I want to calculate all possible combinations of 1:16 with 10 subsets.
combos = combntns(1:16,10)
But the condition is that the returned combos should have minimum 1 member of the following vectors:
V1=1:4,V2=5:8,V3=9:12,V4=13:16,
Any solution?
With that problem size you can afford to generate all combinations and then select those that meet the requirements:
n = 16; %// number of elements to choose from
c = 10; %// combination size
s = 4; %// size of each group (size of V1, V2 etc)
combos = nchoosek(1:n, c);
ind = all(any(any(bsxfun(#eq, combos, reshape(1:n, 1,1,s,[])),2),3),4);
combos = combos(ind,:);
This can be generalized for generic elements and arbitrary condition vectors, assuming all vectors are the same size:
elements = 1:16; %// elements to choose from
c = 10; %// combination size
vectors = {1:4, 5:8, 9:12, 13:16}; %// cell array of vectors
s = numel(vectors{1});
combos = nchoosek(elements, c);
ind = all(any(any(bsxfun(#eq, combos, reshape(cat(1,vectors{:}).', 1,1,s,[])),2),3),4); %'
combos = combos(ind,:);
Related
I have:
A matrix 3D: A = (m, n, k).
An array of choices for the third dimension corresponding to each index of the first dimension. idn = (m, 1) (wherein the value of any idn is a random integer in [1,k].
I need to capture the 2D matrix B (m,n) wherein the referred third dimension to A is taken from the corresponding choice. For example:
idn(1) = 1;
idn(2) = k;
idn(j) = k-1;
Then:
B(1,:) = A(1,:,idn(1)) = A(1,:,1);
B(2,:) = A(2,:,idn(2)) = A(2,:,k);
B(j,:) = A(j,:,idn(j)) = A(j,:,k-1);
Since idn is not constant, a simple squeeze could not help.
I have also tried the below code, but it does not work either.
B = A(:,:,idn(:));
It is very much appreciated if anyone could give me a solution.
This could be done with sub2ind and permute, but the simplest way I can think of is using linear indexing manually:
A = rand(3, 4, 5); % example data
idn = [5; 1; 2]; % example data
ind = (1:size(A,1)).' + size(A,1)*size(A,2)*(idn(:)-1); % 1st and 3rd dimensions
ind = ind + size(A,1)*(0:size(A,2)-1); % include 2nd dimension using implicit expansion
B = A(ind); % index into A to get result
I am trying to generate a matrix, that has all unique combinations of [0 0 1 1], I wrote this code for this:
v1 = [0 0 1 1];
M1 = unique(perms([0 0 1 1]),'rows');
• This isn't ideal, because perms() is seeing each vector element as unique and doing:
4! = 4 * 3 * 2 * 1 = 24 combinations.
• With unique() I tried to delete all the repetitive entries so I end up with the combination matrix M1 →
only [4!/ 2! * (4-2)!] = 6 combinations!
Now, when I try to do something very simple like:
n = 15;
i = 1;
v1 = [zeros(1,n-i) ones(1,i)];
M = unique(perms(vec_1),'rows');
• Instead of getting [15!/ 1! * (15-1)!] = 15 combinations, the perms() function is trying to do
15! = 1.3077e+12 combinations and it's interrupted.
• How would you go about doing in a much better way? Thanks in advance!
You can use nchoosek to return the indicies which should be 1, I think in your heart you knew this must be possible because you were using the definition of nchoosek to determine the expected final number of permutations! So we can use:
idx = nchoosek( 1:N, k );
Where N is the number of elements in your array v1, and k is the number of elements which have the value 1. Then it's simply a case of creating the zeros array and populating the ones.
v1 = [0, 0, 1, 1];
N = numel(v1); % number of elements in array
k = nnz(v1); % number of non-zero elements in array
colidx = nchoosek( 1:N, k ); % column index for ones
rowidx = repmat( 1:size(colidx,1), k, 1 ).'; % row index for ones
M = zeros( size(colidx,1), N ); % create output
M( rowidx(:) + size(M,1) * (colidx(:)-1) ) = 1;
This works for both of your examples without the need for a huge intermediate matrix.
Aside: since you'd have the indicies using this approach, you could instead create a sparse matrix, but whether that's a good idea or not would depend what you're doing after this point.
Consider n row vectors in Matlab, each of size 1xU. For example,
U=20;
n=3;
sU=[U U U];
vectors = arrayfun(#(x) {1:x}, sU);
where vector{1} is the first row vector, vector{2} is the second row vector,..., vector{n} is the last row vector.
We create the matrix Tcoord of size U^n x n reporting all the possible n-tuples from the n row vectors. For each row i of Tcoord, Tcoord(i,1) is an element of the first row vector, Tcoord(i,2) is an element of the second row vector, ... , Tcoord(i,n) is an element of the last row vector.
