I have the following variables and code in MATLAB:
A = [ 2 2 2 2 2 ]
B = [ 1 2 3 4 5 ]
B / A % = 1.5
The answer, 1.5 is really the average of the values in B divided by 2, but why does MATLAB do this with these arrays and the / operator?
EDIT:
Looking at the documentation here near the bottom of the page for matrix right-division:
x = B/A where xA = B
...but multiplying x * A above results in [ 3 3 3 3 3 ]. This does not seem to make sense either. Am I missing something here as well?
The right division operator / (or the mrdivide function) will solve a systems of linear equations xA = B for x. From the above documentation link:
If A is a rectangular m-by-n matrix with m ~= n, and B is a matrix with n columns, then x = B/A returns a least-squares solution of the system of equations x*A = B.
It's computing a value for x that best approximates a solution in a least-squares sense, since in this case you have an underdetermined system of equations. There is no exact solution, so x*A doesn't exactly reproduce B.
c = a/b perform right-matrix division. It is not the same as the division of each element.
Related
I'm struggling to vectorise a function which performs a somewhat pairwise difference between two vectors x = 2xN and v = 2xM, for some arbitrary N, M. I have this to work when N = 1, although, I would like to vectorise this function to apply to inputs with N arbitrary.
Indeed, what I want this function to do is for each column of x find the normed difference between x(:,column) (a 2x1) and v (a 2xM).
A similar post is this, although I haven't been able to generalise it.
Current implementation
function mat = vecDiff(x,v)
diffVec = bsxfun(#minus, x, v);
mat = diffVec ./ vecnorm(diffVec);
Example
x =
1
1
v =
1 3 5
2 4 6
----
vecDiff(x,v) =
0 -0.5547 -0.6247
-1.0000 -0.8321 -0.7809
Your approach can be adapted as follows to suit your needs:
Permute the dimensions of either x or v so that its number of columns becomes the third dimension. I'm choosing v in the code below.
This lets you exploit implicit expansion (or equivalently bsxfun) to compute a 2×M×N array of differences, where M and N are the numbers of columns of x and v.
Compute the vector-wise (2-)norm along the first dimension and use implicit expansion again to normalize this array:
x = [1 4 2 -1; 1 5 3 -2];
v = [1 3 5; 2 4 6];
diffVec = x - permute(v, [1 3 2]);
diffVec = diffVec./vecnorm(diffVec, 2, 1);
You may need to apply permute differently if you want the dimensions of the output in another order.
Suppose your two input matrices are A (a 2 x N matrix) and B (a 2 x M matrix), where each column represents a different observation (note that this is not the traditional way to represent data).
Note that the output will be of the size N x M x 2.
out = zeros(N, M, 2);
We can find the distance between them using the builtin function pdist2.
dists = pdist2(A.', B.'); (with the transpositions required for the orientation of the matrices)
To get the individual x and y distances, the easiest way I can think of is using repmat:
xdists = repmat(A(1,:).', 1, M) - repmat(B(1,:), N, 1);
ydists = repmat(A(2,:).', 1, M) - repmat(B(2,:), N, 1);
And we can then normalise this by the distances found earlier:
out(:,:,1) = xdists./dists;
out(:,:,2) = ydists./dists;
This returns a matrix out where the elements at position (i, j, :) are the components of the normed distance between A(:,i) and B(:,j).
Consider three row vectors in Matlab, A, B, C, each with size 1xJ. I want to construct a matrix D of size Kx3 listing every triplets (a,b,c) such that:
a is the position in A of A(a).
b is the position in B of B(b).
A(a)-B(b) is an element of C.
c is the position in C of A(a)-B(b).
A(a) and B(b) are different from Inf, -Inf.
For example,
A=[-3 3 0 Inf -Inf];
B=[-2 2 0 Inf -Inf];
C=[Inf -Inf -1 1 0];
D=[1 1 3; %-3-(-2)=-1
2 2 4; % 3-2=1
3 3 5]; % 0-0=0
I would like this code to be efficient, because in my real example I have to repeat it many times.
This question relates to my previous question here, but now I'm looking for the positions of the elements.
You can use combvec (or any number of alternatives) to get all pairings of indices a and b for the corresponding arrays A and B. Then it's simply a case of following your criteria
Find the differences
Check which differences are in C
Remove elements you don't care about
Like so:
% Generate all index pairings
D = combvec( 1:numel(A), 1:numel(B) ).';
% Calculate deltas
delta = A(D(:,1)) - B(D(:,2));
delta = delta(:); % make it a column
% Get delta index in C (0 if not present)
[~,D(:,3)] = ismember(delta,C);
% If A or B are inf then the delta is Inf or NaN, remove these
idxRemove = isinf(delta) | isnan(delta) | D(:,3) == 0;
D(idxRemove,:) = [];
For your example, this yields the expected results from the question.
