I have an adjacency matrix(n*n) of 1's and 0's extracted from an unweighted and undirected graph, my goal is to remove all-zeros columns from this matrix and their corresponding rows which are not connected to any node from the graph.
I want to apply one algorithm by using this adjacency matrix but sadly NaN produces because of some columns in this matrix are all 0's. So, i only need the connected components.
fid= fopen('file.txt','rt');
format = repmat('%q',[1 2]);
filee= textscan(fid,format,'Delimiter', '\t');
fclose(fid);
AA2= [filee{:, 1} , filee{:, 2}];
[nodenames, ~, id] = unique(AA2(:));
Adjacency_Matrix= accumarray(reshape(id, size(AA2)), 1, [numel(nodenames), numel(nodenames)]);
Adjoint2 = sum(Adjacency_Matrix~=0,1);
https://drive.google.com/file/d/0B6u8fZadKIp2OFd2X1NrZEdIclU/view
By this command Adjoint2 = sum(Adjacency_Matrix3~=0,1); I can know how many 1's i have in every column. In this matrix, some columns had no 1's, and so don't want them.
As the matrix is (NN) i want to get a (mm) matrix with columns that has only the 1's.
You can use any function:
If the matrix is symmetric you can do this to remove both zero columns and rows:
idx = any(Adjacency_Matrix);
result = Adjacency_Matrix(idx,idx);
else you can generate indexes of both columns and rows:
idx_column = any(Adjacency_Matrix);
idx_row = any(Adjacency_Matrix,2);
Here you can remove both columns and rows
result = Adjacency_Matrix(idx_row, idx_column)
If you want to remove only columns use this:
result = Adjacency_Matrix(:, idx_column)
If you want to remove only rows use this:
result = Adjacency_Matrix(idx_row, :)
This is going to be a brute force solution, since you are using an adjacency matrix. You will have to loop through all of the rows of the matrix and determine which ones are empty, and create a new adjacency matrix from these rows, omitting the correct columns.
You can use the following code:
i = 1;
while(i<=length(A))
if(sum(A(i,:)) == 0 && sum(A(:,i)) == 0)
% node i is isolated
A(:,i) = []; % remove its related column
A(i,:) = []; % remove its related row
else
i = i + 1;
end
end
If sum of elements of a row and respected column is zero, it means this node is isolated. Therefore, you can remove its related row and column.
Related
I have a 102-by-102 matrix. I want to select square sub-matrices of orders from 2 up to 8 using random column numbers. Here is what I have done so far.
matt is the the original matrix of size 102-by-102.
ittr = 30
cols = 3;
for i = 1:ittr
rr = randi([2,102], cols,1);
mattsub = matt([rr(1) rr(2) rr(3)], [rr(1) rr(2) rr(3)]);
end
I have to extract matrices of different orders from 2 to 8. Using the above code I would have to change the mattsub line every time I change cols. I believe it is possible to do with another loop inside but cannot figure out how. How can I do this?
There is no need to extract elements of a vector and concatenate them, just use the vector to index a matrix.
Instead of :
mattsub = matt([rr(1) rr(2) rr(3)], [rr(1) rr(2) rr(3)]);
Use this:
mattsub = matt(rr, rr);
Defining a set of random sizes is pretty easy using the randi function. Once this is done, they can be projected along your iterations number N using arrayfun. Within the iterations, the randperm and sort functions can be used in order to build the random indexers to the original matrix M.
Here is the full code:
% Define the starting parameters...
M = rand(102);
N = 30;
% Retrieve the matrix rows and columns...
M_rows = size(M,1);
M_cols = size(M,2);
% Create a vector of random sizes between 2 and 8...
sizes = randi(7,N,1) + 1;
% Generate the random submatrices and insert them into a vector of cells...
subs = arrayfun(#(x)M(sort(randperm(M_rows,x)),sort(randperm(M_cols,x))),sizes,'UniformOutput',false);
This can work on any type of matrix, even non-squared ones.