Tcoord_temp = cell(1,n);
[Tcoord_temp{:}] = ndgrid(vectors{:});
Tcoord_temp = cat(n+1, Tcoord_temp{:});
Tcoord = reshape(Tcoord_temp,[],n);
Suppose now that I augment each of the n row vectors of 3 elements. For example,
vectors_augmented{1}=[vectors{1} 8 9 10];
vectors_augmented{2}=[vectors{2} 11 12 13];
vectors_augmented{3}=[vectors{3} 14 15 16];
I then create a matrix similar to Tcoord but now using vectors_augmented.
Tcoord_temp = cell(1,n);
[Tcoord_temp{:}] = ndgrid(vectors_augmented{:});
Tcoord_temp = cat(n+1, Tcoord_temp{:});
Tcoord_augmented = reshape(Tcoord_temp,[],n); %(U+3)^nxn
I would like your help to re-order the rows of the matrix Tcoord_augmented in a matrix Tcoord_augmented_reshape such that
Tcoord_augmented_reshape(1:U^n,:) is equal to Tcoord.
The remaining rows of Tcoord_augmented_reshape contains the other left rows of Tcoord_augmented.
The simplest approach is to build an auxiliary zero-one matrix the same size as Tcoord_augmented and sort rows based on that:
aug_size = [3 3 3]; % augment size of each vector. Not necessarily equal
vectors_aux = arrayfun(#(a) {[false(1,U) true(1, a)]}, aug_size);
T_aux = cell(1,n);
[T_aux{:}] = ndgrid(vectors_aux{:});
T_aux = cat(n+1, T_aux{:});
T_aux = reshape(T_aux,[],n);
[~, ind] = sortrows(T_aux, n:-1:1); % indices of stably sorting the rows.
% Most significant column is rightmost, as per your code
Tcoord_augmented_reorder = Tcoord_augmented(ind, :);
Is it possible in matlab/octave to use the sort function to sort an array based on the relative frequency of their elements?
For example the array
m= [4,4,4,10,10,10,4,4,5]
should result in this array:
[5,10,10,10,4,4,4,4,4]
5 is the less frequent element and is on the top while 4 is the most frequent and it's on bottom.
Should one use the indices provided by histcount?
The following code first calculates how often each element occurs and then uses runLengthDecode to expand the unique elements.
m = [4,4,4,10,10,10,4,4,5];
u_m = unique(m);
elem_count = histc(m,u_m);
[elem_count, idx] = sort(elem_count);
m_sorted = runLengthDecode(elem_count, u_m(idx));
The definition of runLengthDecode is copied from this answer:
For MATLAB R2015a+:
function V = runLengthDecode(runLengths, values)
if nargin<2
values = 1:numel(runLengths);
end
V = repelem(values, runLengths);
end
For versions before R2015a:
function V = runLengthDecode(runLengths, values)
%// Actual computation using column vectors
V = cumsum(accumarray(cumsum([1; runLengths(:)]), 1));
V = V(1:end-1);
%// In case of second argument
if nargin>1
V = reshape(values(V),[],1);
end
%// If original was a row vector, transpose
if size(runLengths,2)>1
V = V.'; %'
end
end
One way would be to use accumarray to find the count of each number (I suspect you can use histcounts(m,max(m))) but then you have to clear all the 0s).
m = [4,4,4,10,10,10,4,4,5];
[~,~,subs]=unique(m);
freq = accumarray(subs,subs,[],#numel);
[~,i2] = sort(freq(subs),'descend');
m(i2)
By combinging my approach with that of m.s. you can get a simpler solution:
m = [4,4,4,10,10,10,4,4,5];
[U,~,i1]=unique(m);
freq= histc(m,U);
[~,i2] = sort(freq(i1),'descend');
m(i2)
You could count the number of repetitions with bsxfun, sort that, and apply that sorting to m:
[~, ind] = sort(sum(bsxfun(#eq,m,m.')));
result = m(ind);
I have a logical array B and a matrix A of size nx2 containing n couples of start/stop indexes
A= [start1, stop1; start2, stop2; start3, stop3];
How can I select the subarrays of array B based on start/stop couples contained in array A?
I'm doing it by:
for i=1:1:size(A,1)
B(A(i,1):A(i,2)) = true;
end
Is there any way of doing it in a more elegant way without using the for cycle (even conveniently rearranging array A)?
you can
n = numel(B);
iA = zeros( 1, n+1 ); %// +1 for boundary case
iA( A(:,1) ) = 1;
iA( A(:,2)+1 ) = -1;
iA(end) = []; %// discard boundary entry
B( cumsum(iA) > 0 ) = true;
Assumptions made
A(:,1) is always >= 1
A(:,2) is always <= n (number of elements in B)
the sections defined in A are non overlapping
If each stop is assured to be smaller than the next start (the index ranges don't overlap), another approach is
B(mod(sum(bsxfun(#le, 1:numel(B), [A(:,1)-1; A(:,2)])),2)>0) = true;