You said that A and B are at most 7 elements long, so you have up to 49 pairings to check. This isn't too bad, but readers should be careful that the pairings can grow quickly for larger inputs.
Suppose I have a matrix A [m x 1], where m is not necessarily even. I to create a matrix B also [m x 1] which tells me the decile of the elements in A (i.e. matrix B has numbers from 1 to 10).
I know I can use the function sort(A) to get the position of the elements in A and from there I can manually get deciles. Is there another way of doing it?
I think one possibility would be B = ceil(10 * tiedrank(A) / length(A) . What do you think? Are there any issues with this?
Also, more generally, if I have a matrix A [m x n] and I want to create a matrix B also [m x n], in which each column of B should have the decile of the corresponding column in A , is there a way of doing it without a for loop through the columns?
Hope the problem at hand is clear. So far I have been doing it using the sort function and then manually assigning the deciles, but it is very inefficient.
This is how I would do it:
N = 10;
B = ceil(sum(bsxfun(#le, A(:), A(:).'))*N/numel(A));
This counts, for each element, how many elements are less than or equal to it; and then rounds the results to 10 values.
Depending on how you define deciles, you may want to change #le to #lt, or ceil to floor. For numel(A) multiple of N, the above definition gives exactly numel(A)/N values in each of the N quantiles. For example,
>> A = rand(1,8)
A =
0.4387 0.3816 0.7655 0.7952 0.1869 0.4898 0.4456 0.6463
>> N = 4;
>> B = ceil(sum(bsxfun(#le, A(:), A(:).'))*N/numel(A))
B =
2 1 4 4 1 3 2 3
This question already has answers here:
Create a zero-filled 2D array with ones at positions indexed by a vector
(4 answers)
Closed 6 years ago.
I have a vector v of size (m,1) whose elements are integers picked from 1:n. I want to create a matrix M of size (m,n) whose elements M(i,j) are 1 if v(i) = j, and are 0 otherwise. I do not want to use loops, and would like to implement this as a simple vector-matrix manipulation only.
So I thought first, to create a matrix with repeated elements
M = v * ones(1,n) % this is a (m,n) matrix of repeated v
For example v=[1,1,3,2]'
m = 4 and n = 3
M =
1 1 1
1 1 1
3 3 3
2 2 2
then I need to create a comparison vector c of size (1,n)
c = 1:n
1 2 3
Then I need to perform a series of logical comparisons
M(1,:)==c % this results in [1,0,0]
.
M(4,:)==c % this results in [0,1,0]
However, I thought it should be possible to perform the last steps of going through each single row in compact matrix notation, but I'm stumped and not knowledgeable enough about indexing.
The end result should be
M =
1 0 0
1 0 0
0 0 1
0 1 0
A very simple call to bsxfun will do the trick:
>> n = 3;
>> v = [1,1,3,2].';
>> M = bsxfun(#eq, v, 1:n)
M =
1 0 0
1 0 0
0 0 1
0 1 0
How the code works is actually quite simple. bsxfun is what is known as the Binary Singleton EXpansion function. What this does is that you provide two arrays / matrices of any size, as long as they are broadcastable. This means that they need to be able to expand in size so that both of them equal in size. In this case, v is your vector of interest and is the first parameter - note that it's transposed. The second parameter is a vector from 1 up to n. What will happen now is the column vector v gets replicated / expands for as many values as there are n and the second vector gets replicated for as many rows as there are in v. We then do an eq / equals operator between these two arrays. This expanded matrix in effect has all 1s in the first column, all 2s in the second column, up until n. By doing an eq between these two matrices, you are in effect determining which values in v are equal to the respective column index.
Here is a detailed time test and breakdown of each function. I placed each implementation into a separate function and I also let n=max(v) so that Luis's first code will work. I used timeit to time each function:
function timing_binary
n = 10000;
v = randi(1000,n,1);
m = numel(v);
function luis_func()
M1 = full(sparse(1:m,v,1));
end
function luis_func2()
%m = numel(v);
%n = 3; %// or compute n automatically as n = max(v);
M2 = zeros(m, n);
M2((1:m).' + (v-1)*m) = 1;
end
function ray_func()
M3 = bsxfun(#eq, v, 1:n);
end
function op_func()
M4= ones(1,m)'*[1:n] == v * ones(1,n);
end
t1 = timeit(#luis_func);
t2 = timeit(#luis_func2);
t3 = timeit(#ray_func);
t4 = timeit(#op_func);
fprintf('Luis Mendo - Sparse: %f\n', t1);
fprintf('Luis Mendo - Indexing: %f\n', t2);
fprintf('rayryeng - bsxfun: %f\n', t3);
fprintf('OP: %f\n', t4);
end
This test assumes n = 10000 and the vector v is a 10000 x 1 vector of randomly distributed integers from 1 up to 1000. BTW, I had to modify Luis's second function so that the indexing will work as the addition requires vectors of compatible dimensions.
