How to convert adjacency list to adjacency matrix via matab
For example: Here is the adjacency list(undirected), the third column is the weight.
1 2 3
1 3 4
1 4 5
2 3 4
2 5 8
2 4 7
++++++++++++++++++++++
that should be converted to:
1 2 3 4 5
1 0 4 5 0
2 3 4 7 8
3 4 7 0 0
4 0 7 0 0
5 0 8 0 0
You can use sparse matrix. Let rows be the first column, cols the second, and s the weight.
A = sparse([rows; cols],[cols; rows],[s; s]);
If you want to see the matrix. use full().
UPDATE:
I made the answer a bit simpler (everything in one line, instead of adding the transposed, and included explanations, as requested:
list = [1 2 3
1 3 4
1 4 5
2 3 4
2 5 8
2 4 7];
rows = list(:,1)
cols = list(:,2)
s = list(:,3)
Now, rows, cols and s contains the needed information. Sparse matrices need three vectors. Each row of the two first vectors, rows and cols is the index of the value given in the same row of s (which is the weight).
The sparse command assigns the value s(k) to the matrix element adj_mat(rows(k),cols(k)).
Since an adjacency matrix is symmetric, A(row,col) = A(col,row). Instead of doing [rows; cols], it is possible to first create the upper triangular matrix, and then add the transposed matrix to complete the symmetric matrix.
A = sparse([rows; cols],[cols; rows],[s; s]);
full(A)
A =
0 3 4 5 0
3 0 4 7 8
4 4 0 0 0
5 7 0 0 0
0 8 0 0 0
It's really hard to tell what your'e asking. Is this right?
list = [1 2 3
1 3 4
1 4 5
2 3 4
2 5 8
2 4 7];
matrix = zeros(max(max(list(:, 1:2)))); %// Or just zeros(5) if you know you want a 5x5 result
matrix(sub2ind(size(matrix), list(:,1), list(:,2))) = list(:,3); %// Populate the upper half
matrix = matrix + matrix' %'// Find the lower half via symmetry
matrix =
0 3 4 5 0
3 0 4 7 8
4 4 0 0 0
5 7 0 0 0
0 8 0 0 0
Related
[1 2 3 4 5 6 7 8 9 ;
9 8 7 6 5 4 3 2 1 ;
1 2 0 0 1 0 0 0 1 ]
The last row has five columns with zeros. I would like to keep only one column per zero crossing.
like this
[1 2 3 5 8 9 ;
9 8 7 5 2 1 ;
1 2 0 1 0 1 ]
Is this possible with fast Matlab functions or do I have to write some slow complicated for loop ?
You can create a logical array many different ways to find the columns to remove. Something like this would work
% Find the zeros that are not the first zero
cols_to_remove = data(end,:) == 0 & ~diff([false, data(end,:) == 0]) == 1;
% Now remove them
data(:, cols_to_remove) = [];
I have a two long vector. Vector one contains values of 0,1,2,3,4's, 0 represent no action, 1 represent action 1 and 2 represent the second action and so on. Each action is 720 sample point which means that you could find 720 consecutive twos then 720 consecutive 4s for example. Vector two contains raw data corresponding to each action. I need to create a matrix for each action ( 1, 2, 3 and 4) which contains the corresponding data of the second vector. For example matrix 1 should has all the data (vector 2 data) which occurred at the same indices of action 1. Any Help??
Example on small amount of data:
Vector 1: 0 0 1 1 1 0 0 2 2 2 0 0 1 1 1 0 0 2 2 2
Vector 2: 6 7 5 6 4 6 5 9 8 7 9 7 0 5 6 4 1 5 8 0
Result:
Matrix 1:
5 6 4
0 5 6
Matrix 2:
9 8 7
5 8 0
Here is one approach. I used a cell array to store the output matrices, hard-coding names for such variables isn't a good plan.
V1=[0 0 1 1 1 0 0 2 2 2 0 0 1 1 1 0 0 2 2 2]
V2=[6 7 5 6 4 6 5 9 8 7 9 7 0 5 6 4 1 5 8 0]
%// Find length of sequences of 1's/2's
len=find(diff(V1(find(diff(V1)~=0,1)+1:end))~=0,1)
I=unique(V1(V1>0)); %// This just finds how many matrices to make, 1 and 2 in this case
C=bsxfun(#eq,V1,I.'); %// The i-th row of C contains 1's where there are i's in V1
%// Now pick out the elements of V2 based on C, and store them in cell arrays
Matrix=arrayfun(#(m) reshape(V2(C(m,:)),len,[]).',I,'uni',0);
%// Note, the reshape converts from a vector to a matrix
%// Display results
Matrix{1}
Matrix{2}
Since, there is a regular pattern in the lengths of groups within Vector 1, that could be exploited to vectorize many things while proposing a solution. Here's one such implementation -
%// Form new vectors out of input vectors for non-zero elements in vec1
vec1n = vec1(vec1~=0)
vec2n = vec2(vec1~=0)
%// Find positions of group shifts and length of groups
df1 = diff(vec1n)~=0
grp_change = [true df1]
grplen = find(df1,1)
%// Reshape vec2n, so that we end up with N x grplen sized array
vec2nr = reshape(vec2n,grplen,[]).' %//'
%// ID/tag each group change based on their unique vector 2 values
[R,C] = sort(vec1n(grp_change))
%// Re-arrange rows of reshaped vector2, s.t. same ID rows are grouped succesively
vec2nrs = vec2nr(C,:)
%// Find extents of each group & use those extents to have final cell array output
grp_extent = diff(find([1 diff(R) 1]))
out = mat2cell(vec2nrs,grp_extent,grplen)
Sample run for the given inputs -
>> vec1
vec1 =
0 0 1 1 1 0 0 2 2 2 ...
