Say I have a nxm matrix and want to treat each row as vectors in a function. So, if I have a function that adds vectors, finds the Cartesian product of vectors or for some reason takes the input of several vectors, I want that function to treat each row in a matrix as a vector.
This sounds like a very operation in Matlab. You can access the ith row of a matrix A using A(i, :). For example, to add rows i and j, you would do A(i, :) + A(j, :).
Given an nxm matrix A:
If you want to edit a single column/row you could use the following syntax: A(:, i) for the ith-column and A(i, :) for ith-row.
If you want to edit from a column/row i to a column/row j, you could use that syntax: A(:, i:j) or A(i:j, :)
If you want to edit (i.e.) from the penultimate column/row to the last one, you could you: A(:, end-1:end) or A(end-1:end, :)
EDIT:
I can't add a comment above because I don't have 50 points, but you should post the function setprod. I think you should be able to do what you want to do, by iterating the matrix you're passing as an argument, with a for-next statement.
I think you're going to have to loop:
Input
M = [1 2;
3 4;
5 6];
Step 1: Generate a list of all possible row pairs (row index numbers)
n = size(M,1);
row_ind = nchoosek(1:n,2)
Step 2: Loop through these indices and generate the product set:
S{n,n} = []; //% Preallocation of cell matrix
for pair = 1:size(row_ind,1)
p1 = row_ind(pair,1);
p2 = row_ind(pair,2);
S{p1,p2} = setprod(M(p1,:), M(p2,:))
end
Transform the matrix into a list of row vectors using these two steps:
Convert the matrix into a cell array of the matrix rows, using mat2cell.
Generate a comma-separated list from the cell array, using linear indexing of the cell contents.
Example: let
v1 = [1 2];
v2 = [10 20];
v3 = [11 12];
M = [v1; v2; v3];
and let fun be a function that accepts an arbitrary number of vectors as its input. Then
C = mat2cell(M, ones(1,size(M,1)));
result = fun(C{:});
is the same as result = fun(v1, v2, v3).
Related
Sorry about the bad title, I'm struggling to word this question well. Basically what I want to do is extract elements from a 2d matrix, from row by row, taking out a number of elements (N) starting at a particular column (k). In for loops, this would look like.
A = magic(6);
k = [2,2,3,3,4,4]; % for example
N = 3;
for j = 1:length(A)
B(j,:) = A(j,k(j):k(j)+N-1);
end
I figure there must be a neater way to do it than that.
You could use bsxfun to create an array of indices to use. Then combine this with the row numbers and pass it to sub2ind.
inds = sub2ind(size(A), repmat(1:size(A, 1), 3, 1), bsxfun(#plus, k, (0:(N-1))')).';
B = A(inds);
Or alternately without sub2ind (but slightly more cryptic).
B = A(bsxfun(#plus, 1:size(A,1), ((bsxfun(#plus, k, (0:(N-1)).')-1) * size(A,1))).');
Here's one approach using bsxfun's masking capability and thus logical indexing -
C = (1:size(A,2))';
At = A.';
B = reshape(At(bsxfun(#ge,C,k) & bsxfun(#lt,C,k+N)),N,[]).';
I have the following vector u:
u=[a1,a2,a3,a4,b1,b2,b3,b4,c1,c2,c3,c4];
I want to permute the elements of of u to make the following vector, uNew:
uNew=[a1,b1,c1,a2,b2,c2,a3,b3,c3,a4,b4,c4];
I can think of no way of doing this other than with a for loop:
uNew=[];
for i=1:4
uNew=[uNew,u(i:4:end)];
end
But I'm hoping that a built-in function exits? Thanks!
Reshape it to a matrix which, read column by column, contains the order you want:
n=3 % number of categories (a,b,c)
u2=reshape(u,[],n).'
then convert it back to a vector:
u2=u2(:);
I'd use Daniel's approach. But just to provide an alternative:
m = 4; %// size of each initial "block"
[~, ind] = sort(mod(0:numel(u)-1,m)+1);
uNew = u(ind);
Note that this works because, as per sort's documentation,
The ordering of the elements in B (output) preserves the order of any equal elements in A (input)
Another approach:
N = 4;
uNew = u(mod(0:numel(u)-1, N) * N + floor((0:numel(u)-1)/N) + 1);
I have a 1-by-4 cell array, D. Each of the cell elements contains 2-by-2 double matrices. I want to do random permutation over each matrix independently which in result I will have the same size cell array as D but its matrices' elements will be permuted and then the inverse in order to obtain the original D again.
for a single matrix case I have the code and it works well as follows:
A=rand(3,3)
p=randperm(numel(A));
A(:)=A(p)
[p1,ind]=sort(p);
A(:)=A(ind)
but it doesn't work for a cell array.
The simplest solution for you is to use a loop:
nd = numel(D);
D_permuted{1,nd} = [];
D_ind{1,nd} = [];
for d = 1:nd)
A=D{d};
p=randperm(numel(A));
A(:)=A(p)
[~,ind]=sort(p);
D_permuted{d} = A;
D_ind{d} = ind;
end
Assuming your D matrix is just a list of identically sized (e.g. 2-by-2) matrices, then you could avoid the loop by using a 3D double matrix instead of the cell-array.
For example if you hade a D like this:
n = 5;
D = repmat([1,3;2,4],1,1,n)*10 %// Example data
Then you can do the permutation like this
m = 2*2; %// Here m is the product of the dimensions of each matrix you want to shuffle
[~,I] = sort(rand(m,n)); %// This is just a trick to get the equivalent of a vectorized form of randperm as unfortunately randperm only accepts scalars
idx = reshape(I,2,2,n);
D_shuffled = D(idx);
I have an array with n dimensions, and I have a sequence along one dimension at a certain location on all other dimensions. How do I find the location of this sequence? Preferably without loops.