Running this code, we get:
>> timing_binary
Luis Mendo - Sparse: 0.015086
Luis Mendo - Indexing: 0.327993
rayryeng - bsxfun: 0.040672
OP: 0.841827
Luis Mendo's sparse code wins (as I expected), followed by bsxfun, followed by indexing and followed by your proposed approach using matrix operations. The timings are in seconds.
Assuming n equals max(v), you can use sparse:
v = [1,1,3,2];
M = full(sparse(1:numel(v),v,1));
What sparse does is build a sparse matrix using the first argument as row indices, the second as column indices, and the third as matrix values. This is then converted into a full matrix with full.
Another approach is to define the matrix containing initially zeros and then use linear indexing to fill in the ones:
v = [1,1,3,2];
m = numel(v);
n = 3; %// or compute n automatically as n = max(v);
M = zeros(m, n);
M((1:m) + (v-1)*m) = 1;
I think I've also found a way to do it, and it would be nice if somebody could tell me which of the methods shown is faster for very large vectors and matrices. The additional method I thought of is the following
M= ones(1,m)'*[1:n] == v * ones(1,n)
I have 2 matrices A (nxm) and B (nxd) and want to multiply element-wise each column of A with a row of B. There are m columns in A and n 1xd vectors in B so the results are m nxd matrices. Then I want to sum(result_i, 1) to get m 1xd vectors, which I want to apply vertcat to get a mxd matrix. I'm doing this operations using for loop and it is slow because n and d are big. How can I vectorize this in matlab to make it faster? Thank you.
EDIT:
You're all right: I was confused by my own question. What I meant by "multiply element-wise each column of A with a row of B" is to multiply n elements of a column in A with the corresponding n rows of B. What I want to do with one column of A is as followed (and I repeat this for m columns of A, then vertcat the C's vector together to get an mxd matrix):
column_of_A =
3
3
1
B =
3 1 3 3
2 2 1 2
1 3 3 3
C = sum(diag(column_of_A)*B, 1)
16 12 15 18
You can vectorize your operation the following way. Note, however, that vectorizing comes at the cost of higher memory usage, so the solution may end up not working for you.
%# multiply nxm A with nx1xd B to create a nxmxd array
tmp = bsxfun(#times,A,permute(B,[1 3 2]));
%# sum and turn into mxd
out = squeeze(sum(tmp,1));
You may want to do everything in one line, which may help the Matlab JIT compiler to save on memory.
EDIT
Here's a way to replace the first line if you don't have bsxfun
[n,m] = size(A);
[n,d] = size(B);
tmp = repmat(A,[1 1 d]) .* repmat(permute(B,[1 3 2]),[1,m,1]);
It's ugly, but as far as I can see, it works. I'm not sure it will be faster than your loop though, plus, it has a large memory overhead. Anyway, here goes:
A_3D = repmat(reshape(A, size(A, 1), 1, size(A, 2)), 1, size(B, 2));
B_3D = repmat(B, [ 1 1 size(A, 2)]);
result_3D = sum(A_3D .* B_3D, 1);
result = reshape(result_3D, size(A, 2), size(B, 2))
What it does is: make A into a 3D matrix of size n x 1 x m, so one column in each index of the 3rd dimension. Then we repeat the matrix so we get an n x d x m matrix. We repeat B in the 3rd dimension as well. We then do a piecewise multiplication of all the elements and sum them. The resulting matrix is a 1 x d x m matrix. We reshape this into a m x d matrix.
I'm pretty sure I switched around the size of the dimensions a few times in my explanation, but I hope you get the general gist.
Multiplying with a diagonal matrix seems at least twice as fast, but I couldn't find a way to use diag, since it wants a vector or 2D matrix as input. I might try again later tonight, I feel there must be a faster way :).
[Edit] Split up the command in parts to at least make it a little bit readable.
This is the way I would do this:
sum(repmat(A,1,4).*B)
If you don't know the number of columns of B:
sum(repmat(A,1,size(B,2)).*B)