I have a matrix M(x,y). I want to apply a threshold in all values in x, such that if x
Example:
M = 1, 2;
3, 4;
5, 6;
If t = 5 is applied on the 1st dimension, the result will be
R = 0, 2;
0, 4;
5, 6;
One way (use M(:,1) to select the first column; M(:,1)<5 returns row indices for items in the first column that are lest than 5))-
> R = M;
> R(M(:,1)<5,1) = 0
R =
0 2
0 4
5 6
Another -
R = M;
[i,j]=find(M(:,1)<5); % locate rows (i) and cols (j) where M(:,1) < 5
% so j is just going to be all 1
% and i has corresponding rows
R(i,1)=0;
To do it in a matrix of arbitrary dimensions:
thresh_min = 5;
M(M < thresh_min) = 0;
The statement M < thresh_min returns indices of M that are less than thresh_min. Then, reindexing into M with these indices, you can set all of these valuse fitting your desired criterion to 0 (or whatever else).
Related
I just wanted to simply find the index of (row, col) that is a minimum point of a matrix A. I can use
[minval, imin] = min( A(:) )
and MATLAB built in function
[irow, icol] = ind2sub(imin);
But for efficiency reason, where matrix A is trigonal, i wanted to implement the following function
function [i1, i2] = myind2ind(ii, N);
k = 1;
for i = 1:N
for j = i+1:N
I(k, 1) = i; I(k, 2) = j;
k = k + 1;
end
end
i1 = I(ii, 1);
i2 = I(ii, 2);
this function returns 8 and 31 for the following input
[irow, icol] = myind2ind(212, 31); % irow=8, icol = 31
How can I implement myind2ind function more efficient way without using the internal "I"?
The I matrix can be generated by nchoosek.
For example if N = 5 we have:
N =5
I= nchoosek(1:N,2)
ans =
1 2
1 3
1 4
1 5
2 3
2 4
2 5
3 4
3 5
4 5
so that
4 repeated 1 times
3 repeated 2 times
2 repeated 3 times
1 repeated 4 times
We can get the number of rows of I with the Gauss formula for triangular number
(N-1) * (N-1+1) /2 =
N * (N -1) / 2 =
10
Given jj = size(I,1) + 1 - ii as a row index I that begins from the end of I and using N * (N -1) / 2 we can formulate a quadratic equation:
N * (N -1) / 2 = jj
(N^2 -N)/2 =jj
So
N^2 -N - 2*jj = 0
Its root is:
r = (1+sqrt(8*jj))/2
r can be rounded and subtracted from N to get the first element (row number of triangular matrix) of the desired output.
R = N + 1 -floor(r);
For the column number we find the index of the first element idx_first of the current row R:
idx_first=(floor(r+1) .* floor(r)) /2;
The column number can be found by subtracting current linear index from the linear index of the first element of the current row and adding R to it.
Here is the implemented function:
function [R , C] = myind2ind(ii, N)
jj = N * (N - 1) / 2 + 1 - ii;
r = (1 + sqrt(8 * jj)) / 2;
R = N -floor(r);
idx_first = (floor(r + 1) .* floor(r)) / 2;
C = idx_first-jj + R + 1;
end
I have the code below for oppositely ordering two vectors. It works, but I want to specify the line
B_diff(i) = B(i) - B(i+1);
to hold true not just for only
B_diff(i) = B(i) - B(i+1); but for
B_diff(i) = B(i) - B(i+k); where k can be any integer less than or equal to n. The same applies to "A". Any clues as to how I can achieve this in the program?
