Suppose I have a sparse matrix Sparstica that is a vertical concatenation of several other sparse matrices. When I type Sparstica(:), I get a list of the nonzero elements. In the left column, will be the index of the element, in the right column will be the nonzero element.
How can I manipulate the i-th and j-th non-zero element of every other sparse block matrix in the middle n-2 blocks (n sparse block matrices in total)?
Appended: To clarify what I mean by the i-th and j-th element of every other sparse matrix, suppose I have
Sparstica = [A_1; A_2; A_3; ... ; A_n]
This was created from vertcat. Now I need to take the i-th and j-th, say the 3rd and 5th, nonzero element of every other sparse matrix from A_2 to A_{N-1} (I know the notation for this actually isn't allowed, but just for demonstrative purposes). I'd like to accomplish this without using for-loops if possible.
You can find the non-zero elements using find:
>> A = speye(3)
A =
(1,1) 1
(2,2) 1
(3,3) 1
>> I = find(A ~= 0)
I =
1
5
9
If you need the indices in row/column format, use ind2sub:
>> [X, Y] = ind2sub(size(A), I)
X =
1
2
3
Y =
1
2
3
Related
I have a line of code as follows
nrma2 = full(abs(sum(A.*A',1)));
but sum(...,1) is a row. abs() returns only a row with absolute numbers
So, what will full() do?
full() will make a full matrix out of a sparse one. A sparse matrix can be a row vector, that's just a 1 -by- N matrix. The code you posted presumably has A as a sparse matrix, or includes the full() in case it is not a full matrix already to ensure it becomes a full vector.
A = sparse([1,0;1,0])
A =
(1,1) 1
(2,1) 1
nrma2 = full(abs(sum(A.*A',1))) % Full row-matrix
nrma2 =
1 0
abs(sum(A.*A',1)) % Sparse row-matrix
ans =
(1,1) 1
Specifically, from the documentation on full():
full
Convert sparse matrix to full storage
(...)
S — Sparse matrix to convert
Sparse matrix to convert, specified as a matrix. If S is already a
full matrix, then A is identical to S.
I am looking for an easy way to obtain the column-wise average of a subset of values in a matrix (indexed by a logical matrix), preferably without having to use a loop. The problem that I am experiencing is that because the number of values in each column is different, matlab collapses the values-matrix and its column-wise mean becomes the total mean (of the entire matrix). Is there a specific function or easy workaround for this problem? See below for an example.
%% define value matrix and logical indexing matrix
values=[1 2 3 4; 5 6 7 8; 9 10 11 12];
indices=[1 0 0 1; 0 1 0 1; 1 1 0 0];
indices=indices==1; %convert to logical
%% calculate column-wise average
mean(values(indices),1)
accumarray-based approach
Use the column index as a grouping variable for accumarray:
[~, col] = find(indices);
result = accumarray(col(:), values(indices), [size(values,2) 1], #mean, NaN).';
Note that:
The second line uses (:) to force the first input to be a column vector. This is needed because the first line may produce col as a row or column vector, depending on the size of indices.
NaN is used as fill value. This is specified as the fifth input to accumarray.
The third input to accumarray defines output size. It is necessary to specify it explicitly (as opposed to letting accumarray figure it out) in case the last columns of indices only contain false.
Hacky approach
Multiply and divide element-wise by indices. That will turn unwanted entries into NaN, which can then be ignored by mean using the 'omitnan' option:
result = mean(values.*indices./indices, 1, 'omitnan');
Manual approach
Multiply element-wise by indices, sum each column, and divide element-wise by the sum of each column of indices:
result = sum(values.*indices, 1) ./ sum(indices, 1);
I have a network with edges and points. The FID shows the ID number of each edge that is created by Start_Point and End_Point. For example edge 3 is between the two points of 2 and 3 in the network.
