Given two parameters:
n %number of repetitions per value
k %max value to repeat
I would like to create a vector of size n*k, which is a concatenation of k vectors of size n, such that the i'th vector contains the value i at each coordinate.
Example:
n = 5;
k = 9;
Desired result:
[1,1,1,1,1,2,2,2,2,2,3,3,3,3,3,4,4,4,4,4,5,5,5,5,5,6,6,6,6,6,7,7,7,7,7,8,8,8,8,8,9,9,9,9,9];
Is there an elegant way to achieve this?
Thanks!
quite a few ways to do it:
method 1:
A=1:k
repelem(A',n,1)'
method 2:
A=1:k
kron(A', ones(n,1))'
method 3:
A=1:k
B=repmat(A, n, 1)
B(:)'
method 4:
A=1:k
B=ones(n,1)*A
B(:)'
Here is an alternative method
A = reshape(mtimes((1:k).',ones(1,n)).',1,n*k)
A =
Columns 1 through 22
1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4 4 4 5 5
Columns 23 through 44
5 5 5 6 6 6 6 6 7 7 7 7 7 8 8 8 8 8 9 9 9 9
Column 45
9
It multiplies each element by ones n times
>> mtimes((1:k).',ones(1,5)).'
ans =
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9
and then reshapes the whole matrix to one vector
Related
I have a matrix A ,and vector x as following (left side)
where S0, H0,...is row number of each block. I want to exchange these blocks such that S0 and S1; H0 and H1 are near together as right side. This is my code
S0=3;
H0=2;
N0=2;
S1=4;
H1=5;
N1=4;
Cols=5;
Rows=S0+H0+N0+S1+H1+N1;
A=randi(10,[ Rows Cols]);
x=randi(10,[Rows 1]);
%% Exchange two block
temp=A(S0+H0+1:S0+H0+N0,1:end);
A(S0+H0+1:S0+H0+H1,1:end)=A(S0+H0+N0+S1+1:S0+H0+N0+S1+H1,1:end);
A(S0+H0+N0+S1+1:S0+H0+N0+S1+H1,1:end)=temp;
%% How exchange x
The above code is not work. How can I fixed it in MATLAB? Thank in advance.
One approach with mat2cell and cell2mat -
grps = [S0,H0,N0,S1,H1,N1]
new_pattern = [1 4 2 5 3 6]
celldata_roworder = mat2cell((1:size(A,1))',grps); %//'
newx = cell2mat(celldata_roworder(new_pattern)).'; %//'
newA = A(newx,:)
Sample run -
Input :
A =
6 8 9 8 7
4 8 8 3 4
3 8 2 1 10
5 2 6 8 3
5 7 4 7 7
4 5 6 8 7
6 3 4 7 4
8 1 5 5 2
5 9 2 4 1
5 2 3 9 5
2 2 1 4 2
1 7 10 9 8
3 9 7 8 4
4 6 10 9 9
7 8 2 6 8
10 2 10 7 6
10 10 8 10 2
5 6 6 5 10
3 7 5 1 3
8 1 3 9 10
grps =
3 2 2 4 5 4
new_pattern =
1 4 2 5 3 6
Output:
newx =
1 2 3 8 9 10 11 4 5 12 ...
13 14 15 16 6 7 17 18 19 20
newA =
3 3 2 5 8
4 3 3 7 7
1 5 2 8 1
4 6 4 1 4
7 1 5 8 8
4 9 10 10 8
7 10 10 4 3
7 3 1 6 9
2 9 2 6 10
1 1 7 10 3
10 10 10 4 7
9 1 8 9 5
8 7 4 5 7
9 8 7 5 3
1 10 7 6 8
8 1 10 6 1
4 6 3 3 2
7 9 3 2 9
6 9 7 4 8
6 7 6 8 10
I assume you are using a 2-dimensional matrix with Row rows and Cols columns.