For example, I want to rearrange the first column of the matrix
A =
1 4
6 9
3 8
4 2
such that, the condition should hold true not only for
(a11-a12)(a21-a22)<=0;
but also for all
(a11-a13)(a21-a23)<=0;
(a11-a14)(a21-a24)<=0;
(a12-a13)(a22-a23)<=0;
(a12-a14)(a22-a24)<=0; and
(a13-a14)(a23-a24)<=0;
## MATLAB CODE ##
A = xlsread('column 1');
B = xlsread('column 2');
n = numel(A);
B_diff = zeros(n-1,1); %Vector to contain the differences between the elements of B
count_pos = 0; %To count the number of positive entries in B_diff
for i = 1:n-1
B_diff(i) = B(i) - B(i+1);
if B_diff(i) > 0
count_pos = count_pos + 1;
end
end
A_desc = sort(A,'descend'); %Sort the vector A in descending order
if count_pos > 0 %If B_diff contains positive entries, divide A_desc into two vectors
A_less = A_desc(count_pos+1:n);
A_great = sort(A_desc(1:count_pos),'ascend');
A_new = zeros(n,1); %To contain the sorted elements of A
else
A_new = A_desc; %This is then the sorted elements of A
end
if count_pos > 0
A_new(1) = A_less(1);
j = 2; %To keep track of the index for A_less
k = 1; %To keep track of the index for A_great
for i = 1:n-1
if B_diff(i) <= 0
A_new(i+1) = A_less(j);
j = j + 1;
else
A_new(i+1) = A_great(k);
k = k + 1;
end
end
end
A_diff = zeros(n-1,1);
for i = 1:n-1
A_diff(i) = A_new(i) - A_new(i+1);
end
diff = [A_diff B_diff]
prod = A_diff.*B_diff
The following code orders the first column of A opposite to the order of the second column.
A= [1 4; 6 9; 3 8; 4 2]; % sample matrix
[~,ix]=sort(A(:,2)); % ix is the sorting permutation of A(:,2)
inverse=zeros(size(ix));
inverse(ix) = numel(ix):-1:1; % the un-sorting permutation, reversed
B = sort(A(:,1)); % sort the first column
A(:,1)=B(inverse); % permute the first column according to inverse
Result:
A =
4 4
1 9
3 8
6 2
I have a symmetric m-by-m matrix A. Each element has a value between 0 and 1. I now want to choose n rows / columns of A which form an n-by-n sub-matrix B.
The criteria for choosing these elements, is that the sum of all elements of B must be the minimum out of all possible n-by-n sub-matrices of A.
For example, suppose that A is a 4-by-4 matrix:
A = [0 0.5 1 0; 0.5 0 0.5 0; 1 0.5 1 1; 0 0 1 0.5]
And n is set to 3. Then, the best B is the one taking the first, second and fourth rows / columns of A:
B = [0 0.5 0; 0.5 0 0; 0 0 0.5]
Where the sum of these elements is 0 + 0.5 + 0 + 0.5 + 0 + 0 + 0 + 0 + 0.5 = 1.5, which is smaller than another other possible 3-by-3 sub-matrices (e.g. using the first, third and fourth rows / columns).
How can I do this?
This is partly a mathematics question, and partly a Matlab one. Any help with either would be great!
Do the following:
m = size(A,1);
n=3;
sub = nchoosek(1:m,n); % (numCombinations x n)
subR = permute(sub,[2,3,1]); % (n x 1 x numCombinations), row indices
subC = permute(sub,[3,2,1]); % (1 x n x numCombinations), column indices
lin = bsxfun(#plus,subR,m*(subC-1)); % (n x n x numCombinations), linear indices
allB = A(lin); % (n x n x numCombinations), all possible Bs
sumB = sum(sum(allB,1),2); % (1 x 1 x numCombinations), sum of Bs
sumB = squeeze(sumB); % (numCombinations x 1), sum of Bs
[minB,minBInd] = min(sumB);
fprintf('Indices for minimum B: %s\n',mat2str(sub(minBInd,:)))
fprintf('Minimum B: %s (Sum: %g)\n',mat2str(allB(:,:,minBInd)),minB)
This looks only for submatrices where the row indices are the same as the column indices, and not necessarily consecutive. That is how I understood the question.
This is a bit brute force, but should work
A = [0 0.5 1 0; 0.5 0 0.5 0; 1 0.5 1 1; 0 0 1 0.5];
sizeA = size(A,1);
size_sub=3;
idx_combs = nchoosek(1:sizeA, size_sub);
for ii=1:size(idx_combs,1)
sub_temp = A(idx_combs(ii,:),:);
sub = sub_temp(:,idx_combs(ii,:));
sum_temp = sum(sub);
sums(ii) = sum(sum_temp);
end
[min_set, idx] = min(sums);
sub_temp = A(idx_combs(idx,:),:);
sub = sub_temp(:,idx_combs(idx,:))
Try to convolve the matrix A with a smaller matrix M. Eg if you is interested in finding the 3x3 submatrix then let M be ones(3). This code shows how it works.
