [];
Output:
>> A
A =
1 2 3 4
13 14 15 16
17 18 19 20
Note: Here (in the alternate method), the output replaces the original A. So you need to back up A if you need it afterwards. You could do this before discarding operation to backup Input matrix
%// Input A is backed up in B
B = A;
You can select the indices of the rows you want to keep:
A([1,3,5],:)
ans =
1 2 3 4
9 10 11 12
17 18 19 20
I have two 3D matrices A(kl,1,r) and B(1,rs,r). kl=rs.
I need to get a new matrix C(kl,rs,r) which should have the product of column vector of A(kl,1) by the row vector of B(1,rs) for every page r without for loop
C=zeros(size(A,1),size(B,2),r);
for rr=1:size(A,3)
dummy=squeeze(A(:,:,rr))*squeeze(B(:,:,rr))';
C(:,:,rr)=dummy;
end
can anyone help with that? :)
Using bsxfun, you could do that directly in one line
out = bsxfun(#times, A, B);
Sample Inputs:
>> A
A(:,:,1) =
6
10
3
A(:,:,2) =
2
2
1
>> B
B(:,:,1) =
5 5 4
B(:,:,2) =
8 7 8
Results:
out(:,:,1) =
30 30 24
50 50 40
15 15 12
out(:,:,2) =
16 14 16
16 14 16
8 7 8