You don't need another loop, one is enough. If you use randi to get a random integer as size of your submatrix, and then use those to get random column and row indices you can easily get a random submatrix. Do note that the ouput is a cell, as the submatrices won't all be of the same size.
N=102; % Or substitute with some size function
matt = rand(N); % Initial matrix, use your own
itr = 30; % Number of iterations
mattsub = cell(itr,1); % Cell for non-uniform output
for ii = 1:itr
X = randi(7)+1; % Get random integer between 2 and 7
colr = randi(N-X); % Random column
rowr = randi(N-X); % random row
mattsub{ii} = matt(rowr:(rowr+X-1),colr:(colr+X-1));
end
So I want to concatenate an m x n matrix to obtain a 1 x mn matrix. The matrix I want to concatenate are generated from a while loop. Although the number of columns will always be 3, I however cannot tell how many rows there will be for each iteration. Also, the row sizes for each iteration may not always be the same.
The code runs in cases where the row sizes were all equal to 6, but in cases where they aren't equal I get an error:
Error using vertcat Dimensions of matrices being concatenated are not consistent.
parts of the code are as follows:
A = [];
B = [];
searchArea = 2;
for ii = 1: numel(velocity)
Do ....
while area(ii,:) < searchArea
Do ....
% COLLATE vectors for A
A = [A; [Ax(ii), Ay(ii), Az(ii)]];
Do ...
end
%# Copy the A into new variable (B) and Reshape into row vector so as to associate each row to its corresponding velocity
B = [B; reshape(A.',1,[])];
A = [];
end
Could someone please advice me on what I am doing wrong here. I would clarify further if there be need. Thanks guys!
If it's your intent that B ends up being a row vector, then you need to change this:
B = [B; reshape(A.',1,[])]; % Does vertical concatenation
to this:
B = [B reshape(A.',1,[])]; % Does horizontal concatenation (note there's no semicolon)
so that each row vector gotten from reshaping A gets added to the end of the row instead of as a new row (as the semicolon indicates).
I have a quite big (107 x n) matrix X. Within these n columns, each three columns belong to each other. So, the first three columns of matrix X build a block, then columns 4,5,6 and so on.
Within each block, the first 100 row elements of the first column are important X(1:100,1:3:end). Whenever in this first column the number of zeros or NaNs is greater or equal 20, it should delete the whole block.
Is there a way to do this without a loop?
Thanks for any advice!
Assuming the number of columns of the input to be a multiple of 3, there could be two approaches here.
Approach #1
%// parameters
rl = 100; %// row limit
cl = 20; %// count limit
X1 = X(1:rl,1:3:end) %// Important elements from input
match_mat = isnan(X1) | X1==0 %// binary array of matches
match_blk_id = find(sum(match_mat)>=cl) %// blocks that satisfy requirements
match_colstart = (match_blk_id-1).*3+1 %// start column indices that satisfy
all_col_ind = bsxfun(#plus,match_colstart,[0:2]') %//'columns indices to be removed
X(:,all_col_ind)=[] %// final output after removing to be removed columns
Or if you prefer "compact" codes -
X1 = X(1:rl,1:3:end);
X(:,bsxfun(#plus,(find(sum(isnan(X1) | X1==0)>=cl)-1).*3+1,[0:2]'))=[];
Approach #2
X1 = X(1:rl,1:3:end)
match_mat = isnan(X1) | X1==0 %// binary array of matches
X(:,repmat(sum(match_mat)>=cl,[3 1]))=[] %// Find matching blocks, replicate to
%// next two columns and remove them from X
Note: If X is not a multiple of 3, use this before using the codes - X = [X zeros(size(X,1) ,3 - mod(size(X,2),3))].
Is there any way that I can sum up columns values for each group of three rows in a matrix?
I can sum three rows up in a manual way.
For example
% matrix is the one I wanna store the new data.