I use matlab. I know what dimension it should be in, but the sequence isnt necessarily there. Find and == dont work. I could make an nd find function using crosscorrelation but Im guessing this is already implemented and I just dont know what function to call.
example:
ND = rand(10,10,10,10);
V = ND(randi(10),randi(10),randi(10),:);
[I1, I2, I3] = find(ND==V);
Edit: The sequence to be found spans the entire dimension it is on, I did not mention this in my original formulation of the problem. Knedlsepp`s solution solves exactly the problem I had, but Luis' solution solves a more general problem for when the sequence doesn't necessarily span the entire dimension.
As there are multiple ways to interpret your question, I will clarify: This approach assumes a 1D sequence of size: numel(V) == size(ND, dimToSearch). So, for V = [1,2] and ND = [1,2,1,2] it is not applicable. If you want this functionality go with Luis Mendo's answer, if not this will likely be faster.
This will be a perfect opportunity to use bsxfun:
We start with some example data:
ND = rand(10,10,10,10);
V = ND(3,2,:,3);
If you don't have the vector V given in the correct dimension (in this case [1,1,10,1]) you can reshape it in the following way:
dimToSearch = 3;
Vdims = ones(1, ndims(ND));
Vdims(dimToSearch) = numel(V);
V = reshape(V, Vdims);
Now we generate a cell that will hold the indices of the matches:
I = cell(1, ndims(ND));
At this point we compute the size of ND if it were collapsed along the dimension dimToSearch (we compute dimToSearch according to V, as at this point it will have the correct dimensions):
dimToSearch = find(size(V)>1);
collapsedDims = size(ND);
collapsedDims(dimToSearch) = 1;
Finally the part where we actually look for the pattern:
[I{:}] = ind2sub(collapsedDims, find(all(bsxfun(#eq, ND, V), dimToSearch)));
This is done in the following way: bsxfun(#eq, ND, V) will implicitly repmat the array V so it has the same dimensions as ND and do an equality comparison. After this we do a check with all to see if all the entries in the dimension dimToSearch are equal. The calls to find and ind2sub will then generate the correct indices to your data.
Let d be the dimension along which to search. I'm assuming that the sought sequence V may be shorter than size(ND,d). So the sequence may appear once, more than once, or never along each dimension-d- "thread".
The following code uses num2cell to reshape ND into a cell array such that each dimension-d-thread is in a different cell. Then strfind is applied to each cell to determine matches with V, and the result is a cell array with the same dimensions as ND, but where the dimension d is a singleton. The contents of each cell tell the d-dimension-positions of the matches, if any.
Credit goes to #knedlsepp for his suggestion to use num2cell, which greatly simplified the code.
ND = cat(3, [1 2 1 2; 3 4 5 6],[2 1 0 5; 0 0 1 2] ); %// example. 2x4x2
V = 1:2; %// sought pattern. It doesn't matter if it's a row, or a column, or...
d = 2; %// dimension along which to search for pattern V
result = cellfun(#(x) strfind(x(:).', V(:).'), num2cell(ND,d), 'UniformOutput', 0);
This gives
ND(:,:,1) =
1 2 1 2
3 4 5 6
ND(:,:,2) =
2 1 0 5
0 0 1 2
V =
1 2
result{1,1,1} =
1 3 %// V appears twice (at cols 1 and 3) in 1st row, 1st slice
result{2,1,1} =
[] %// V doesn't appear in 2nd row, 1st slice
result{1,1,2} =
[] %// V appears appear in 1st row, 2nd slice
result{2,1,2} =
3 %// V appears once (at col 3) in 2nd row, 2nd slice
One not very optimal way of doing it:
dims = size(ND);
Vrep = repmat(V, [dims(1), dims(2), dims(3), 1]);
ND_V_dist = sqrt(sum(abs(ND.^2-Vrep.^2), 4));
iI = find(ND_V_dist==0);
[I1, I2, I3] = ind2sub([dims(1), dims(2), dims(3)], iI);
I have a cell array of 53 different (40,000 x 2000) sparse matrices. I need to take the mean over the third dimension, so that for example element (2,5) is averaged across the 53 cells. This should yield a single (33,000 x 2016) output. I think there ought to be a way to do this with cellfun(), but I am not able to write a function that works across cells on the same within-cell indices.
You can convert from sparse matrix to indices and values of nonzeros entries, and then use sparse to automatically obtain the sum in sparse form:
myCell = {sparse([0 1; 2 0]), sparse([3 0; 4 0])}; %// example
C = numel(myCell);
M = cell(1,C); %// preallocate
N = cell(1,C);
V = cell(1,C);
for c = 1:C
[m n v] = find(myCell{c}); %// rows, columns and values of nonzero entries
M{c} = m.';
N{c} = n.';
V{c} = v.';
end
result = sparse([M{:}],[N{:}],[V{:}])/C; %'// "sparse" sums over repeated indices
This should do the trick, just initialize an empty array and sum over each element of the cell array. I don't see any way around using a for loop without concatenating it into one giant 3D array (which will almost definitely run out of memory)
running_sum=zeros(size(cell_arr{1}))
for i=1:length(cell_arr)
running_sum=running_sum+cell_arr{i};
end
means = running_sum./length(cell_arr);