I have a matrix of zeros and ones. Any given column of the matrix is either full of zeros or has a single one.
E.g.:
A = [0 0 0 0 0;
1 0 0 0 0
0 0 0 0 0
0 0 1 0 0
0 0 0 0 1];
I want to get vector B that gives me the line position of each 1. If there is no 1 on the column it should give me the maximum number of rows. Eg:
B = [2 5 4 5 5];
Any easy way of getting this?
A possible solution with matrix multiplication:
A = [0 0 0 0 0;
1 0 0 0 0
0 0 0 0 0
0 0 1 0 0
0 0 0 0 1];
[r ,~] = size(A);
B = (1:r) * A;
B(B==0)=r;
comparison with other method:
n = 9000;
ro = randperm(n,4000);
co = randperm(n , 4000);
A = accumarray([ro(:) co(:)],1);
disp('------matrix multiplication---------:')
tic
[r ,~] = size(A);
B = (1:r) * A;
B(B==0)=r;
toc
disp('------find---------:')
tic
[r,~]=find(A);
B = double(any(A));
B(B==1)= r; B(B==0)=n;
toc
result:
------matrix multiplication---------:
Elapsed time is 0.0569789 seconds.
------find---------:
Elapsed time is 0.252345 seconds.
You can use the two-output version of max, which gives the position of each maximum. For columns consisting of only zeros the maximum will be in the first row, so you need to correct this by checking if the found maximum was 0 or 1
[m, result] = max(A, [], 1); % maximum of each column, and its row index
result(~m) = size(A, 1); % if the maximum was 0: modify result
Related
I am trying to create a neighbourhood graph from a given binary matrix B. Neighbourhood graph (A) is defined as an adjacency matrix such that
(A(i,j) = A(j,i) = 1)
if the original matrix B(i) = B(j) = 1 and i and j are adjacent to each (left, right, up, down or diagonal). Here I used the linear subscript to access the original matrix B. For example, consider the below matrix
B = [ 0 1 0;
0 1 1;
0 0 0 ];
My A will be a 9 * 9 graph as given below
A = [ 0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0;
0 0 0 0 1 0 0 1 0;
0 0 0 1 0 0 0 1 0;
0 0 0 0 0 0 0 0 0;
0 0 0 0 0 0 0 0 0;
0 0 0 1 1 0 0 0 0;
0 0 0 0 0 0 0 0 0 ];
Since in the original B matrix, B(4), B(5) and B(8) are adjacent with corresponding entries 1, the adjacency matrix A has 1 at A(4,5), A(5,4), A(4,8), A(8,4), A(5,8) and A(8,5).
How can I create such an adjacency matrix A given the matrix B in an efficient way?
This doesn't require any toolbox, and works for square or rectangular matrices. It uses array operations with complex numbers.
Consider a binary matrix B of size M×N.
Create an M×N matrix, t, that contains the complex coordinates of each nonzero entry of B. That is, entry t(r,c) contains r+1j*c if B(r,c) is nonzero, and NaN otherwise.
Compute an M*N×M*N matrix, d, containing the absolute difference for each pair of entries of B. Pairs of entries of B that are nonzero and adjacent will produce 1 or sqrt(2) in matrix d.
Build the result matrix, A, such that it contains 1 iff the corresponding entry in d equals 1 or sqrt(2). Equivalently, and more robust to numerical errors, iff the corresponding entry in d is between 0 and 1.5.
Code:
B = [0 1 0; 0 1 1; 0 0 0]; % input
t = bsxfun(#times, B, (1:size(B,1)).') + bsxfun(#times, B, 1j*(1:size(B,2)));
t(t==0) = NaN; % step 1
d = abs(bsxfun(#minus, t(:), t(:).')); % step 2
A = d>0 & d<1.5; % step 3
To get B back from A:
B2 = zeros(sqrt(size(A,1)));
B2(any(A,1)) = 1;
Here is a solution using image processing toolbox* that creates sparse matrix representation of the adjacency matrix:
B = [ 0 1 0;
0 1 1;
0 0 0 ]
n = numel(B);
C = zeros(size(B));
f = find(B);
C(f) = f;
D = padarray(C,[1 1]);
%If you don't have image processing toolbox
%D = zeros(size(C)+2);
%D(2:end-1,2:end-1)=C;
E = bsxfun(#times, im2col(D,[3 3]) , reshape(B, 1,[]));
[~ ,y] = find(E);
result = sparse(nonzeros(E),y,1,n,n);
result(1:n+1:end) = 0;
*More efficient implementation of im2col can be found here.