0 0 1 1 1 0 0 2 2 2
>> vec2
vec2 =
6 7 5 6 4 6 5 9 8 7 ...
9 7 0 5 6 4 1 5 8 0
>> celldisp(out)
out{1} =
5 6 4
0 5 6
out{2} =
9 8 7
5 8 0
Here is another solution:
v1 = [0 0 1 1 1 0 0 2 2 2 0 0 1 1 1 0 0 2 2 2];
v2 = [6 7 5 6 4 6 5 9 8 7 9 7 0 5 6 4 1 5 8 0];
m1 = reshape(v2(v1 == 1), 3, [])'
m2 = reshape(v2(v1 == 2), 3, [])'
EDIT: David's solution is more flexible and probably more efficient.
given matrix A of size 6 by 6 contain blocks of numbers,each block of size 2 by 2, and outher reference matrix R of size 2 by 12 also contain blocks of numbers, each block of size 2 by 2. the perpse of the whole process is to form a new matrix, called the Index matrix, contain index's that refer to the position of the blocks within the matrix A based on the order of the blocks within the reference matrix R. and here is an exemple
matrix A:
A =[1 1 2 2 3 3;
1 1 2 2 3 3;
1 1 3 3 4 4;
1 1 3 3 4 4;
4 4 5 5 6 6;
4 4 5 5 6 6 ]
matrix R:
R=[1 1 2 2 3 3 4 4 5 5 6 6;
1 1 2 2 3 3 4 4 5 5 6 6 ]
the new matrix is:
Index =[1 2 3;
1 3 4;
4 5 6]
any ideas ?
With my favourite three guys - bsxfun, permute, reshape for an efficient and generic solution -
blksz = 2; %// blocksize
num_rowblksA = size(A,1)/blksz; %// number of blocks along rows in A
%// Create blksz x blksz sized blocks for A and B
A1 = reshape(permute(reshape(A,blksz,num_rowblksA,[]),[1 3 2]),blksz^2,[])
R1 = reshape(R,blksz^2,1,[])
%// Find the matches with "bsxfun(#eq" and corresponding indices
[valid,idx] = max(all(bsxfun(#eq,A1,R1),1),[],3)
%// Or with PDIST2:
%// [valid,idx] = max(pdist2(A1.',reshape(R,blksz^2,[]).')==0,[],2)
idx(~valid) = 0
%// Reshape the indices to the shapes of blocked shapes in A
Index = reshape(idx,[],num_rowblksA).'
Sample run with more random inputs -
>> A
A =
2 1 1 2
1 2 2 1
1 1 1 1
2 2 2 2
1 2 2 1
1 2 1 1
>> R
R =
2 1 1 1 1 2 2 2 1 1 1 1
2 1 2 1 1 2 2 1 2 2 2 1
>> Index
Index =
0 0
5 5
3 0
Suppose an index vector in binary like below
Input
1 1 0 0 1 0 1
1 2 3 4 5 6 7
Intended output
1 2 5 7
which denotes nth number to be chosen. So I want to change 1 1 0 0 1 0 1 to 1 2 5 7, is there some easy way for this?
If you actually want to use your output to index another vector, do it directly.
You just need to transform your binary vector to logical
A = [1 1 0 0 1 0 1]; %assuming its double
B = [1 2 3 4 5 6 7];
C = B( logical(A) )
C =
1 2 5 7
The solution is using the function find(indicesBinary)
In Matlab, how can I remove spesific rows from a matrix I require? If for example I would like to remove all rows from a matrix which contain a spesific value (like 0 or NaN)?
Let's say you have A
A = [1 2 3;4 5 0; 7 8 9; 10 NaN 12]
A =
1 2 3
4 5 0
7 8 9
10 NaN 12
Then, you can choose the rows as follows:
any(isnan(A'))
ans =
0 0 0 1
To delete those NaN-containing rows, you can do:
A(any(isnan(A')),:) = []
A =
1 2 3
4 5 0
7 8 9
You can choose 0-containing rows by any(A' == 0). If you want all elements to be 0s or NaNs, then you can use all instead of any.