I'm using MATLAB. I have a matrix with N elements filled with numbers [1; N], but there are only K unique number between them (K is much less than N). What is an efficient way to relabel the matrix so that it contains only numbers [1; K]? Equal numbers should become equal, and not equal should become not equal.
Example for N = 10, K = 4:
[1 4 8 9 4 1 8 9 4 1] -> [1 2 3 4 2 1 3 4 2 1]
N ~ 1 000 000
K ~ 10 000
Use the third output argument of unique:
a=[1 4 8 9 4 1 8 9 4 1];
[~, ~, b] = unique(a)
b =
1 2 3 4 2 1 3 4 2 1
Related
How can I generate the following matrix having m rows and n columns?
1st row 1 2 3 L n-1 n
2nd 2 3 4 L n 1
3rd 3 4 5 n-1 1 2
Nth M M M ....
N-1 m-1 m m+1 L m-3 m-2
last m m+1 m+2 l m-2 m-1
It's difficult to tell from your description, but it appears you want to create a matrix where the first row is 1:n and each successive row is a circular shift to the left of the previous row. If so, you can still use hankel for this (as Dev-iL mentions in their answer), but incorporate a remainder operation like so:
n = 5;
m = 8;
mat = rem(hankel(0:(m-1), (m-1):(m+n-2)), n)+1
mat =
1 2 3 4 5
2 3 4 5 1
3 4 5 1 2
4 5 1 2 3
5 1 2 3 4
1 2 3 4 5
2 3 4 5 1
3 4 5 1 2
This looks like a Hankel matrix. You should use the 2-input syntax for it,
H = hankel(c,r)
So for example, with m = 4 and n = 5 we get:
m = 4; n = 5;
X = hankel( 1:m, m:m+n-1 )
X =
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
The Mean Square Error(MSE), is a method used to define the difference in between two blocks, and can be calculated as follow:
a and b two blocks equal size
MSE = sqrt(sum(sum((a-b).^2)))/size(a or b)
If the MSE is less than a given threshold, than the two blocks are not diffrent.
Given two matrix A and B, the purpose is to divide the two matrix to blocks of a given size, then extract the first block from A and let it be a, then search for a block b from B where the Mean Square Error between a and b is less then a given threshold, then return the position of the block b from the matrix B. and so on. and here is an example:
Given two matrix A and B where:
A= [1 1 4 4 2 2
1 1 4 4 2 2
2 2 9 9 5 5
2 2 9 9 5 5
3 3 4 4 9 9
3 3 4 4 9 9];
B = [ 2 2 4 4 9 9
2 2 4 4 9 9];
the threshold is 2
The first block a obtained from the matrix A is:
1 1
1 1
The block b obtained from the matrix B that MSR between a and b is less than the threshold is:
2 2
2 2
Therefore we return the position of the block b in the matrix B which is 1
The second block a obtained from the matrix A is:
4 4
4 4
The block b obtained from the matrix B where MSR between a and b is less than threshold is:
4 4
4 4
Therefore we return the position of the block b in the matrix B which is 2. and so on.
The final result should be as follow
RES= [1 2 1
1 3 2
1 2 3];
Is there a faster way?
Here's a vectorized approach with bsxfun.
First define data:
A = [1 1 4 4 2 2
1 1 4 4 2 2
2 2 9 9 5 5
2 2 9 9 5 5
3 3 4 4 9 9
3 3 4 4 9 9]; %// data: A
B = [2 2 4 4 9 9
2 2 4 4 9 9]; %// data: B
m = 2; %// number of rows per block
n = 2; %// number of cols per block
Then apply the following steps:
Reshape matrices so that each block is a row (inspired by this great answer).
Compute the MSE for all pairs of blocks.
Find the argmin of the MSE with respect to blocks of B (for each block of A). Note that if there are several minimizing blocks in B this finds the first.
Reshape result into a matrix.
Code:
A2 = reshape(permute(reshape(A, size(A, 1), n, []), [2 1 3]), n*m, []); %// step 1
B2 = reshape(permute(reshape(B, size(B, 1), n, []), [2 1 3]), n*m, []);
mse = squeeze(sum(bsxfun(#minus, A2, permute(B2,[1 3 2])).^2, 1)); %// step 2
[~, result] = min(mse, [], 2); %// step 3
result = reshape(result, size(A,1)/m, size(A,2)/n); %// step 4
This question already has answers here:
Got confused with a vector indexed by a matrix, in Matlab
(2 answers)
Closed 8 years ago.
Suppose:
a =
1 2 3
4 5 6
2 3 4
and
b =
1 3 2
6 4 8
In MATLABa(b) gives:
>> a(b)
ans =
1 2 4
3 2 6
What is the reason for this output?
when you have a matrix a:
a =
1 2 3
4 5 6
7 8 9
and b:
b =
1 3 4
3 2 6
then a(b) is a way of adressing items in a and gives you:
>> a(b)
ans =
1 7 2
7 4 8
to understand this you have to think of a als a single column vector
>> a(:)
ans =
1
4
7
2
5
8
3
6
9
now the first row of b (1 3 4) addresses elements in this vector so the first, the 3rd and the forth element of that single column vector which are 1 7 and 2 are adressed. Next the secound row of b is used as adresses for a secound line in the output so the 3rd, the 2nd and the 6th elements are taken from a, those are 7 4 and 8.