FID Start_Point End_Point
1 1 2
2 1 4
3 2 3
4 2 4
I want to create a 4-by-4 matrix of these points. if there is an edge between 2 points the value is 1 else is inf:
[inf, 1, inf, 1;
1, inf, 1, 1;
inf, 1, inf, inf;
1, 1, inf, inf]
How can I create such a matrix in MATLAB?
You can convert it to a sparse matrix and then use full command to obtain adjacency matrix.
edges= [1 2;
3 4;
3 1
2 3];
n=size(edges,1);
% create sparse matrix with given edges and their reverse direction
A = sparse([edges(:,1); edges(:,2)],[edges(:,2); edges(:,1)],[ones(n,1); ones(n,1)]);
% create adjacency matrix
B=full(A);
% set zeros to inf
B(B==0)=inf;
and this is the result :
A =
(2,1) 1
(3,1) 1
(1,2) 1
(3,2) 1
(1,3) 1
(2,3) 1
(4,3) 1
(3,4) 1
>> B
B =
Inf 1 1 Inf
1 Inf 1 Inf
1 1 Inf 1
Inf Inf 1 Inf
Edit :
the sparse command create a sparse matrix with addressing values of its elements. One prototype for this command is as follow :
A=sparse(rows,cols,values);
for example A=sparse([1;2],[1,3],[10,5]) is a matrix which A(1,1)=10 and A(2,3)=5 and other elements are zero:
A=sparse([1;2],[1,3],[10,5]);
>> full(A)
ans =
10 0 0
0 0 5
In your case you have to add two directions to sparse matrix (symmetric) and all values are one. So you need to construct sparse matrix as :
A = sparse([edges(:,1); edges(:,2)],[edges(:,2); edges(:,1)],[ones(n,1); ones(n,1)]);
full command convert a sparse matrix to a dense one.
So you basically want to create an adjacency matrix from an adjacency list of edges? The number of your edges (i.e. your FID column) is irrelevant so I'm assuming your input data is of the form
edges = [1 2
1 4
2 3
2 4]
Now the first column of edges is the rows of your adjacency matrix and the second is the columns (or vice versa, it doesn't matter since your matrix is symmetrical)
The simplest solution is to use linear index which you would get via the sub2ind function:
adj = inf(size(edges,2));
idx = sub2ind(size(adj),edges(:,1), edges(:,2))
adj(idx) = 1;
I suspect your edges matrix will already be symmetrical but if it's not then just use
edges_sym = [edges; fliplr(edges)]
instead of edges
You can use accumarray:
edges1 = accumarray([startpoint endpoint]),1);
edges2 = edges1.'; % transpose your matrix, to obtain both edges
edges = edges1+edges2;
edges(edges==0)=inf;
accumarray gathers all points with common indices, pastes the value 1 on those indices. edges1 is the transpose of edges2, thus transpose, then add the two together. Find all indices on which the matrix is 0, then fill those values with inf.
Alternative:
edges= [1 2;
3 4;
3 1
2 3];
matrix = accumarray([edges;fliplr(edges)],1,[],[],inf);
fliplr flips your matrix left to right, to get all the desired combinations of indices. Then use accumarray to set a 1 on all locations specified by edges and put inf at the other locations.
If you are sure your matrix is symmetric, don't use fliplr, if you sure your matrix is non-symmetric, use fliplr and if you are not sure use this:
matrix = accumarray([edges;fliplr(edges)],1,[],#mean,inf);
where the #mean makes sure to set double entries to 1 anyway. For weighted edges do the following, where weights is an Nx1 array containing the weights and N is the number of edges.
matrix = accumarray([edges;fliplr(edges)],weights,[],#mean,inf);
I have a matrix A in Matlab of dimension mxn. I want to construct a vector B of dimension mx1 such that B(i)=1 if all elements of A(i,:) are equal and 0 otherwise. Any suggestion? E.g.