A = toeplitz(10:-1:1) % Create a to eplitz matrix (example matrix)
m = 3; % Submatrix size
mC = ceil(m/2); % Distance to center of submatrix
M = ones(m);
Aconv = conv2(A,M); % Do the convolution.
[~,minColIdx] = min(min(Aconv(1+mC:end-mC,1+mC:end-mC))); % Find column center with smallest sum
[~,minRowIdx] = min(min(Aconv(1+mC:end-mC,minColIdx+mC),[],2)); % Find row center with smlest sum
minRowIdx = minRowIdx+mC-1 % Convoluted matrix is larger than A
minColIdx = minColIdx+mC-1 % Convoluted matrix is larger than A
range = -mC+1:mC-1
B = A(minRowIdx+range, minColIdx+range)
The idea is to imitate a fir filter y(n) = 1*x(n-1)+1*x(n)+1*x(n+1). For now it only finds the first smallest matrix though. Notice the +1 adjustment because first matrix element is 1. Then notice the the restoration right below.
I have a matrix M that looks similar to this:
M = [ 1, 2, 3, 0, 0;
1, 2, 0, 0, 0;
2, 3, 4, 5, 0;
4, 5, 6, 0, 0;
1, 2, 3, 4, 5;
]
I'm trying to get a column vector with the rightmost non-zero value of each row in A, but ONLY for the rows that have the first column == 1.
I'm able to calculate a filter for the rows:
r = M( :, 1 ) == 1;
> r = [ 1; 1; 0; 0; 1 ]
And I have a set of indices for "the rightmost non-zero value of each row in M":
> c = [ 3, 2, 4, 3, 5 ]
How do I combine these in a slicing of A in order to get what I'm looking for? I'm looking for something like:
A( r, c )
> ans = [ 3; 2; 5 ]
But doing this gets me a 3x3 matrix, for some reason.
The shortest way I can think of is as follows:
% Get the values of the last non-zero entry per row
v = M(sub2ind(size(M), 1:size(M,1), c))
% Filter out the rows that does not begin with 1.
v(r == 1)
This seems to work (I assume other operations defining r,c have been performed):
M(sub2ind(size(A),find(r==1).',c(r==1))).'
Short interpretation of the problem and solution:
M( r, c )
gives a 3 x 5 matrix (not 3 x 1 as desired) due to mixing of logical and subscript indices. The logical indices in r pick out rows in A with r==1. Meanwhile row array c picks out elements from each row according to the numeric index:
ans =
3 2 0 3 0
0 2 0 0 0
3 2 4 3 5
What you really want are indices into the rightmost nonzero elements in each row starting with 1. The solution uses linear indices (numeric) to get the correct elements from the matrix.
I think this should do the trick. I wonder if there is more elegant way of doing this though.
% get only rows u want, i.e. with first row == 1
M2 = M(r,:);
% get indices of
% "the rightmost non-zero value of each row in M"
% for the rows u want
indicesOfinterest = c(r==1);
noOfIndeciesOfinterest = numel(indicesOfinterest);
% desired output column vector
output = zeros(noOfIndeciesOfinterest, 1);
% iterate through the indeces and select element in M2
% from each row and column indicated by the indice.
for idx = 1:noOfIndeciesOfinterest
output(idx) = M2(idx, indicesOfinterest(idx));
end
output % it is [3; 2 ; 5]
You can use
arrayfun(#(x) M(x,c(x)), find(r))
But unless you need r and c for other purposes, you can use
arrayfun(#(x) M(x,find(M(x,:),1,'last')), find(M(:,1)==1))
Here is a way to do it using linear indexing:
N = M';
lin_index = (0:size(N,1):prod(size(N))-1) + c;
v = N(lin_index);
v(r)
Is there a vectorized, automated way to fill a row vector l times with repeating numbers x such that x is increased by y after a certain number k of elements? k, l, x, and y are given.
Two examples:
(k = 4, l = 4, x = 0, y = 1): $A = [0 0 0 0; 1 1 1 1; 2 2 2 2; 3 3 3 3];$
(k = 2, l = 3, x = 0, y = 0.1): $B = [0 0; 0.1 0.1; 0.2 0.2]$
You can use repmat together with a:b
This way your fist example would look like this:
repmat((0:3)', 1,4)
The second one:
repmat((0:0.1:0.2)', 1,2)
You can also try linspace or similar functions to be as close to what you want as possible