I want to obtain all the possible permutations of one vector elements by another vector elements. For example one vector is A=[0 0 0 0] and another is B=[1 1]. I want to replace the elements of A by B to obtain all the permutations in a matrix like this [1 1 0 0; 1 0 1 0; 1 0 0 1; 0 1 1 0; 0 1 0 1; 0 0 1 1]. The length of real A is big and I should be able to choose the length of B_max and to obtain all the permutations of A with B=[1], [1 1], [1 1 1],..., B_max.
Thanks a lot
Actually, since A and B are always defined, respectively, as a vector of zeros and a vector of ones, this computation is much easier than you may think. The only constraints you should respect concerns B, which shoud not be empty and it's elements cannot be greater than or equal to the number of elements in A... because after that threshold A will become a vector of ones and calculating its permutations will be just a waste of CPU cycles.
Here is the core function of the script, which undertakes the creation of the unique permutations of 0 and 1 given the target vector X:
function p = uperms(X)
n = numel(X);
k = sum(X);
c = nchoosek(1:n,k);
m = size(c,1);
p = zeros(m,n);
p(repmat((1-m:0)',1,k) + m*c) = 1;
end
And here is the full code:
clear();
clc();
% Define the main parameter: the number of elements in A...
A_len = 4;
% Compute the elements of B accordingly...
B_len = A_len - 1;
B_seq = 1:B_len;
% Compute the possible mixtures of A and B...
X = tril(ones(A_len));
X = X(B_seq,:);
% Compute the unique permutations...
p = [];
for i = B_seq
p = [p; uperms(X(i,:).')];
end
Output for A_len = 4:
p =
1 0 0 0
0 1 0 0
0 0 1 0
0 0 0 1
1 1 0 0
1 0 1 0
1 0 0 1
0 1 1 0
0 1 0 1
0 0 1 1
1 1 1 0
1 1 0 1
1 0 1 1
0 1 1 1
I have a cell array (11000x500) with three different type of elements.
1) Non-zero doubles
2) zero
3) Empty cell
I would like to find all occurances of a non-zero number between two zeros.
E.g. A = {123 13232 132 0 56 0 12 0 0 [] [] []};
I need the following output
out = logical([0 0 0 0 1 0 1 0 0 0 0 0]);
I used cellfun and isequal like this
out = cellfun(#(c)(~isequal(c,0)), A);
and got the follwoing output
out = logical([1 1 1 0 1 0 1 0 0 1 1 1]);
I need help to perform the next step where i can ignore the consecutive 1's and only take the '1's' between two 0's
Could someone please help me with this?
Thanks!
Here is a quick way to do it (and other manipulations binary data) using your out:
out = logical([1 1 1 0 1 0 1 0 0 1 1 1]);
d = diff([out(1) out]); % find all switches between 1 to 0 or 0 to 1
len = 1:length(out); % make a list of all indices in 'out'
idx = [len(d~=0)-1 length(out)]; % the index of the end each group
counts = [idx(1) diff(idx)]; % the number of elements in the group
elements = out(idx); % the type of element (0 or 1)
singles = idx(counts==1 & elements==1)
and you will get:
singles =
5 7
from here you can continue and create the output as you need it:
out = false(size(out)); % create an output vector
out(singles) = true % fill with '1' by singles
and you get:
out =
0 0 0 0 1 0 1 0 0 0 0 0
You can use conv to find the elements with 0 neighbors (notice that the ~ has been removed from isequal):
out = cellfun(#(c)(isequal(c,0)), A); % find 0 elements
out = double(out); % cast to double for conv
% elements that have more than one 0 neighbor
between0 = conv(out, [1 -1 1], 'same') > 1;
between0 =
0 0 0 0 1 0 1 0 0 0 0 0
(Convolution kernel corrected to fix bug found by #TasosPapastylianou where 3 consecutive zeros would result in True.)