It's just a kind of matrix indexing.
Matrix indexes numeration in 'a' matrix is:
1 4 7
2 5 8
3 6 9
This is a possible duplicate to this post where I gave an answer: Got confused with a vector indexed by a matrix, in Matlab
However, I would like to duplicate my answer here as I think it is informative.
That's a very standard MATLAB operation that you're doing. When you have a vector or a matrix, you can provide another vector or matrix in order to access specific values. Accessing values in MATLAB is not just limited to single indices (i.e. A(1), A(2) and so on).
For example, let's say we had a vector a = [1 2 3 4]. Let's also say we had b as a matrix such that it was b = [1 2 3; 1 2 3; 1 2 3]. By doing a(b) to access the vector, what you are essentially doing is a lookup. The output is basically the same size as b, and you are creating a matrix where there are 3 rows, and each element accesses the first, second and third element. Not only can you do this for a vector, but you can do this for a matrix as well.
Bear in mind that when you're doing this for a matrix, you access the elements in column major format. For example, supposing we had this matrix:
A = [1 2
3 4
5 6
7 8]
A(1) would be 1, A(2) would be 3, A(3) would be 5 and so on. You would start with the first column, and increasing indices will traverse down the first column. Once you hit the 5th index, it skips over to the next column. So A(5) would be 2, A(6) would be 4 and so on.
Here are some examples to further your understanding. Let's define a matrix A such that:
A = [5 1 3
7 8 0
4 6 2]
Here is some MATLAB code to strengthen your understanding for this kind of indexing:
A = [5 1 3; 7 8 0; 4 6 2]; % 3 x 3 matrix
B = [1 2 3 4];
C = A(B); % C should give [5 7 4 1]
D = [5 6 7; 1 2 3; 4 5 6];
E = A(D); % E should give [8 6 3; 5 7 4; 1 8 6]
F = [9 8; 7 6; 1 2];
G = A(F); % G should give [2 0; 3 6; 5 7]
As such, the output when you access elements this way is whatever the size of the vector or matrix that you specify as the argument.
In order to be complete, let's do this for a vector:
V = [-1 9 7 3 0 5]; % A 6 x 1 vector
B = [1 2 3 4];
C = V(B); % C should give [-1 9 7 3]
D = [1 3 5 2];
E = V(D); % E should give [-1 7 0 9]
F = [1 2; 4 5; 6 3];
G = V(F); % G should give [-1 9; 3 0; 5 7]
NB: You have to make sure that you are not providing indexes that would make the accessing out of bounds. For example if you tried to specify the index of 5 in your example, it would give you an error. Also, if you tried anything bigger than 9 in my example, it would also give you an error. There are 9 elements in that 3 x 3 matrix, so specifying a column major index of anything bigger than 9 will give you an out of bounds error.
Is there an efficient way (e.g., without using for loops manually) to create a matrix in Matlab that enumerates the 2-D coordinates of a matrix of a given size?
For example, if I'm given an m x n matrix, I want the resulting mn x 2 matrix to be as follows:
1 1
1 2
1 3
...
1 n
2 1
2 2
...
m 1
m 2
...
m n
Thanks in advance!
mat = [1 2;3 4;5 6;7 8;9 10];
[m,n] = size(mat);
vec = [kron(1:m,ones(1,n)); kron(ones(1,m),1:n)]'
1 1
1 2
2 1
2 2
3 1
3 2
4 1
4 2
5 1
5 2
Robert P. has a correct (and elegant) answer with a nifty use of kron, but just for fun here's the alternative with ndgrid,
>> mat=zeros(5,2);
>> [nn,mm] = ndgrid(1:size(mat,2),1:size(mat,1))
>> vec = [mm(:) nn(:)]
vec =
1 1
1 2
2 1
2 2
3 1
3 2
4 1
4 2
5 1
5 2
This problem is probably less to do with Matlab and more to do with matrix algebra (which I mostly forget from my college courses). Say I have a m x n matrix X and a m x 1 matrix B. How would I divide the X by B such that all the elements of the ith row of X are piecewise divided by the ith row of B, resulting in another m x n matrix Y?
E.g.
X = [2 4 8; 3 9 27; 4 16 64]
B = [2; 3; 4]
X ? B = [2/2 4/2 8/2; 3/3 9/3 27/3; 4/4 16/4 64/4]
ans =
1 2 4
1 3 9
1 4 16
Better not use repmat - it is slow and allocates additional memory for the workspace. You can use bsxfun, which is an inbuilt function, so it is faster and avoids the extra workspace:
X = [2 4 8; 3 9 27; 4 16 64]
B = [2; 3; 4]
bsxfun(#rdivide, X, B)
ans =
1 2 4
1 3 9
1 4 16
Junuxx's comment pointed me in the right direction. The solution I used to get what I wanted is:
B_prime = repmat(B,1,3)
X ./ B_prime
ans =
1 2 4
1 3 9
1 4 16
I'd still like to know what this kind of operation is called (if it even has a formal name).
X is m x n and B is m x 1
size(X,2) gives the value of n i.e. number of columns
So, you need to do:
X./repmat(B,1,size(X,2))
X = [2 4 8; 3 9 27; 4 16 64]
B = [2; 3; 4]
Result= X./B(:,ones(1,3)) %is faster then repmat
Result =
1 2 4
1 3 9
1 4 16