You can use the colon : as a second index to address a full row, e.g. for the third row:
A(3, :)
(equal to A(3, 1:end) but little bit clearer).
So you could split your matrix into lines and re-arrange them like this (putting back together the lines to a two-dimensional matrix):
A = [ A(3:4, :); A(1:2, :); A(5:end, :) ]
This moves rows 3 and 4 at the beginning, then old lines 1 and 2 and then all the rest. Does this help you?
Hint: you can use eye for experimenting.
for eg if i have a matrix
4 5 9 8 3 8
3 2 4 10 1 3
1 9 9 6 7 7
2 1 7 4 6 7
2 6 3 5 4 2
7 2 2 9 3 4
How do I calculate the sum of the diagonal of the element 10 if I have its row and column indices?
So the output should be 9 + 10 + 7 + 7.
Thanks!
column = 4;
row = 2;
output = sum(diag(A, column - row));
Here you go:
>> x = [4,5,9,8,3 ,8
3,2,4,10,1, 3
1,9,9,6,7 ,7
2,1,7,4,6 ,7
2,6,3,5,4 ,2
7,2,2,9,3 ,4]
x =
4 5 9 8 3 8
3 2 4 10 1 3
1 9 9 6 7 7
2 1 7 4 6 7
2 6 3 5 4 2
7 2 2 9 3 4
>> xsum = sum(diag(x,4-2));
>> xsum
xsum = 33
parameterize the indices in case you need to use it more than once.
I have a 3D matrix and I want to store each 2D component of this in the row of another 2D matrix which has many rows as the 3rd dimension of the 3D matrix.
How can I do this?
With permute & reshape -
reshape(permute(A,[3 2 1]),size(A,3),[])
Sample run -
>> A
A(:,:,1) =
7 1 7 5
3 4 8 5
9 4 2 6
A(:,:,2) =
7 7 2 4
7 6 5 6
3 2 9 3
A(:,:,3) =
7 7 5 3
3 9 2 8
5 9 2 3
>> reshape(permute(A,[3 2 1]),size(A,3),[])
ans =
7 1 7 5 3 4 8 5 9 4 2 6
7 7 2 4 7 6 5 6 3 2 9 3
7 7 5 3 3 9 2 8 5 9 2 3
If you don't mind a little indexing madness...
You can build a linear index with the appropriate shape, which applied on the original array will give the desired result:
B = A(bsxfun(#plus, (1:L*M:L*M*N).', reshape(bsxfun(#plus, (0:L:L*M-1).', 0:L-1),1,[])));
Example:
>> A = randi(10,2,3,4)-1; %// example array; size 2x3x4
>> A
A(:,:,1) =
5 3 2
9 8 9
A(:,:,2) =
8 7 4
9 8 6
A(:,:,3) =
3 4 8
0 4 4
A(:,:,4) =
2 8 8
4 6 7
Result:
>> B
B =
5 3 2 9 8 9
8 7 4 9 8 6
3 4 8 0 4 4
2 8 8 4 6 7
That is easily done with MATLABs matrix unrolling syntax:
A=ones(N,M,O);
B=zeros(O,N*M);
for ii=1:size(A,3)
aux=A(:,:,ii); % aux is NxM
B(ii,:)=aux(:); % unroll!
end
(note I called O the thing you call N in your pictures)
I have a matrix for which I extract each column and do repmat function for each of them to build another matrix. Since i I have to do this for a large number of vectors(each column of my first matrix) it takes so long(relative to which I expect). If I do this for the whole matrix and then do something to build them, does it takes less time?