That's if you want a logical vector. If you want the indices, just add find:
between0 = find(conv(out, [1 -1 1], 'same') > 1);
between0 =
5 7
Another solution, this completely avoids your initial logical matrix though, I don't think you need it.
A = {123 13232 132 0 56 0 12 0 0 [] [] []};
N = length(A);
B = A; % helper array
for I = 1 : N
if isempty (B{I}), B{I} = nan; end; % convert empty cells to nans
end
B = [nan, B{:}, nan]; % pad, and collect into array
C = zeros (1, N); % preallocate your answer array
for I = 1 : N;
if ~any (isnan (B(I:I+2))) && isequal (logical (B(I:I+2)), logical ([0,1,0]))
C(I) = 1;
end
end
C = logical(C)
C =
0 0 0 0 1 0 1 0 0 0 0 0
In each iteration I want to add 1 randomly to binary vector,
Let say
iteration = 1,
k = [0 0 0 0 0 0 0 0 0 0]
iteration = 2,
k = [0 0 0 0 1 0 0 0 0 0]
iteration = 3,
k = [0 0 1 0 0 0 0 1 0 0]
, that goes up to length(find(k)) = 5;
Am thinking of for loop but I don't have an idea how to start.
If it's important to have the intermediate vectors (those with 1, 2, ... 4 ones) as well as the final one, you can generate a random permutation and, in your example, use the first 5 indices one at a time:
n = 9; %// number of elements in vector
m = 5; %// max number of 1's in vector
k = zeros(1, n);
disp(k); %// output vector of all 0's
idx = randperm(n);
for p = 1:m
k(idx(p)) = 1;
disp(k);
end
Here's a sample run:
0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 1
1 0 0 1 0 0 0 0 1
1 0 0 1 1 0 0 0 1
1 1 0 1 1 0 0 0 1
I wouldn't even use a loop. I would generate a random permutation of indices that sample from a vector going from 1 up to the length of k without replacement then just set these locations to 1. randperm suits the task well:
N = 10; %// Length 10 vector
num_vals = 5; %// 5 values to populate
ind = randperm(N, num_vals); %// Generate a vector from 1 to N and sample num_vals values from this vector
k = zeros(1, N); %// Initialize output vector k to zero
k(ind) = 1; %// Set the right values to 1
Here are some sample runs when I run this code a few times:
k =
0 0 1 1 0 1 1 0 0 1
k =
1 0 0 0 1 0 1 1 0 1
k =
1 0 0 0 1 0 1 1 0 1
k =
0 1 1 1 0 0 1 0 0 1
However, if you insist on using a loop, you can generate a vector from 1 up to the desired length, randomly choose an index in this vector then remove this value from the vector. You'd then use this index to set the location of the output:
N = 10; %// Length 10 vector
num_vals = 5; %// 5 values to populate
vec = 1 : N; %// Generate vector from 1 up to N
k = zeros(1, N); %// Initialize output k
%// Repeat the following for as many times as num_vals
for idx = 1 : num_vals
%// Obtain a value from the vector
ind = vec(randi(numel(vec), 1));
%// Remove from the vector
vec(ind) = [];
%// Set location in output to 1
k(ind) = 1;
end
The above code should still give you the desired effect, but I would argue that it's less efficient.
i have a vector a = [1; 6; 8]
and want to create a matrix with n columns and size(a,1) rows.
Each i'th row is all zeros but on the a(i) index is one.
>> make_the_matrix(a, 10)
ans =
1 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 0 0 1 0 0 0
use sparse
numCol = 10; % number of colums in output matrix, should not be greater than max(a)
mat = sparse( 1:numel(a), a, 1, numel(a), numCol );
if you want a full matrix just use
full(mat)
Here is my first thought:
a = [1;6;8];
nCols = 10;
nRows = length(a);
M = zeros(nRows,nCols);
M(:,a) = eye(nRows)
Basically the eye is assigned to the right columns of the matrix.