A=[1 2 3; 9 9 9; 2 2 2; 1 1 4]
B=[0;1;1;0]
One way with diff -
B = all(diff(A,[],2)==0,2)
Or With bsxfun -
B = all(bsxfun(#eq,A,A(:,1)),2)
Here's another example that's a bit more obfuscated, but also does the job:
B = sum(histc(A,unique(A),2) ~= 0, 2) == 1;
So how does this work? histc counts the frequency or occurrence of numbers in a dataset. What's cool about histc is that we can compute the frequency along a dimension independently, so what we can do is calculate the frequency of values along each row of the matrix A separately. The first parameter to histc is the matrix you want to compute the frequency of values of. The second parameter denotes the edges, or which values you are looking at in your matrix that you want to compute the frequencies of. We can specify all possible values by using unique on the entire matrix. The next parameter is the dimension we want to operate on, and I want to work along all of the columns so 2 is specified.
The result from histc will give us a M x N matrix where M is the total number of rows in our matrix A and N is the total number of unique values in A. Next, if a row contains all equal values, there should be only one value in this row where all of the values were binned at this location where the rest of the values are zero. As such, we determine which values in this matrix are non-zero and store this into a result matrix, then sum along the columns of the result matrix and see if each row has a sum of 1. If it does, then this row of A qualifies as having all of the same values.
Certainly not as efficient as Divakar's diff and bsxfun method, but an alternative since he took the two methods I would have used :P
Some more alternatives:
B = var(A,[],2)==0;
B = max(A,[],2)==min(A,[],2)
I have a 3D matrix say for eg. A(10x5x8). I need to get a 2D matrix (Boolean) out of it of size 10x5.
True if its elemental 3 Dimensional values are all same. i.e. Result(1,1) = 1 if A(1,1,1) == A(1,1,2) == A(1,1,3) etc..
False if at least one is different.
I expect a vectored approach which is fast and efficient.
Sample input:
A(:,:,1) = 1 2
2 2
A(:,:,2) = 1 1
2 3
Expected Output:
Result = 1 0
1 0
Use bsxfun with the eq function and use the first slice as the first input and compare with the other slices for the second input. Allow the first input to broadcast itself over the multiple slices.
Once you do that, use all and check the third dimension:
ind1 = bsxfun(#eq, A(:,:,1), A(:,:,2:end);
ind2 = all(ind1, 3);
The logic behind the above is very simple. How the first line of code works is that you would create a temporary matrix that would take the first slice of A and let it duplicate itself for as many slices as you have in A, without the first slice. Once you do this, you would do an element-by-element equality with this temporary matrix and the other slices. If you had a 3D column that was all equal, the one element from the first slice would be compared with every single value that corresponds to the same 3D column. Should they all equal to each other, then you would get a 3D column of all logical 1s. Therefore, to have a 3D column that is all equal to each other, all of the values should be 1, which is why all is used - to check if all values in a 3D column are equal to 1. Should all of the 3D column be a logical 1, we have matched your criteria.
Example run
>> A1 = [1 2; 2 2];
>> A2 = [1 1; 2 3];
>> A3 = [1 3; 2 4];
>> A4 = [1 5; 2 6];
>> A = cat(3, A1, A2, A3, A4);
>> ind1 = bsxfun(#eq, A(:,:,1), A(:,:,2:end);
>> ind2 = all(ind1, 3)
ind2 =
1 0
1 0
I made a matrix of 4 slices where the 3D column at the top left corner and the bottom left corner have all of the same values. Once you run through the code at the beginning of the post, we get what you expect.
Here's with short and sweet diff and must be quite memory efficient -
out = ~any(diff(A,[],3),3)
So, basically with diff along the third dimension diff(..[],3), you would calculate differences between the same (i,j) but on different 3D slices. Thus, if all such diff outputs are zeros, that would indicate that all dim3 elements for the same (i,j) are the same. This all zeros information is then picked up by ~any() also along dim3 with ~any(.,3) for the desired 2D array output.