% data is the original dataset.
matrix(1,1:end) = sum(data(1:3, 1:end))
matrix(2,1:end) = sum(data(4:6, 1:end))
...
But if the dataset is huge, this wouldn't work.
Is there any way to do this automatically without loops?
Here are four other ways:
The obligatory for-loop:
% for-loop over each three rows
matrix = zeros(size(data,1)/3, size(data,2));
counter = 1;
for i=1:3:size(data,1)
matrix(counter,:) = sum(data(i:i+3-1,:));
counter = counter + 1;
end
Using mat2cell for tiling:
% divide each three rows into a cell
matrix = mat2cell(data, ones(1,size(data,1)/3)*3);
% compute the sum of rows in each cell
matrix = cell2mat(cellfun(#sum, matrix, 'UniformOutput',false));
Using third dimension (based on this):
% put each three row into a separate 3rd dimension slice
matrix = permute(reshape(data', [], 3, size(data,1)/3), [2 1 3]);
% sum rows, and put back together
matrix = permute(sum(matrix), [3 2 1]);
Using accumarray:
% build array of group indices [1,1,1,2,2,2,3,3,3,...]
idx = floor(((1:size(data,1))' - 1)/3) + 1;
% use it to accumulate rows (appliead to each column separately)
matrix = cell2mat(arrayfun(#(i)accumarray(idx,data(:,i)), 1:size(data,2), ...
'UniformOutput',false));
Of course all the solution so far assume that the number of rows is evenly divisble by 3.
This one-liner reshapes so that all the values needed for a particular cell are in a column, does the sum, and then reshapes the back to the expected shape.
reshape(sum(reshape(data, 3, [])), [], size(data, 2))
The naked 3 could be changed if you want to sum a different number of rows together. It's on you to make sure the number of rows in each group divides evenly.
Slice the matrix into three pieces and add them together:
matrix = data(1:3:end, :) + data(2:3:end, :) + data(3:3:end, :);
This will give an error if size(data,1) is not a multiple of three, since the three pieces wouldn't be the same size. If appropriate to your data, you might work around that by truncating data, or appending some zeros to the end.
You could also do something fancy with reshape and 3D arrays. But I would prefer the above (unless you need to replace 3 with a variable...)
Prashant answered nicely before but I would have a simple amendment:
fl = filterLength;
A = yourVector (where mod(A,fl)==0)
sum(reshape(A,fl,[]),1).'/fl;
There is the ",1" that makes the line run even when fl==1 (original values).
I discovered this while running it in a for loop like so:
... read A ...
% Plot data
hold on;
averageFactors = [1 3 10 30 100 300 1000];
colors = hsv(length(averageFactors));
clear legendTxt;
for i=1:length(averageFactors)
% ------ FILTERING ----------
clear Atrunc;
clear ttrunc;
clear B;
fl = averageFactors(i); % filter length
Atrunc = A(1:L-mod(L,fl),:);
ttrunc = t(1:L-mod(L,fl),:);
B = sum(reshape(Atrunc,fl,[]),1).'/fl;
tB = sum(reshape(ttrunc,fl,[]),1).'/fl;
length(B)
plot(tB,B,'color',colors(i,:) )
%kbhit ()
endfor
I want to efficiently delete a lot of data from the beginning of a matrix of dimension 2*n. The matrix looks like this:
x1 x2
x3 x4
...
...
I want to delete all rows that have the the first element of a row that is smaller than some number and stop when a row isn't smaller (the elements are in numerical order)
What I do at the moment is slow:
while 1
if list{i}(1) <= someNumber
list{i}(1,:) = []
else
break;
end
end
There must be a neat way of doing this quickly in MATLAB?
Thank you.
One way is to just compare the entire first column in one go and then delete, i.e.
rows2delete = list{i}(:,1) <= someNumber; %# creates logical array with 1 for deletion
list{i}(rows2delete,:) = []; %# delete some rows, all corresponding cols