Consider this as an example:
A=[1 4 7;2 5 8;3 6 9]
I want to produce these
A1=[1 2 3 1 2 3 1 2 3
1 2 3 1 2 3 1 2 3
1 2 3 1 2 3 1 2 3]
A2=[4 5 6 4 5 6 4 5 6
4 5 6 4 5 6 4 5 6
4 5 6 4 5 6 4 5 6]
A3=[7 8 9 7 8 9 7 8 9
7 8 9 7 8 9 7 8 9
7 8 9 7 8 9 7 8 9]
As an alternative to #thewaywewalk's answer and using kron and repmat:
clear
A=[1 4 7;2 5 8;3 6 9];
B = repmat(kron(A',ones(3,1)),1,3);
A1 = B(1:3,:)
A2 = B(4:6,:)
A3 = B(7:end,:)
Which results in the following:
A1 =
1 2 3 1 2 3 1 2 3
1 2 3 1 2 3 1 2 3
1 2 3 1 2 3 1 2 3
A2 =
4 5 6 4 5 6 4 5 6
4 5 6 4 5 6 4 5 6
4 5 6 4 5 6 4 5 6
A3 =
7 8 9 7 8 9 7 8 9
7 8 9 7 8 9 7 8 9
7 8 9 7 8 9 7 8 9
Or as #Divakar pointed out, it would be advisable to create a single 3D array and store all your data in it (general solution):
n = 3; %// # of times you want to repeat the arrays.
A=[1 4 7;2 5 8;3 6 9];
B = repmat(kron(A',ones(n,1)),1,n);
C = zeros(n,n*size(A,2),3);
C(:,:,1) = B(1:n,:);
C(:,:,2) = B(n+1:2*n,:);
C(:,:,3) = B(2*n+1:end,:);
Try if this fits your needs:
A = [1 4 7;2 5 8;3 6 9];
n = 3; %// size(A,1)
cellArrayOutput = arrayfun(#(x) repmat( A(:,x).',n,n ), 1:size(A,2), 'uni',0)
instead of different variable names, everything is stored in a cell array.
if you insist on different names, I'd recommend to use structs:
A = [1 4 7;2 5 8;3 6 9];
n = 3;
structOutput = struct;
for ii = 1:size(A,2)
structOutput.(['A' num2str(ii)]) = repmat( A(:,ii).', n, n );
end
which gives you:
>> structOutput.A1
ans =
1 2 3 1 2 3 1 2 3
1 2 3 1 2 3 1 2 3
1 2 3 1 2 3 1 2 3
and so on.
I don't expect to much performance plus, you should share your full code for further help.
Anybody know a fast way to produce a matrix consisting of a linspace for each row? For example, the sort of pattern I'm looking for in this matrix is:
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
...
1 2 3 4 5 6 7 8 9 10
Anyone know any fast tricks to produce this WITHOUT using a for loop?
I just figured this out, so just in case anyone else was troubled by this, we can achieve this exact pattern by:
a=linspace(1,10,10);
b=ones(3,1)*a;
This will give:
>> a = 1 2 3 4 5 6 7 8 9 10
>> b = 1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
You need to use repmat.
Example:
>> B = repmat(1:10,[3 1])
B =
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
You can vary the value of 3 there. You can change it accordingly.
Another shortcut I can recommend is similar to repmat, but you specify a base array first of a = 1:10;. Once you do this, you specify a series of 1s in the first dimension when indexing which should produce a matrix of the same vectors with many rows as you want, where each row consists of the base array a. As such:
%// number of times to replicate
n = 4;
a = 1:10;
a = a(ones(1,n),:);
Result:
a =
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
Insert this command: transpose(ndgrid(1:10,1:n));, where n is the number of rows desired in the result.
You can consider these solutions:
With basic matrix indexing (taken from here)
b=a([1:size(a,1)]' * ones(1,NumToReplicate), :) %row-wise replication
b=a(:, ones(NumToReplicate, 1)) %column-wise replication
With bsxfun:
bsxfun(#times,a,(ones(1,NumToReplicate))') %row-wise replication
bsxfun(#times,a',(ones(1,NumToReplicate))) %column-wise replication
You are welcome to benchmark above two solutions with repmat.